Simplifying Trig Function
1. Check identities
2. Algebra (factoring, combining like terms, and fraction)
3. Check identites
Some Proofs to help
-cotx= cosx/sinx
-tanx= sinx/cosx
-1+cot^2x=csc^2x
-1+tan^2x=sec^2x
sin^2+cos^2=1
The way to simplify is to find and rearrange the functions in a way the makes them resemble one of the proofs. After you have done that then you use the proofs to replace something in the equation. And then you keep repeating those steps until you can not do it anymore without making the equation bigger.
Ex. Prove sec^4x-tan^4x/sec^2x
(sec^2x-tan^2x)(sec^2x+tan^2x)/sec^2x
1=sec^2x-tan^2x
1(sec^2x+tan^2x)/sec^2x
sin^2x/cos^2x/1
=1+sin/2x
Happy New Year
Thursday, December 31, 2009
Tuesday, December 29, 2009
Alicia's Reflection # 19 (christmas holiday)
Okay so I hope everyone is enjoying their holidays.... I know I am. I am going to reflect on some old things we learned way back towards the beginning of the year.
**Logs
Condense:
Ex) logm + log7 + 4logn
= log7mn^4
Ex) 5loga + logd + log6
= log6da^5
Ex) 4logt - logc
= t^4/c
Ex) logn - 3logh -logy
= n/yh^3
Expand:
Ex) log5gh^2
= log5 + 2logh +logg
Ex) m^3b^7/f
= 3logm + 7logb - logf
**The Unit Circle
90 degrees, (0,1), pi/2
180 degrees, (-1,0), pi
270 degrees, (0,-1), 3pi/2
360 degrees, (1,0), 2pi
**6 Trig Functions
sin = y/r
cos = x/r
tan = y/x
csc = r/y
sec = r/x
cot = x/y
**Degrees & Radians
Degrees to radians= Degree * pi/180
Radians to degrees= Radians * 180/pi
**To solve coterminal angles, either add or subtract 360 to the angle.
Happy New Year!!
**Logs
Condense:
Ex) logm + log7 + 4logn
= log7mn^4
Ex) 5loga + logd + log6
= log6da^5
Ex) 4logt - logc
= t^4/c
Ex) logn - 3logh -logy
= n/yh^3
Expand:
Ex) log5gh^2
= log5 + 2logh +logg
Ex) m^3b^7/f
= 3logm + 7logb - logf
**The Unit Circle
90 degrees, (0,1), pi/2
180 degrees, (-1,0), pi
270 degrees, (0,-1), 3pi/2
360 degrees, (1,0), 2pi
**6 Trig Functions
sin = y/r
cos = x/r
tan = y/x
csc = r/y
sec = r/x
cot = x/y
**Degrees & Radians
Degrees to radians= Degree * pi/180
Radians to degrees= Radians * 180/pi
**To solve coterminal angles, either add or subtract 360 to the angle.
Happy New Year!!
Amy's Reflection #19 (Christmas Blog)
ok here is a review on some of the things we learned...
1.) area of a non right triangle = 1/2 (leg)(leg)SIN(angle b/w)
Example: non-right triangle: HIJ (left to right)H = 65 degrees, j = 2, i = 6. Find the area.
A = (1/2)(2)(6)sin(65)
A = 5.438
2.) Law of Sines(used to non-right triangles):
Sin A/a = Sin B/b= Sin C/c
Example: you have a triangle with the sides 4 and 5 & you also have an angle of 30 degrees.
A = 1/2 (4) (5) Sin 30 degrees
A = 10 Sin 30 degrees which is aproximately = 5
3.) Law of Cosines (used when you can't use Law of Sines):
(opposite leg)^2 = (adjacent leg)^2 + (other adjacent leg)^2 - 2(adjacent leg) (adjacent leg) cos (angle between)
Example: you have a triangle with the sides of 5, 6, and 7. find the angle between 5 and 6.
7^2=6^2+5^2-2(5)(6)
cos a7^2-6^2-5^2= 2(5)(6)
cos acos a= 7^2-6^2-5^2 / -2(6)(5)
a= cos-1 ((7^2-^6^2-5^2)/(-2(5)(6))
a= 78.463 degrees
4.) For any line : m = tan (alpha)
**m = slope , (alpha) = angle of inclination
5.) For a conic: tan 2 (alpha) = B/A-C
**if A=C then pie/4 (always)
**A = coefficient of x^2, B = coefficient of xy, C = coefficient of y^2
Examples:
1. Find the angle of inclination of x^2 - 2xy + 3y^2 = 1.
tan 2 (alpha) = B/A-C
A = 1 , B = -2 , C = 3
tan 2 (alpha) = -2/1 -3 = 1
tan 2 (alpha) = 1
2A = tan^-1 (1)
2 (alpha) = 45 , 225
alpha = 45/2 , 225/2
alpha = 22.5 , 112.5
2. x^2 + y^2 - 3xy + 4x - sqrt.
x = 1alpha = 1 (because A = 1 & C = 1 so A = C)
alrighty then, i hope this helped to refreshen your minds...see y'all in a week..
1.) area of a non right triangle = 1/2 (leg)(leg)SIN(angle b/w)
Example: non-right triangle: HIJ (left to right)H = 65 degrees, j = 2, i = 6. Find the area.
A = (1/2)(2)(6)sin(65)
A = 5.438
2.) Law of Sines(used to non-right triangles):
Sin A/a = Sin B/b= Sin C/c
Example: you have a triangle with the sides 4 and 5 & you also have an angle of 30 degrees.
A = 1/2 (4) (5) Sin 30 degrees
A = 10 Sin 30 degrees which is aproximately = 5
3.) Law of Cosines (used when you can't use Law of Sines):
(opposite leg)^2 = (adjacent leg)^2 + (other adjacent leg)^2 - 2(adjacent leg) (adjacent leg) cos (angle between)
Example: you have a triangle with the sides of 5, 6, and 7. find the angle between 5 and 6.
7^2=6^2+5^2-2(5)(6)
cos a7^2-6^2-5^2= 2(5)(6)
cos acos a= 7^2-6^2-5^2 / -2(6)(5)
a= cos-1 ((7^2-^6^2-5^2)/(-2(5)(6))
a= 78.463 degrees
4.) For any line : m = tan (alpha)
**m = slope , (alpha) = angle of inclination
5.) For a conic: tan 2 (alpha) = B/A-C
**if A=C then pie/4 (always)
**A = coefficient of x^2, B = coefficient of xy, C = coefficient of y^2
Examples:
1. Find the angle of inclination of x^2 - 2xy + 3y^2 = 1.
tan 2 (alpha) = B/A-C
A = 1 , B = -2 , C = 3
tan 2 (alpha) = -2/1 -3 = 1
tan 2 (alpha) = 1
2A = tan^-1 (1)
2 (alpha) = 45 , 225
alpha = 45/2 , 225/2
alpha = 22.5 , 112.5
2. x^2 + y^2 - 3xy + 4x - sqrt.
x = 1alpha = 1 (because A = 1 & C = 1 so A = C)
alrighty then, i hope this helped to refreshen your minds...see y'all in a week..
Monday, December 28, 2009
Stephanie's Reflection
Graphing Parabolas
~Descriminate tells you how many intercepts a graph has
b^2-4ac
if +ve 2 x-intercepts
if -ve no x-intercepts
if 0 1 x-intercept
~Axis of Symmetry x=-b/2a if non-standard form
~Vertex (-b/2a, f(-b/2a)) for non-standard form
~to find the intersections
solve for y
set equal
solve for x
plug back in
The number one thing you should know when solving a log is the basic formula (exponential form) which happens to be b^a=x while log for would be log b^x=a. When a problem does not have anything on the opposite side of the equal sign (opposite of log), put an x. If a log problem doesn't have a base, it is understood to be 10 because 10 is the default base.
CIRCLES
The equation of a circle in standard form is (x-h)^2-(y-h)^2=r^2 with the center being (h,k) and r being the radius.
Finding the intersection of a line and a circle:
1) solve linear equation for y
2) substitute in circle equation
3) solve for x
4) plug x in to get y value
(if x happens to be imaginary, there is no point of intersection)
ELLIPSES
1) (x-h)^2/(length of x/2)^2 + (y-k)^2/(length of y/2)^2 =1
2)center is (h,k)
3) major axis has larger denominator
4) vertex is on major axis
5) focus is smaller denom squared = larger denom squared - focus squared
focus is on major axis
Graphing:
1) find center
2) major axis = plus or minus the square root of the bigger denom
3) vertex
4) other intercepts
5) focus
6) length of major axis = 2 square root of
7) length of minor axis = 2 square root of
8) graph
HYPERBOLAS
1) (x+h)^2/(length/2)^2 - (y-k)^2/(length/2)^2 =1
OR
-(x-h)^2/(length/2)^2 + (y-k)^2/(length/2)^2 =1
2) center (h,k)
3) major axis is non-negative
4) vertex is the square root of non-negative denom
5) asymptotes y=+/-(square root of y)/(square root of x)x
6) focus^2 = x denom + y denom
focus^2 = vertex^2 + other denom
to sketch:
1) shape
2) center
3) major
4) minor
5) other intercept - none for hyperbolas
6) focus
7) asymptotes y=+/-square root of y/square root of x
8) vertex
9) sketch
A) draw a box using the vertex and +/-sr of other denom
B) draw diagonal through box corners
C) sketch a parabola on each vertex
D) label focus and asymptotes
~Descriminate tells you how many intercepts a graph has
b^2-4ac
if +ve 2 x-intercepts
if -ve no x-intercepts
if 0 1 x-intercept
~Axis of Symmetry x=-b/2a if non-standard form
~Vertex (-b/2a, f(-b/2a)) for non-standard form
~to find the intersections
solve for y
set equal
solve for x
plug back in
The number one thing you should know when solving a log is the basic formula (exponential form) which happens to be b^a=x while log for would be log b^x=a. When a problem does not have anything on the opposite side of the equal sign (opposite of log), put an x. If a log problem doesn't have a base, it is understood to be 10 because 10 is the default base.
CIRCLES
The equation of a circle in standard form is (x-h)^2-(y-h)^2=r^2 with the center being (h,k) and r being the radius.
Finding the intersection of a line and a circle:
1) solve linear equation for y
2) substitute in circle equation
3) solve for x
4) plug x in to get y value
(if x happens to be imaginary, there is no point of intersection)
ELLIPSES
1) (x-h)^2/(length of x/2)^2 + (y-k)^2/(length of y/2)^2 =1
2)center is (h,k)
3) major axis has larger denominator
4) vertex is on major axis
5) focus is smaller denom squared = larger denom squared - focus squared
focus is on major axis
Graphing:
1) find center
2) major axis = plus or minus the square root of the bigger denom
3) vertex
4) other intercepts
5) focus
6) length of major axis = 2 square root of
7) length of minor axis = 2 square root of
8) graph
HYPERBOLAS
1) (x+h)^2/(length/2)^2 - (y-k)^2/(length/2)^2 =1
OR
-(x-h)^2/(length/2)^2 + (y-k)^2/(length/2)^2 =1
2) center (h,k)
3) major axis is non-negative
4) vertex is the square root of non-negative denom
5) asymptotes y=+/-(square root of y)/(square root of x)x
6) focus^2 = x denom + y denom
focus^2 = vertex^2 + other denom
to sketch:
1) shape
2) center
3) major
4) minor
5) other intercept - none for hyperbolas
6) focus
7) asymptotes y=+/-square root of y/square root of x
8) vertex
9) sketch
A) draw a box using the vertex and +/-sr of other denom
B) draw diagonal through box corners
C) sketch a parabola on each vertex
D) label focus and asymptotes
Sunday, December 20, 2009
Amy's Reflection #18
This blog is for this weekend...i'll post the christmas blog later this week..here are some stuff from chapter 6 + examples...
Ellipses
Steps:
1. find the center
2. determine the major axis
3. find the vertex (± √big denom)
4. find the other intercept ( ± √small denom)
5. find the focus (c^2 = a^2 + b^2)
6. determine the length of the major axis (2√big denom)
7. find the length of the minor axis (2√small denom)
8. finally graph
Example 1: Graph the following ellipse. Find its major intercepts, length of the major axis, minor intercepts, length of the minor axis, and foci.
x^2/4 + y^2/9 = 1
This ellipse is centered at (0, 0). Since the larger denominator is with the y variable, the major axis lies along the y-axis.
Since a^2 = 9 then a = 3 & Since b^2 = 4
then b = 2Major intercepts: (0, 3), (0, –3)
Length of major axis: 2 √9 = 6
Minor intercepts: (2, 0), (–2, 0)
Length of minor axis: 2√4 = 4
c^2 = a^2 + b^2
= 9 - 4
= 5
= √5
Foci: (0, √5) , (0, -√5)
then you graph your points..
Parabolas:
how to find the axis of symmetry, vertex, focus, & directrx??
1.) to find the axis of symmetry: x = -b/2a
2.) for the vertex: (-b/2a, f(-b/2a)) or use complete the square:
y = (x+a)^2 + b.....a & b are numbers and (-a,b) = vertex
3.) to find the focus: 1/4p= the coefficient of x^2 and then add p
Note:
*If opens up, add to y value from vertex, if opens down, subtract
*If opens right, add to x value to vertex, if opens left, subtract)
4.) directrix: is p units behind the vertex
Note:
*If opens up, subtract; if opens down, add from y-value of vertex.*If opens right, subtract x-value*If opens left, add x-value
Example: x^2 + 1
~vertex:
x = -b/2a
x = 0/2(1) = 0
0^2 + 1 = 1
(0,1)
~Focus:
1/4p = 1
4p = 1
p = 1/4
(0, 1 + 1/4)
(0, 5/4)
~directrix:
y = 1 - 1/4
y = 3/4
Ellipses
Steps:
1. find the center
2. determine the major axis
3. find the vertex (± √big denom)
4. find the other intercept ( ± √small denom)
5. find the focus (c^2 = a^2 + b^2)
6. determine the length of the major axis (2√big denom)
7. find the length of the minor axis (2√small denom)
8. finally graph
Example 1: Graph the following ellipse. Find its major intercepts, length of the major axis, minor intercepts, length of the minor axis, and foci.
x^2/4 + y^2/9 = 1
This ellipse is centered at (0, 0). Since the larger denominator is with the y variable, the major axis lies along the y-axis.
Since a^2 = 9 then a = 3 & Since b^2 = 4
then b = 2Major intercepts: (0, 3), (0, –3)
Length of major axis: 2 √9 = 6
Minor intercepts: (2, 0), (–2, 0)
Length of minor axis: 2√4 = 4
c^2 = a^2 + b^2
= 9 - 4
= 5
= √5
Foci: (0, √5) , (0, -√5)
then you graph your points..
Parabolas:
how to find the axis of symmetry, vertex, focus, & directrx??
1.) to find the axis of symmetry: x = -b/2a
2.) for the vertex: (-b/2a, f(-b/2a)) or use complete the square:
y = (x+a)^2 + b.....a & b are numbers and (-a,b) = vertex
3.) to find the focus: 1/4p= the coefficient of x^2 and then add p
Note:
*If opens up, add to y value from vertex, if opens down, subtract
*If opens right, add to x value to vertex, if opens left, subtract)
4.) directrix: is p units behind the vertex
Note:
*If opens up, subtract; if opens down, add from y-value of vertex.*If opens right, subtract x-value*If opens left, add x-value
Example: x^2 + 1
~vertex:
x = -b/2a
x = 0/2(1) = 0
0^2 + 1 = 1
(0,1)
~Focus:
1/4p = 1
4p = 1
p = 1/4
(0, 1 + 1/4)
(0, 5/4)
~directrix:
y = 1 - 1/4
y = 3/4
Alicia's Christmas Holiday Reflection
Alrighty so thank god we finally have a break from all the stress of exams. I hope we all made decent grades on our adv. math exam!!! Okay well im going to reflect back to some old stuff that we learned in the beginning of the year and some of the stuff that I remember seeing on the exam that I should have studied better.
Ch. 6 Conics
The standard equation of a circle is (x-h)^2+(y-k)^2 .....the center is (h,k)
To find the intersection of a line and a circle:
1. solve the linear eqn for y.
2. substitute in the circle eqn.
3. solve for x.
4. plug the x value in to get the y value.
If your x value is imaginary, then there is no point of intersection.
EX: find the center and radius.(x-3)^2+(y+7)^2=19
c:(h,k)
center: (3,-7)
radius: square root of 19
--Parabolas have no major axis and no asymptotes.
Axis of symmetry x=-b/2a
Finding the vertex
(-b/2a, f(-b/2a))
or
complete the square to get vertex form
y=(x+a)^2+b a&b are #'s
(-a,b) vertex
focus: 1/4p= coeff of x^2 then add p.
directrix is p units behind vertex. subtract p.
EX: 1/8y^2
v(0,0)
Focus: p=2(2,0)
directrix: x=-2
Ch. 6 Conics
The standard equation of a circle is (x-h)^2+(y-k)^2 .....the center is (h,k)
To find the intersection of a line and a circle:
1. solve the linear eqn for y.
2. substitute in the circle eqn.
3. solve for x.
4. plug the x value in to get the y value.
If your x value is imaginary, then there is no point of intersection.
EX: find the center and radius.(x-3)^2+(y+7)^2=19
c:(h,k)
center: (3,-7)
radius: square root of 19
--Parabolas have no major axis and no asymptotes.
Axis of symmetry x=-b/2a
Finding the vertex
(-b/2a, f(-b/2a))
or
complete the square to get vertex form
y=(x+a)^2+b a&b are #'s
(-a,b) vertex
focus: 1/4p= coeff of x^2 then add p.
directrix is p units behind vertex. subtract p.
EX: 1/8y^2
v(0,0)
Focus: p=2(2,0)
directrix: x=-2
Terrio's 1st Chirstmas Holiday Reflection
So far my holidays have been pretty decent. We didnt win our soccer game Friday but we didnt loose either and that was probably the worst thing that happened as of right now. We have to do three blogs, it does not matter when they are done as long as they are done over the holidays, and here goes the first one of the holidays....
SOHCAHTOA:
sin theta=opposite/hypotenuse
cos theta=adjacent/hypotenuse
tan theta=opposite/adjacent
SOHCAHTOA is used when either you have two sides of a right triangle and you need to find an angle or you have an angle and one side. Here's an example:
A right triangle has 3 angles: 90°, 30°, and 60°. The hypotenuse is x cm. The side opposite the 60° angle is 8 cm. What is the length of the hypotenuse?
You would use the sin formula and the equation would be sin(60)=8/x.
Then you would get .8660=8/x
You divide 8 by .8660 and get 9.2380
So the hypotenuse of the triangle would be 9.2380 cm.
SOHCAHTOA:
sin theta=opposite/hypotenuse
cos theta=adjacent/hypotenuse
tan theta=opposite/adjacent
SOHCAHTOA is used when either you have two sides of a right triangle and you need to find an angle or you have an angle and one side. Here's an example:
A right triangle has 3 angles: 90°, 30°, and 60°. The hypotenuse is x cm. The side opposite the 60° angle is 8 cm. What is the length of the hypotenuse?
You would use the sin formula and the equation would be sin(60)=8/x.
Then you would get .8660=8/x
You divide 8 by .8660 and get 9.2380
So the hypotenuse of the triangle would be 9.2380 cm.
Holiday Post 1
Triangle trigonometry:
sine = opp/hyp
cos = adj/hyp
tan = opp/adj
csc = hyp/opp
sec=hyp/adj
cot=adj/opp
Moving between quadrants:
I to IV = make it negative and add 360º
I to III = add 180º
I to II = make it negative and add 180º
Trigonometric Identites:
csc = 1/sin x
sec = 1/cos x
cot x = 1/tan x
sin (-x) = -sin x
cos (-x) = cos x
csc (-x) = -csc x
sec (-x) = sec x
tan (-x) = -tan x
cot (-x) = -cot x
sin^2 x+cos^2 x = 1
1+tan^2 x = sec^2 x
1+cot^2 x = csc^2 x
sin x = cos(90*-x)
tan x = cot (90*-x)
sec x = csc (90*-x)
cos x = sin (90*-x)
cot x = tan (90*-x)
csc x = sec (90*-x)
tan x = sinx/cosx
cot x = cosx/sinx
sine = opp/hyp
cos = adj/hyp
tan = opp/adj
csc = hyp/opp
sec=hyp/adj
cot=adj/opp
Moving between quadrants:
I to IV = make it negative and add 360º
I to III = add 180º
I to II = make it negative and add 180º
II to IV = move 180º
Law of sines (sinA/a) = (sinB/b) = (sinC/c)
Law of cosines (opp leg)² = (other adj leg)² -2(adj leg) (adj Leg) cos (angle b/w)
sinΘ = y/r
cosΘ = x/r
Law of sines (sinA/a) = (sinB/b) = (sinC/c)
Law of cosines (opp leg)² = (other adj leg)² -2(adj leg) (adj Leg) cos (angle b/w)
sinΘ = y/r
cosΘ = x/r
tanΘ = y/x
cotΘ = x/y
cscΘ = r/y
cotΘ = x/y
cscΘ = r/y
secΘ = r/x
r=√(x² + y²)
Graphing Trig functions:
y=Asin(Bx-h)+C
A = amplitude or height
B determines period(p)
p = (2π/B)
h = ((phase shift-horizontal shift)/(opposite))
C= vertical shift
csc = 1/sin x
sec = 1/cos x
cot x = 1/tan x
sin (-x) = -sin x
cos (-x) = cos x
csc (-x) = -csc x
sec (-x) = sec x
tan (-x) = -tan x
cot (-x) = -cot x
sin^2 x+cos^2 x = 1
1+tan^2 x = sec^2 x
1+cot^2 x = csc^2 x
sin x = cos(90*-x)
tan x = cot (90*-x)
sec x = csc (90*-x)
cos x = sin (90*-x)
cot x = tan (90*-x)
csc x = sec (90*-x)
tan x = sinx/cosx
cot x = cosx/sinx
Devin's Reflection
When ever you are solving a quadratic it is good to know the amounts of answers that you will be looking for.
x^2 + ... + c - Can answer the question of how many answer are in a quadratic.
What ever the highest exponent is, represents the number of answers you will get.
If it is x^4, then you will get 4 answers.
When you take a square root you get to answers.'
You get a positive and a negative answer.
If a + square root of(b) is a root then a - square root of(b) is also a root.
If a + bi is a root then a - bi is also a root.
If you have an odd degree at least one root must be a real root.
The sum of the roots
- 2nd coeff/leading coeff
The product of the roots
- if even constant/leading coeff
- if odd contant/leadin coeff
MERRY CHRISTMAS
x^2 + ... + c - Can answer the question of how many answer are in a quadratic.
What ever the highest exponent is, represents the number of answers you will get.
If it is x^4, then you will get 4 answers.
When you take a square root you get to answers.'
You get a positive and a negative answer.
If a + square root of(b) is a root then a - square root of(b) is also a root.
If a + bi is a root then a - bi is also a root.
If you have an odd degree at least one root must be a real root.
The sum of the roots
- 2nd coeff/leading coeff
The product of the roots
- if even constant/leading coeff
- if odd contant/leadin coeff
MERRY CHRISTMAS
Stephanie's Reflection
Trig Chart:
0°
sin0=0
cos0=1
tan0=0
csc0=undefined
sec0=1
cot0=0
30°
sinπ/6=1/2
cosπ/6=√3/2
tanπ/6=√3/3
cscπ/6=2
secπ/6=2 √3/3
cotπ/6=√3
45°
sinπ/4=√2/2
cosπ/4=√2/2
tanπ/4=1
cscπ/4=√2
secπ/4=√2
cotπ/4=1
60°
sinπ/3=√3/2
cosπ/3=1/2
tanπ/3=√3
cscπ/3=2 √3/3
secπ/3=2
cotπ/3=√3/2
90°
sinπ/2=1
cosπ/2=0
tanπ/2=undefined
cscπ/2=1
secπ/2=undefined
cotπ/2=0
Reference Angles (must be between 0° and 90°)
1)find which quadrant angle is in
2)determine the sign in that quadrant (+ve or -ve)
3)subtract 180° until the angle is between 0° and 90° (0 and π/2)
1)find the reference angle using chart or calculator
2)find what quadrant you need to be in based on the sign of the value
3)use notes to move to that quadrant
To Move:
I to IV = make negative and add 360°
I to III = add 180°
I to II = make negative and add 180°
II to IV = add 180°
Unit Circle
90 pi/2
180 pi
270 3pi/2
360 2pi
AREA OF A NON-RIGHT TRIANGLE
A=1/2(leg)(leg)sin(angle between)
RIGHT TRIANGLES
cos theta=adjacent/hypotenuse
tan theta=opposite/adjacent
LAW OF SINES
sinA/a sinB/b sinC/c
(opposite leg)²=(adjacent leg)² + (other leg)² - 2(adjacent leg)(adjacent leg)cos°
EG: x, 5, 6 angle = 35°
x²=5²+6²-2(5)(6)cos35°
x=√(5²+6²-2(5)(6)cos35°)
x≈3.443
Area of Inscribed Shapes
A=nr²sinΘcosΘ
SOLVING TRIG EQUATIONS
for any line m = tan alpha
m = slope , alpha = angle of inclination
For a conic: tan 2 alpha = B/A-C
if A=C then pi/4
A = coefficient of x^2, B = coefficient of xy, C = coefficient of y^2
y=Asin(Bx-h)+C
amplitude is hight
b is period p=2π/b
h is horizontal shift
c is vertical shift
Identities:
0°
sin0=0
cos0=1
tan0=0
csc0=undefined
sec0=1
cot0=0
30°
sinπ/6=1/2
cosπ/6=√3/2
tanπ/6=√3/3
cscπ/6=2
secπ/6=2 √3/3
cotπ/6=√3
45°
sinπ/4=√2/2
cosπ/4=√2/2
tanπ/4=1
cscπ/4=√2
secπ/4=√2
cotπ/4=1
60°
sinπ/3=√3/2
cosπ/3=1/2
tanπ/3=√3
cscπ/3=2 √3/3
secπ/3=2
cotπ/3=√3/2
90°
sinπ/2=1
cosπ/2=0
tanπ/2=undefined
cscπ/2=1
secπ/2=undefined
cotπ/2=0
Reference Angles (must be between 0° and 90°)
1)find which quadrant angle is in
2)determine the sign in that quadrant (+ve or -ve)
3)subtract 180° until the angle is between 0° and 90° (0 and π/2)
1)find the reference angle using chart or calculator
2)find what quadrant you need to be in based on the sign of the value
3)use notes to move to that quadrant
To Move:
I to IV = make negative and add 360°
I to III = add 180°
I to II = make negative and add 180°
II to IV = add 180°
Unit Circle
sin theta = y/r
cos theta = x/r
tan theta = y/x
csc theta = r/x
sec theta = x/y
cot theta = x/y
r=sqrtx^2+y^2
180 pi
270 3pi/2
360 2pi
sin pos in one and two and neg in three and four
cos pos in one and four and neg in two and three
tan pos in one and three and neg in two and four
cot pos in one and neg in two three and four
sec pos in one and neg in two three and four
csc pos in one and two and neg in three and fourAREA OF A NON-RIGHT TRIANGLE
A=1/2(leg)(leg)sin(angle between)
RIGHT TRIANGLES
- hypotenuse opposite angle
- a=1/2bh
- SOHCAHTOA
cos theta=adjacent/hypotenuse
tan theta=opposite/adjacent
LAW OF SINES
sinA/a sinB/b sinC/c
- used when you know pairs in a non-right triangle
- you are setting up proportions
- A=1/2bh
- SOHCAHTOA
- sinΘ=opposite leg/hypotenuse
- cosΘ=adjacent leg/hypotenuse
- tanΘ=opposite leg/adjacent leg
(opposite leg)²=(adjacent leg)² + (other leg)² - 2(adjacent leg)(adjacent leg)cos°
EG: x, 5, 6 angle = 35°
x²=5²+6²-2(5)(6)cos35°
x=√(5²+6²-2(5)(6)cos35°)
x≈3.443
Area of Inscribed Shapes
A=nr²sinΘcosΘ
SOLVING TRIG EQUATIONS
for any line m = tan alpha
m = slope , alpha = angle of inclination
For a conic: tan 2 alpha = B/A-C
if A=C then pi/4
A = coefficient of x^2, B = coefficient of xy, C = coefficient of y^2
y=Asin(Bx-h)+C
amplitude is hight
b is period p=2π/b
h is horizontal shift
c is vertical shift
Identities:
- check identities
- algebra
- identities
- cscΘ=1/sinΘ
- secΘ=1/cosΘ
- cotΘ=1/tanΘ
- sin -Θ= -sinΘ and cos -Θ= -cosΘ
- csc -Θ= -cscΘ and sec -Θ= -secΘ
- tan -Θ= -tanΘ and cot -Θ= -cotΘ
- sin²Θ+cos²Θ=1
- 1+tan²Θ=sec²Θ
- 1+cot²Θ=csc²Θ
- sinΘ=cos(90°-Θ) and cosΘ=sin(90°-Θ)
- tanΘ=cot(90°-Θ) and cotΘ=tan(90°-Θ)
- secΘ=csc(90°-Θ) and cscΘ=sec(90°-Θ)
Sunday, December 13, 2009
Taylor Reflection uh 17?
The one thing i can think to blog about is one of the very first things we went over in advanced math
that is how to sketch a parabola
**#1you need to see if the parabola will open up or down. think of it this was: if the first thing you see in the equation is a negative sign relate that to which way negative numbers go on a graph or think "if some thing is negative you get a thumbs down" like wise "if something is positive it gets a thumbs up" so first thing you see at the front of the equation is a negative sign? thumbs down therefore the parabola opens down. If the first thing you see is a positive number? thumbs up therefore the parabola opens up.(using analogies like this is good for memory. If you start thinking in terms of analogies you get faster at retaining information)
**#2deciding the number of X intercepts is also an easy remembering problem to fix. first you need to answer the problembsquared - 4(a)(c)as you said you are very good at plugging in this formula because you have remembered it well.look at your answer to that and remember: positive answer is two x interceptsnegative answer is nonezero for an answer is one X interceptits better to have two than none so POSITIVE thing to have TWONEGATIVE thing to have NONE(i dont have a trick to remember zero.. i think its just a process of elimination thing.. if i didnt get a positive answer or a negative anser that means its not two x intercepts nor is it no X intercepts,, well that means its one X intercept)
**#3 to find an x intercept you solve for Xit says that in your question"find X intercept"remember "find X"(dont forget to put answer into point form. when solving for x you will always wind up having to square root. you know this meas the answer will be +/-. be sure to show this when convertine to point form. {I.E. (#,0) & (-#,0)} in many of the problems we had there was also a matter of carring a number to the other side. this is no big deal you just tack it on also. for example... if you ended with X-2= +/- square root 6/2 you would add 2 to both sides and put in point form. therefore you'd have (squareroot 6/2+2,0) & (- squareroot 6/2+2,0))
**#4y- intercept is just taking the 0 in the y spot for the last answer and plugging it into the x spot in the equation. which then leaves you only the Y variable to solve for. Remember: "Find Y intercept""find Y"common sense will tell you the only way to do that is to plug something into the X spot.. and i told you what to plug in
**#5 Axis of semmatry is a simple conversion formula you'll have to memorize the same way you did for the quadratic formula. by writing it down everytime you solve for axis of semmatry until you see the formula in your sleep.the Formula (in case it isnt written down) is X= -b/2(a) (the a and b plug ins of course come from the original equation)your answer will be the point to put your DOTTED LINE on. because this formula solves for X you know it will pass through that point on the X line. You also know its a vertical line. so no worries.
**#6 the vertex is also just a matter of plugging in remember this step follows the step ahead of it so it retains the answer for X that means half of your vertex is solvedyou already have your X point for the vertex answer that answer is also plugged into the original equation which again leaves you to solve for y.this means you now have your vertex point because you solved for X in step 5 and your answer after plugging that in gave you the Y
FINALLY! now that you have turned everything into points its just a matter of locating them and marking them all on your graph.After each point is marked connect the dots.just as a quick check look back and see if your parabola is supposed to open up or down if your graph matches then congratulations! everything seems to have gone well.
i need some help on the translations chapter as well as help with chapter eight
i really need easy to remember steps for working the problems where you simplify using trig functions and the section where you use the unit circle to solve from chapter eight.
thanks!
that is how to sketch a parabola
**#1you need to see if the parabola will open up or down. think of it this was: if the first thing you see in the equation is a negative sign relate that to which way negative numbers go on a graph or think "if some thing is negative you get a thumbs down" like wise "if something is positive it gets a thumbs up" so first thing you see at the front of the equation is a negative sign? thumbs down therefore the parabola opens down. If the first thing you see is a positive number? thumbs up therefore the parabola opens up.(using analogies like this is good for memory. If you start thinking in terms of analogies you get faster at retaining information)
**#2deciding the number of X intercepts is also an easy remembering problem to fix. first you need to answer the problembsquared - 4(a)(c)as you said you are very good at plugging in this formula because you have remembered it well.look at your answer to that and remember: positive answer is two x interceptsnegative answer is nonezero for an answer is one X interceptits better to have two than none so POSITIVE thing to have TWONEGATIVE thing to have NONE(i dont have a trick to remember zero.. i think its just a process of elimination thing.. if i didnt get a positive answer or a negative anser that means its not two x intercepts nor is it no X intercepts,, well that means its one X intercept)
**#3 to find an x intercept you solve for Xit says that in your question"find X intercept"remember "find X"(dont forget to put answer into point form. when solving for x you will always wind up having to square root. you know this meas the answer will be +/-. be sure to show this when convertine to point form. {I.E. (#,0) & (-#,0)} in many of the problems we had there was also a matter of carring a number to the other side. this is no big deal you just tack it on also. for example... if you ended with X-2= +/- square root 6/2 you would add 2 to both sides and put in point form. therefore you'd have (squareroot 6/2+2,0) & (- squareroot 6/2+2,0))
**#4y- intercept is just taking the 0 in the y spot for the last answer and plugging it into the x spot in the equation. which then leaves you only the Y variable to solve for. Remember: "Find Y intercept""find Y"common sense will tell you the only way to do that is to plug something into the X spot.. and i told you what to plug in
**#5 Axis of semmatry is a simple conversion formula you'll have to memorize the same way you did for the quadratic formula. by writing it down everytime you solve for axis of semmatry until you see the formula in your sleep.the Formula (in case it isnt written down) is X= -b/2(a) (the a and b plug ins of course come from the original equation)your answer will be the point to put your DOTTED LINE on. because this formula solves for X you know it will pass through that point on the X line. You also know its a vertical line. so no worries.
**#6 the vertex is also just a matter of plugging in remember this step follows the step ahead of it so it retains the answer for X that means half of your vertex is solvedyou already have your X point for the vertex answer that answer is also plugged into the original equation which again leaves you to solve for y.this means you now have your vertex point because you solved for X in step 5 and your answer after plugging that in gave you the Y
FINALLY! now that you have turned everything into points its just a matter of locating them and marking them all on your graph.After each point is marked connect the dots.just as a quick check look back and see if your parabola is supposed to open up or down if your graph matches then congratulations! everything seems to have gone well.
i need some help on the translations chapter as well as help with chapter eight
i really need easy to remember steps for working the problems where you simplify using trig functions and the section where you use the unit circle to solve from chapter eight.
thanks!
Properties of Triangles
Since our midterm exam is on a lot of trigonometric stuff. I decided that I would go over some properties of triangles and quadrilaterals.
- The sum of the angles in a triangle add up to 180°
- The sum of the angles in any polygon other than a triangle add up to 360°
- Equilateral triangles have equal sides
- All equilateral triangles are equiangular, therefore, their angles are always 60°
- An isosceles triangle has at least two equal sides. Therefore, an equilateral triangle is an isosceles triangle.
- The perpendicular height of an isosceles triangle cuts the base into two equal pieces.
- The perpendicular height of an isosceles triangle cuts the upper angle into two equal angles.
- Two angles opposite any two equal sides are congruent(and vice-versa)
- The altitude of an isosceles triangle cuts the triangle into two right triangles, which can be used with trigonometric properties such as SOH CAH TOA and CHO SHA CAO
- The law of sines and cosines can be applied to a quadrilateral's diagonals to find the total area or perimeter of a quadrilateral.
I am still having trouble with simplifying trigonometric equations and "more difficult trigonometric equations" I seem to have trouble bringing the identities into context with the actual simplification of the equation.
Stephanie's Reflecion
Graphing Parabolas
~Descriminate tells you how many intercepts a graph has
b^2-4ac
if +ve 2 x-intercepts
if -ve no x-intercepts
if 0 1 x-intercept
~Axis of Symmetry x=-b/2a if non-standard form
~Vertex (-b/2a, f(-b/2a)) for non-standard form
~to find the intersections
solve for y
set equal
solve for x
plug back in
The number one thing you should know when solving a log is the basic formula (exponential form) which happens to be b^a=x while log for would be log b^x=a. When a problem does not have anything on the opposite side of the equal sign (opposite of log), put an x. If a log problem doesn't have a base, it is understood to be 10 because 10 is the default base.
CIRCLES
The equation of a circle in standard form is (x-h)^2-(y-h)^2=r^2 with the center being (h,k) and r being the radius.
Finding the intersection of a line and a circle:
1) solve linear equation for y
2) substitute in circle equation
3) solve for x
4) plug x in to get y value
(if x happens to be imaginary, there is no point of intersection)
ELLIPSES
1) (x-h)^2/(length of x/2)^2 + (y-k)^2/(length of y/2)^2 =1
2)center is (h,k)
3) major axis has larger denominator
4) vertex is on major axis
5) focus is smaller denom squared = larger denom squared - focus squared
focus is on major axis
Graphing:
1) find center
2) major axis = plus or minus the square root of the bigger denom
3) vertex
4) other intercepts
5) focus
6) length of major axis = 2 square root of
7) length of minor axis = 2 square root of
8) graph
HYPERBOLAS
1) (x+h)^2/(length/2)^2 - (y-k)^2/(length/2)^2 =1
OR
-(x-h)^2/(length/2)^2 + (y-k)^2/(length/2)^2 =1
2) center (h,k)
3) major axis is non-negative
4) vertex is the square root of non-negative denom
5) asymptotes y=+/-(square root of y)/(square root of x)x
6) focus^2 = x denom + y denom
focus^2 = vertex^2 + other denom
to sketch:
1) shape
2) center
3) major
4) minor
5) other intercept - none for hyperbolas
6) focus
7) asymptotes y=+/-square root of y/square root of x
8) vertex
9) sketch
A) draw a box using the vertex and +/-sr of other denom
B) draw diagonal through box corners
C) sketch a parabola on each vertex
D) label focus and asymptotes
TRIGONOMETRY
Trig Chart:
0°
sin0=0
cos0=1
tan0=0
csc0=undefined
sec0=1
cot0=0
30°
sinπ/6=1/2
cosπ/6=√3/2
tanπ/6=√3/3
cscπ/6=2
secπ/6=2 √3/3
cotπ/6=√3
45°
sinπ/4=√2/2
cosπ/4=√2/2
tanπ/4=1
cscπ/4=√2
secπ/4=√2
cotπ/4=1
60°
sinπ/3=√3/2
cosπ/3=1/2
tanπ/3=√3
cscπ/3=2 √3/3
secπ/3=2
cotπ/3=√3/2
90°
sinπ/2=1
cosπ/2=0
tanπ/2=undefined
cscπ/2=1
secπ/2=undefined
cotπ/2=0
Reference Angles (must be between 0° and 90°)
1)find which quadrant angle is in
2)determine the sign in that quadrant (+ve or -ve)
3)subtract 180° until the angle is between 0° and 90° (0 and π/2)
1)find the reference angle using chart or calculator
2)find what quadrant you need to be in based on the sign of the value
3)use notes to move to that quadrant
To Move:
I to IV = make negative and add 360°
I to III = add 180°
I to II = make negative and add 180°
II to IV = add 180°
Unit Circle
90 pi/2
180 pi
270 3pi/2
360 2pi
~Descriminate tells you how many intercepts a graph has
b^2-4ac
if +ve 2 x-intercepts
if -ve no x-intercepts
if 0 1 x-intercept
~Axis of Symmetry x=-b/2a if non-standard form
~Vertex (-b/2a, f(-b/2a)) for non-standard form
~to find the intersections
solve for y
set equal
solve for x
plug back in
The number one thing you should know when solving a log is the basic formula (exponential form) which happens to be b^a=x while log for would be log b^x=a. When a problem does not have anything on the opposite side of the equal sign (opposite of log), put an x. If a log problem doesn't have a base, it is understood to be 10 because 10 is the default base.
CIRCLES
The equation of a circle in standard form is (x-h)^2-(y-h)^2=r^2 with the center being (h,k) and r being the radius.
Finding the intersection of a line and a circle:
1) solve linear equation for y
2) substitute in circle equation
3) solve for x
4) plug x in to get y value
(if x happens to be imaginary, there is no point of intersection)
ELLIPSES
1) (x-h)^2/(length of x/2)^2 + (y-k)^2/(length of y/2)^2 =1
2)center is (h,k)
3) major axis has larger denominator
4) vertex is on major axis
5) focus is smaller denom squared = larger denom squared - focus squared
focus is on major axis
Graphing:
1) find center
2) major axis = plus or minus the square root of the bigger denom
3) vertex
4) other intercepts
5) focus
6) length of major axis = 2 square root of
7) length of minor axis = 2 square root of
8) graph
HYPERBOLAS
1) (x+h)^2/(length/2)^2 - (y-k)^2/(length/2)^2 =1
OR
-(x-h)^2/(length/2)^2 + (y-k)^2/(length/2)^2 =1
2) center (h,k)
3) major axis is non-negative
4) vertex is the square root of non-negative denom
5) asymptotes y=+/-(square root of y)/(square root of x)x
6) focus^2 = x denom + y denom
focus^2 = vertex^2 + other denom
to sketch:
1) shape
2) center
3) major
4) minor
5) other intercept - none for hyperbolas
6) focus
7) asymptotes y=+/-square root of y/square root of x
8) vertex
9) sketch
A) draw a box using the vertex and +/-sr of other denom
B) draw diagonal through box corners
C) sketch a parabola on each vertex
D) label focus and asymptotes
TRIGONOMETRY
Angles
- measured in degrees
- to find minutes, multiply what is behind the decimal by 60
- to find seconds, multiply what is behind the decimal by 0 and divide by 300 to get decimal
- angles are measured in degrees and radian
degrees = radians times 180 over pi
- to find coterminal angles, add or subtract 360 degrees or 2 pi
- must use degrees symbol if in degree or its wrong
Trig Chart:
0°
sin0=0
cos0=1
tan0=0
csc0=undefined
sec0=1
cot0=0
30°
sinπ/6=1/2
cosπ/6=√3/2
tanπ/6=√3/3
cscπ/6=2
secπ/6=2 √3/3
cotπ/6=√3
45°
sinπ/4=√2/2
cosπ/4=√2/2
tanπ/4=1
cscπ/4=√2
secπ/4=√2
cotπ/4=1
60°
sinπ/3=√3/2
cosπ/3=1/2
tanπ/3=√3
cscπ/3=2 √3/3
secπ/3=2
cotπ/3=√3/2
90°
sinπ/2=1
cosπ/2=0
tanπ/2=undefined
cscπ/2=1
secπ/2=undefined
cotπ/2=0
Reference Angles (must be between 0° and 90°)
1)find which quadrant angle is in
2)determine the sign in that quadrant (+ve or -ve)
3)subtract 180° until the angle is between 0° and 90° (0 and π/2)
1)find the reference angle using chart or calculator
2)find what quadrant you need to be in based on the sign of the value
3)use notes to move to that quadrant
To Move:
I to IV = make negative and add 360°
I to III = add 180°
I to II = make negative and add 180°
II to IV = add 180°
Unit Circle
sin theta = y/r
cos theta = x/r
tan theta = y/x
csc theta = r/x
sec theta = x/y
cot theta = x/y
r=sqrtx^2+y^2
180 pi
270 3pi/2
360 2pi
sin pos in one and two and neg in three and four
cos pos in one and four and neg in two and three
tan pos in one and three and neg in two and four
cot pos in one and neg in two three and four
sec pos in one and neg in two three and four
csc pos in one and two and neg in three and four
AREA OF A NON-RIGHT TRIANGLE
A=1/2(leg)(leg)sin(angle between)
RIGHT TRIANGLES
cos theta=adjacent/hypotenuse
tan theta=opposite/adjacent
LAW OF SINES
sinA/a sinB/b sinC/c
(opposite leg)²=(adjacent leg)² + (other leg)² - 2(adjacent leg)(adjacent leg)cos°
EG: x, 5, 6 angle = 35°
x²=5²+6²-2(5)(6)cos35°
x=√(5²+6²-2(5)(6)cos35°)
x≈3.443
Area of Inscribed Shapes
A=nr²sinΘcosΘ
SOLVING TRIG EQUATIONS
for any line m = tan alpha
m = slope , alpha = angle of inclination
For a conic: tan 2 alpha = B/A-C
if A=C then pi/4
A = coefficient of x^2, B = coefficient of xy, C = coefficient of y^2
y=Asin(Bx-h)+C
amplitude is hight
b is period p=2π/b
h is horizontal shift
c is vertical shift
Identities:
AREA OF A NON-RIGHT TRIANGLE
A=1/2(leg)(leg)sin(angle between)
RIGHT TRIANGLES
- hypotenuse opposite angle
- a=1/2bh
- SOHCAHTOA
cos theta=adjacent/hypotenuse
tan theta=opposite/adjacent
LAW OF SINES
sinA/a sinB/b sinC/c
- used when you know pairs in a non-right triangle
- you are setting up proportions
- A=1/2bh
- SOHCAHTOA
- sinΘ=opposite leg/hypotenuse
- cosΘ=adjacent leg/hypotenuse
- tanΘ=opposite leg/adjacent leg
(opposite leg)²=(adjacent leg)² + (other leg)² - 2(adjacent leg)(adjacent leg)cos°
EG: x, 5, 6 angle = 35°
x²=5²+6²-2(5)(6)cos35°
x=√(5²+6²-2(5)(6)cos35°)
x≈3.443
Area of Inscribed Shapes
A=nr²sinΘcosΘ
SOLVING TRIG EQUATIONS
for any line m = tan alpha
m = slope , alpha = angle of inclination
For a conic: tan 2 alpha = B/A-C
if A=C then pi/4
A = coefficient of x^2, B = coefficient of xy, C = coefficient of y^2
y=Asin(Bx-h)+C
amplitude is hight
b is period p=2π/b
h is horizontal shift
c is vertical shift
Identities:
- check identities
- algebra
- identities
- cscΘ=1/sinΘ
- secΘ=1/cosΘ
- cotΘ=1/tanΘ
- sin -Θ= -sinΘ and cos -Θ= -cosΘ
- csc -Θ= -cscΘ and sec -Θ= -secΘ
- tan -Θ= -tanΘ and cot -Θ= -cotΘ
- sin²Θ+cos²Θ=1
- 1+tan²Θ=sec²Θ
- 1+cot²Θ=csc²Θ
- sinΘ=cos(90°-Θ) and cosΘ=sin(90°-Θ)
- tanΘ=cot(90°-Θ) and cotΘ=tan(90°-Θ)
- secΘ=csc(90°-Θ) and cscΘ=sec(90°-Θ)
Stephen's reflection
Ok so the exam is on wednesday and these study guides take forever to do. Since the exam is on all 9 chapters, this test is gonna be pretty difficult. But im going to touch up on stuff that is pretty important. Logs. im basically going to describe the log properties. There are five log properties:
1. logbMN=logbM+logbN
2. logbM/N=logbM-logbN
3. logbM^k=KlogbM
4. log(b)B^k=k
5. b^logbK=k
If a question tells you to expand logbMN^2, then the answer is logbM+2logbN. If a problem tells you to condense log45-2log3 then you make it log45/9 which = log5. Expand means to make it bigger and condense means to make it smaller.
There are a few things i dont understand. One the trig chart, i cant remember it so if anyone can teach me tricks id be thankful. Two i have trouble remembering the simple stuff in chap 1 like the quad formula and how to complete the square.
1. logbMN=logbM+logbN
2. logbM/N=logbM-logbN
3. logbM^k=KlogbM
4. log(b)B^k=k
5. b^logbK=k
If a question tells you to expand logbMN^2, then the answer is logbM+2logbN. If a problem tells you to condense log45-2log3 then you make it log45/9 which = log5. Expand means to make it bigger and condense means to make it smaller.
There are a few things i dont understand. One the trig chart, i cant remember it so if anyone can teach me tricks id be thankful. Two i have trouble remembering the simple stuff in chap 1 like the quad formula and how to complete the square.
Alicia's Reflection #17
Alrighty well since our exam is on all of chapters 1-9, im going to reflect on some old stuff we learned in the beginning of the year.
CIRCLES:
The standard equation of a circle is (x-h)^2+(y-k)^2 .....the center is (h,k)
If the equation is not in standard form, you must complete the square to put it in standard form.
If you are given a center and a point, you can use the distance formula to find the radius.
To find the intersection of a line and a circle:
1. solve the linear eqn for y.
2. substitute in the circle eqn.
3. solve for x.
4. plug the x value in to get the y value.
***Reminder. If your x value is imaginary, then there is no point of intersection.
EX: find the center and radius.
(x-3)^2+(y+7)^2=19
c:(h,k)
center: (3,-7)
radius: square root of 19
EX: find the eqn of the circle with the center (1,4) through (3,7)
--in the problem you are given a center and a point so you would plug into the distance formula.
square root of (3-1)^2+(7-4)^2= square root of 4+9=square root of 13.
**13 has no root.
Your answer should be (x-1)^2+(y-4)^2=13
I also can use some help with finding integral coefficients. I had no clue how to do it on the study guides. Thankss!!
CIRCLES:
The standard equation of a circle is (x-h)^2+(y-k)^2 .....the center is (h,k)
If the equation is not in standard form, you must complete the square to put it in standard form.
If you are given a center and a point, you can use the distance formula to find the radius.
To find the intersection of a line and a circle:
1. solve the linear eqn for y.
2. substitute in the circle eqn.
3. solve for x.
4. plug the x value in to get the y value.
***Reminder. If your x value is imaginary, then there is no point of intersection.
EX: find the center and radius.
(x-3)^2+(y+7)^2=19
c:(h,k)
center: (3,-7)
radius: square root of 19
EX: find the eqn of the circle with the center (1,4) through (3,7)
--in the problem you are given a center and a point so you would plug into the distance formula.
square root of (3-1)^2+(7-4)^2= square root of 4+9=square root of 13.
**13 has no root.
Your answer should be (x-1)^2+(y-4)^2=13
I also can use some help with finding integral coefficients. I had no clue how to do it on the study guides. Thankss!!
Ok this weekend SUCKED! I left all my Advanced math study guides at the school because I was in a rush to the soccer game on Friday. Then i had to get a total of 15 service hours at the bonfire festival for religion. I had to go to a graduation party for my cousin, and I had to do a Christmas thing with my cousins in Laplace. I did not get too much time to do any of the homework that I should have been doing over the weekend, but anyway here's something that I just want to show off:
The Trig Chart =] I still remember this!!!
0° sin0=0 cos0=1 tan0=0 csc0=undefined sec0=1 cot0=undefined
30° sinπ/6=1/2 cosπ/6=√3/2 tanπ/6=√3/3 cscπ/6=2 secπ/6=2√3/3 cotπ/6=√3
45° sinπ/4=√2/2 cosπ/4=√2/2 tanπ/4=1 cscπ/4=√2/2 secπ/4=√2/2 cotπ/4=1
60° sinπ/3=√3/2 cosπ/3=1/2 tanπ/3=√3 cscπ/3=2√3/3 secπ/3=2 cotπ/3=√3/3
90° sinπ/2=1 cosπ/2=0 tanπ/2=undefined cscπ/2=1 secπ/2=undefined cotπ/2=0
Oh yeah I just did that.
One thing I still don't know how to do is writing the equations to shifts. I don't even know where to start with this...someone please help me with this
The Trig Chart =] I still remember this!!!
0° sin0=0 cos0=1 tan0=0 csc0=undefined sec0=1 cot0=undefined
30° sinπ/6=1/2 cosπ/6=√3/2 tanπ/6=√3/3 cscπ/6=2 secπ/6=2√3/3 cotπ/6=√3
45° sinπ/4=√2/2 cosπ/4=√2/2 tanπ/4=1 cscπ/4=√2/2 secπ/4=√2/2 cotπ/4=1
60° sinπ/3=√3/2 cosπ/3=1/2 tanπ/3=√3 cscπ/3=2√3/3 secπ/3=2 cotπ/3=√3/3
90° sinπ/2=1 cosπ/2=0 tanπ/2=undefined cscπ/2=1 secπ/2=undefined cotπ/2=0
Oh yeah I just did that.
One thing I still don't know how to do is writing the equations to shifts. I don't even know where to start with this...someone please help me with this
Saturday, December 12, 2009
Amy's Reflection #17
ok since we're having exams i think im gonna post something from the begininng of the year..ya know things we might have forgotten to do...
Completing the Square:
You can use completing the square to solve a quadratic equation when factoring doesn’t work. This method can only work when 1 is the coefficient of x².
For example:
x² + 6x - 2 = 0
x² + 6x = 2
x² + 6x + 9 = -2 + 9
(x + 3)² = 7
x + 3 = √7
x = -3 ± √7
(-3 + √7,0) (-3 -√7,0)
Rational Root therom:
Example: f(x)= 2x^3 + 3x^2 - 8 + 3
Step 1: find all possible roots..
p: factors of 3: 1, -1, 3, -3
q: factors of 2: 1, -1, 2, -2
*p is the leading constant term & q is the leading coefficient
possible roots are (p/q): 1, -1, 1/2, -1/2, 3, -3, 3/2, -3/2
Step 4: slove further (factor): (x - 1) (2x^2 + 5x + 3)= (x - 1) (2x - 1) (x + 3)
Answer: x = 1, 1/2, -3
Domain & Range of functions:
Polynomials-domain of all polynomials is (−∞, ∞).
Fractions-you set the bottom to zero, solve for x, and then set up intervals
Square Roots-domain: set the inside = to zero, then set a # line, try values on either side of each #, and get ride of the negatives-range:graph
Absolute Value-domain: (- ∞ , + ∞)-range: [0 , + ∞)
How to Find the Inverse of a Function:
Completing the Square:
You can use completing the square to solve a quadratic equation when factoring doesn’t work. This method can only work when 1 is the coefficient of x².
For example:
x² + 6x - 2 = 0
x² + 6x = 2
x² + 6x + 9 = -2 + 9
(x + 3)² = 7
x + 3 = √7
x = -3 ± √7
(-3 + √7,0) (-3 -√7,0)
Rational Root therom:
Example: f(x)= 2x^3 + 3x^2 - 8 + 3
Step 1: find all possible roots..
p: factors of 3: 1, -1, 3, -3
q: factors of 2: 1, -1, 2, -2
*p is the leading constant term & q is the leading coefficient
possible roots are (p/q): 1, -1, 1/2, -1/2, 3, -3, 3/2, -3/2
Step 2: plug roots in calc & the zeros will be: 1, 1/2, -3
Step 3: synthetic division: (x - 1) (2x^2 + 5x + 3)Step 4: slove further (factor): (x - 1) (2x^2 + 5x + 3)= (x - 1) (2x - 1) (x + 3)
Answer: x = 1, 1/2, -3
Domain & Range of functions:
Polynomials-domain of all polynomials is (−∞, ∞).
Fractions-you set the bottom to zero, solve for x, and then set up intervals
Square Roots-domain: set the inside = to zero, then set a # line, try values on either side of each #, and get ride of the negatives-range:graph
Absolute Value-domain: (- ∞ , + ∞)-range: [0 , + ∞)
How to Find the Inverse of a Function:
- Replace f(x) with y
- Reverse the roles of x and y
- Solve for y in terms of x
- Replace y with f-1(x)
- check: should equal to x
Example:
f(x) = √x + 4
(x)^2 = (√y + 4)^2
x^2 = y + 4
y = x^2 - 4
f-1(x) = (x^2 - 4)
f(f-1(x)) = f(x^2 - 4) = √(x^2 - 4) + 4 = x
f-1(f(x)) = f-1(√x + 4) = (√x + 4)^2 - 4 = x + 4 - 4 = x
Logarithm Properties:
- logb MN = logb M + logb N
- logb M/N = logb M - logb N
- logb M^K = K logb M
- logb b^k = k
- b^logb^k = k
Changing Bases: (Done when you can't solve a log)
- Rewrite it as an exponential
- Take the log of both sides
- Move the variable to the front
- then solve
Example:
log5 10 = x
5^x = 10
log 5^x = log 10
x log 5 = 1
x = 1/log 5
i hope this helped y'all remember some of this stuff...i still kinda need help with the graphing in chapter eight..like the question on the test we took monday...can someone help me with that??
Friday, December 11, 2009
Devin's Makeup #6
Log properties
-logb MN=logb M+ logb N
-logb M/N= logb M- lob N
-logb M^k= K logb M
-logb b^k=k
=b^logbK=K
Expand logb MN^2
logb M+logbN
Condene log 45-2log3
log(45/9)lo5
Write y in terms of x if lny=1/3
lny=1/3 lnx + ln4
lny=lnx1/3y=4x1/3
Changing the bases
done when you can not solve a log
-rewrite it as an exponential
-take the log of both sides
-move variable to the front
-solve
-use the same steps when solving for x as an exponent when you can not write them as the same base
-logb MN=logb M+ logb N
-logb M/N= logb M- lob N
-logb M^k= K logb M
-logb b^k=k
=b^logbK=K
Expand logb MN^2
logb M+logbN
Condene log 45-2log3
log(45/9)lo5
Write y in terms of x if lny=1/3
lny=1/3 lnx + ln4
lny=lnx1/3y=4x1/3
Changing the bases
done when you can not solve a log
-rewrite it as an exponential
-take the log of both sides
-move variable to the front
-solve
-use the same steps when solving for x as an exponent when you can not write them as the same base
Devin's Makeup #5
Domain of all polynomials is -oo,oo
Range of all odds is -oo,oo
Range of all quadratics is vertex,oo or -oo,vertex
With fractions
-set bottom equal to zero
-solve for x
-set up intervals
(-oo, )( , )...( ,oo)
With Absolute Value
Domain - (-oo,oo)
RAnge [shift,oo) or (-oo,shift]
With Square roots
-set inside = 0
-set up a # line
-try values in either side
-eliminate anything -ve
-set up intervals
Domain-x-value
Range-y-values
-() means not included
-[] means included
Ex- y=x^2+5x+6
domain- -oo,oo
range- -5/2,oo
Ex. x+4/x2-25
domain -00,-5U-5,oo
Ex. x+4-1
domain -oo,oo
range [-1,oo)
Range of all odds is -oo,oo
Range of all quadratics is vertex,oo or -oo,vertex
With fractions
-set bottom equal to zero
-solve for x
-set up intervals
(-oo, )( , )...( ,oo)
With Absolute Value
Domain - (-oo,oo)
RAnge [shift,oo) or (-oo,shift]
With Square roots
-set inside = 0
-set up a # line
-try values in either side
-eliminate anything -ve
-set up intervals
Domain-x-value
Range-y-values
-() means not included
-[] means included
Ex- y=x^2+5x+6
domain- -oo,oo
range- -5/2,oo
Ex. x+4/x2-25
domain -00,-5U-5,oo
Ex. x+4-1
domain -oo,oo
range [-1,oo)
Devin's Makeup #4
To sketch polynomids function do the following
-factor completely
-set up a # line
-plug in either sid of your roots
-+ve above x-axis
--ve below x-axis
-check in calc
-max and min in calc. only
Quadratic max and min
-x=-b/2a te vertex is always your min and max
Inequalities
-chang sign when you multiply or divide by a negative
-if (x+4)(x-1)>0
then break it ip and solve
(x+4)>0
x>-4
(x-1)>0
x>1
Ex. 3x-9>4
3x<5
x<5/3
3x>3
x>1
-factor completely
-set up a # line
-plug in either sid of your roots
-+ve above x-axis
--ve below x-axis
-check in calc
-max and min in calc. only
Quadratic max and min
-x=-b/2a te vertex is always your min and max
Inequalities
-chang sign when you multiply or divide by a negative
-if (x+4)(x-1)>0
then break it ip and solve
(x+4)>0
x>-4
(x-1)>0
x>1
Ex. 3x-9>4
3x<5
x<5/3
3x>3
x>1
Devin's Makeup #3
To solve anything bigger than a quadratic you can use grahing, you can use the quadratic form, or you can use the rational root theroem.
The first way to solve things bigger than a quadratic is to use factoring.
-to factor there must be an even number of terms
Ex. x^3+5x^2-4x-20=0
(x^3+5x^2)-(4x+20=0
x^2(x+5)-4(x+5)=0
(x^2-4)(x+5)=0
x^2-4=0
x^2=4
x=+-2
x+5=0
x=-5
(2,0)(-2,0)(-5,0)
The second way to solve equations bigger than a quadratic is the quadratic form.
-must have only 3 terms
-1st term must exponent must = 2nd exponent x2
-Last term must be a nmber
-do quadratic formula, factor, complete the square
-plug back in for g
Ex. 2x^4-x^2-3=0
g=x^4/2 g=x^2
(2g^2-3g)+(2g-2)
g(2g-3)1(2g-3)
(g+1)(2g-3)=0
g+1=0
g=x^2
x^2=-1
x=+-i
2g-3=0
g=3/2
The rational root thereom is the final way to solve equations bigger than an quadratic.
What to do is take the roots of the equation and use synthetic division to solve.
The first way to solve things bigger than a quadratic is to use factoring.
-to factor there must be an even number of terms
Ex. x^3+5x^2-4x-20=0
(x^3+5x^2)-(4x+20=0
x^2(x+5)-4(x+5)=0
(x^2-4)(x+5)=0
x^2-4=0
x^2=4
x=+-2
x+5=0
x=-5
(2,0)(-2,0)(-5,0)
The second way to solve equations bigger than a quadratic is the quadratic form.
-must have only 3 terms
-1st term must exponent must = 2nd exponent x2
-Last term must be a nmber
-do quadratic formula, factor, complete the square
-plug back in for g
Ex. 2x^4-x^2-3=0
g=x^4/2 g=x^2
(2g^2-3g)+(2g-2)
g(2g-3)1(2g-3)
(g+1)(2g-3)=0
g+1=0
g=x^2
x^2=-1
x=+-i
2g-3=0
g=3/2
The rational root thereom is the final way to solve equations bigger than an quadratic.
What to do is take the roots of the equation and use synthetic division to solve.
Thursday, December 10, 2009
make up blog #3
I'm amazing at chapter 9!
law of sins :)
sinA/a=sinB/b=sinC/c
*only used when you have pairs, an angle and the side opposite of it.
*setting up a proportion.
Ex: a civil engineer wants to determine the distance from points A and B to an inaccessable point C. from direct measurement -- AB=25m,
first you would draw a diagram and lable EVERYTHING. Then, choose your pairs. Finally set up a proportion.
(sin50/25)=(sin20/B)
cross multiply--Bsin50=25sin20
divide by sin50
B=11.162m
you would follow the same process to find side "a".
I still don't understand integral coefficients if anyone wants to help.
law of sins :)
sinA/a=sinB/b=sinC/c
*only used when you have pairs, an angle and the side opposite of it.
*setting up a proportion.
Ex: a civil engineer wants to determine the distance from points A and B to an inaccessable point C. from direct measurement -- AB=25m,
first you would draw a diagram and lable EVERYTHING. Then, choose your pairs. Finally set up a proportion.
(sin50/25)=(sin20/B)
cross multiply--Bsin50=25sin20
divide by sin50
B=11.162m
you would follow the same process to find side "a".
I still don't understand integral coefficients if anyone wants to help.
make up blog #2
i'm going over old stuff, and for the most part, i remember everything. so i'm explaining how to complete the square.
when given an equation (ax^2+bx+c=0) where "c" is the constant, first add or subtract the constant over (ax^2+bx+_=-c). Second, divide the linear term (b) by two and then square it (b/2)^2. Your would then add your squared, lets call it "d", term to both sides of the equal sign (ax^2+bx+d=-c+d). You would then factor to get (ax+b/2)^2=-c+d. After factoring, you would take the square root of both sides (ax+b/2)=+/- √-c+d. Then, you would add or subtract (b/2) to both sides. And finally, you would divide by "a" to get "x" by itself. You would then write in coordinate form.
What I don't understand from the exam review is how to get an equation from INTEGRAL COEFFICIENTS!!!!! help.
when given an equation (ax^2+bx+c=0) where "c" is the constant, first add or subtract the constant over (ax^2+bx+_=-c). Second, divide the linear term (b) by two and then square it (b/2)^2. Your would then add your squared, lets call it "d", term to both sides of the equal sign (ax^2+bx+d=-c+d). You would then factor to get (ax+b/2)^2=-c+d. After factoring, you would take the square root of both sides (ax+b/2)=+/- √-c+d. Then, you would add or subtract (b/2) to both sides. And finally, you would divide by "a" to get "x" by itself. You would then write in coordinate form.
What I don't understand from the exam review is how to get an equation from INTEGRAL COEFFICIENTS!!!!! help.
Make up blon #1
Okay so I guess i'm going to talk about what is going on in chapter 8-4. So, here goes.
I really think that I got most of this chapter.
IDENTITIES AND EQUATIONS
rules to simplify: 1. identities, 2. algebra, 3. identities.
Pretty much all this section is, is knowing your identities.
secθ=1/cosθ
cscθ=1/sinθ
cotθ=1/tanθ OR sinθ/cosθ
^^I think these three are you most important, besideds *polynomial _*
So pretty much, what you do is---
(tanA)(cotA)
(sinA/cosA )(cosA/sinA )
sinAcosA/sinAcosA =1
What I don't really understand in chapter 8 is graphing. If anyone really knows it, that would help a lot.
I really think that I got most of this chapter.
IDENTITIES AND EQUATIONS
rules to simplify: 1. identities, 2. algebra, 3. identities.
Pretty much all this section is, is knowing your identities.
secθ=
cscθ=
cotθ=
^^I think these three are you most important, besideds *polynomial _*
So pretty much, what you do is---
(tanA)(cotA)
(
What I don't really understand in chapter 8 is graphing. If anyone really knows it, that would help a lot.
Reflection 3
In this reflection I will help explain how to solve trig equations with any line in the equation.
For these equations you would use m = tanZ. Where as m would be the slope and Z would be the angle of inclination. If A would happen to equal C in the equation then the angle of inclination would always be pie over four.
An example of this problem would be to find the angle of 2X + 5Y = 15. For this equation m would be -2/5 which would make the equation tanZ = -2/5. You would then have to take the inverse of tan to make change it to Z = tan^-1(-2/5). The answer would come out to be 21.801 degrees. Since tan is positive in quadrant one and three it would be negative in quadrant two and four. So we would have to take 21.801 degrees, make it negative and add 180 degrees which then makes it 158.199 degrees. Now we take the 21.801 degrees and to change it to quadrant four so we make it negative and add 360 degrees which then comes out to 338.199 degrees. So therefore the answer would be Z = 158.199 degrees, and 338.199 degrees.
For these equations you would use m = tanZ. Where as m would be the slope and Z would be the angle of inclination. If A would happen to equal C in the equation then the angle of inclination would always be pie over four.
An example of this problem would be to find the angle of 2X + 5Y = 15. For this equation m would be -2/5 which would make the equation tanZ = -2/5. You would then have to take the inverse of tan to make change it to Z = tan^-1(-2/5). The answer would come out to be 21.801 degrees. Since tan is positive in quadrant one and three it would be negative in quadrant two and four. So we would have to take 21.801 degrees, make it negative and add 180 degrees which then makes it 158.199 degrees. Now we take the 21.801 degrees and to change it to quadrant four so we make it negative and add 360 degrees which then comes out to 338.199 degrees. So therefore the answer would be Z = 158.199 degrees, and 338.199 degrees.
Reflection 2
This reflection is about solving trig equations. A few things to know about solving these equations are that sin is positive on the y-axis, cos is positive on the x-axis, and that tan is positive in quadrant one and three. To move from quadrant to quadrant is quite easy. To move from quadrant one to four u make the first point negative and add 360 degrees, from one to three you add 180 degrees, one to two you make negative and add 180 degrees, and to move from two to four you add 180 degrees.
An example of one is find the value of X between 0 and 2 pie for which sinX = 0.6. You would set it up as sinX 0.6. Then you would take an inverse of sin so that you can find the value of X. So this then makes it X = sin^-1(0.6). So then you would put that into your calculator and you should get the answer 36.870 degrees. Since sin is positive on the y-axis, you would have to find the points in quadrant one and two. Since 36.870 degrees is already in quadrant one, so that is our first answer. Now to find the point in quadrant two we would have to make our first point negative and add 180 degrees. After doing that we would get the point 143.130 degrees. So therefore the answer would be 36.870 degrees and 143.140 degrees.
An example of one is find the value of X between 0 and 2 pie for which sinX = 0.6. You would set it up as sinX 0.6. Then you would take an inverse of sin so that you can find the value of X. So this then makes it X = sin^-1(0.6). So then you would put that into your calculator and you should get the answer 36.870 degrees. Since sin is positive on the y-axis, you would have to find the points in quadrant one and two. Since 36.870 degrees is already in quadrant one, so that is our first answer. Now to find the point in quadrant two we would have to make our first point negative and add 180 degrees. After doing that we would get the point 143.130 degrees. So therefore the answer would be 36.870 degrees and 143.140 degrees.
Devin's Makeup #2
Polynomids and Function Notation
a degree is the highest exponent.
-0-constant
-1-linear
-2-quadratic
-3-cubic
-4-quartic
-5-quintic
-a root, a zero, and x-intercepts are the same thing. A polynomial is an equation with only addition and subtraction of terms. The leading term of the function is the term with the highest exponent.
The function notation is f(x). Plug in what is in the parenthesis for x. To factor a polynomial use synthetic division.
x^9+5x^7+4x^10+9x^2+4
There are a few questions that can be used to help solve polynomials.
a. is it a polynomial?
yes
b. what is the degree?
10
c. what is the leading term?
4x^10
d. what is the leading coefficient
4
a degree is the highest exponent.
-0-constant
-1-linear
-2-quadratic
-3-cubic
-4-quartic
-5-quintic
-a root, a zero, and x-intercepts are the same thing. A polynomial is an equation with only addition and subtraction of terms. The leading term of the function is the term with the highest exponent.
The function notation is f(x). Plug in what is in the parenthesis for x. To factor a polynomial use synthetic division.
x^9+5x^7+4x^10+9x^2+4
There are a few questions that can be used to help solve polynomials.
a. is it a polynomial?
yes
b. what is the degree?
10
c. what is the leading term?
4x^10
d. what is the leading coefficient
4
Devin's Makeup #1
Graphing parabolas can be very difficult but when one knows the main information about graphing parabolas it can be very simple. One key to graphing parabolas is knowing the amount of intercepts.
The discriminate tells how many intercepts are affiliated with the parabola. To use the discriminate you must know the formula.
b^2-4ac
if=+ve_2 x intercept
if=-ve_no x intercept
if=0_1 x intercept
To find the axis of symmetry use the formula x=-b/2a
To find the vertex use the same formula that was usedd to find the axis of symmetry
With the axis of symmetry the answer would be x=a
With the vertex the answer would be (a,0)
To find the intersection do the following
-solve for y
-set them equal
-solve for x
-plug back in
The discriminate tells how many intercepts are affiliated with the parabola. To use the discriminate you must know the formula.
b^2-4ac
if=+ve_2 x intercept
if=-ve_no x intercept
if=0_1 x intercept
To find the axis of symmetry use the formula x=-b/2a
To find the vertex use the same formula that was usedd to find the axis of symmetry
With the axis of symmetry the answer would be x=a
With the vertex the answer would be (a,0)
To find the intersection do the following
-solve for y
-set them equal
-solve for x
-plug back in
Reflection 1
In this reflection I am going to talk about how to find relationships among trig functions. To simplify these you would first you would check identities. cscX = 1/sinX, secX = 1/cosX, cotX = 1/tanX, tanX = sinX/cosX, cotX = cosX/sinX. Other identities are sin^2X + cos^2X = 1, 1 + tan^2X = sec^2X, 1 + cot^2X = csc^2X. There are also identities for negative relationships: sin(-X) = -sinX and cos(-X) = cosX, csc(-X) = -cscX and sec(-X) = secX, tan(-X) = -tanX and cot(-X) = -cotX. And now for the identities of the cofunction realationships: sinX = cos( 90 degrees-X) and cosX = sin(90 degrees-X), tanX = cot(90 degrees-X) and cotX = tan(90 degrees-X), secX = csc(90 degrees-X) and cscX = sec(90 degrees-X). Second you would see if you could use simple algebra, as in factoring or completing the square. Then u would check your identities and keep this process up until u come up with a simple answer.
So in the example: secX - sinXtanX
You would switch out the secX with 1/cosX and switch the tanX with sinX/cosX. Now it would look like 1/cosX - sinx(sinX/cosX)
Then you would distribute the sinX and get 1 - sin^2X/cosX
Then you would look at the sin^2X + cos^2X = 1. Now you would have to try to get 1 - sin^2X on one side so you would subtract the sin^2X. Then you would replace 1 - sin^2X in the equation with cos^2X to get cos^2X/cosX
Now you would use simple algebra and divide to get the answer of cosX
So in the example: secX - sinXtanX
You would switch out the secX with 1/cosX and switch the tanX with sinX/cosX. Now it would look like 1/cosX - sinx(sinX/cosX)
Then you would distribute the sinX and get 1 - sin^2X/cosX
Then you would look at the sin^2X + cos^2X = 1. Now you would have to try to get 1 - sin^2X on one side so you would subtract the sin^2X. Then you would replace 1 - sin^2X in the equation with cos^2X to get cos^2X/cosX
Now you would use simple algebra and divide to get the answer of cosX
Wednesday, December 9, 2009
Alicia's comments
I dont understand how to determine when to solve trig equations using the identities or algebra. I always get confused on if i completely solved it using algebra or identities. I never know when its completely solved.... any help???
Monday, December 7, 2009
Stephen's Reflection
Ok so this week we learned about trig functions. I think the easiest thing to talk about is figuring out which quadrants they are in. Things to remember about trig functions are that sin is positive in quadrant 1 and 2 which is y, cosine is positive in 1 and 4 which is x, and tangent is positive in 1 and 3 which is both + x/y and -x, -y.
Ex: find the values of x between 0 greater than or equal to x greater than or equal to 360 for sinx=.6
you first have to solve for x so you take the inverse of sin on both sides which is sin^-1 so it will look like x=sin^-1(.6)
then you solve sin^-1(.6) which = 36.870. That is in the first quadrant so that is the first answer out of 2.
To move to quadrant two to find the 2nd answer, you make it negative and add 180.
180-36.870=143.130 and that is your second answer.
I understand most of this section but there is one thing i could use help on. When turning a trig function into another, i always do the more complicated for some reason. I dont know if there is an easier way to do it so i use the one im most certain of. So i need advice with that if you know how to give it. And another question, will i always get the same answer as the easier way if i do it the harder way cause that always worries me.
Ex: find the values of x between 0 greater than or equal to x greater than or equal to 360 for sinx=.6
you first have to solve for x so you take the inverse of sin on both sides which is sin^-1 so it will look like x=sin^-1(.6)
then you solve sin^-1(.6) which = 36.870. That is in the first quadrant so that is the first answer out of 2.
To move to quadrant two to find the 2nd answer, you make it negative and add 180.
180-36.870=143.130 and that is your second answer.
I understand most of this section but there is one thing i could use help on. When turning a trig function into another, i always do the more complicated for some reason. I dont know if there is an easier way to do it so i use the one im most certain of. So i need advice with that if you know how to give it. And another question, will i always get the same answer as the easier way if i do it the harder way cause that always worries me.
Sunday, December 6, 2009
This week we learned how to simplify and prove trig equations by using a mixture of both identities and algebra. Here are some of the relationships you need to look for in order to solve the equations.
Reciprocal Relationships:
cscΘ=1/sinΘ
secΘ=1/cosΘ
cotΘ=1/tanΘ
Relationships with Negatives:
sin -Θ= -sinΘ and cos -Θ= -cosΘ
csc -Θ= -cscΘ and sec -Θ= -secΘ
tan -Θ= -tanΘ and cot -Θ= -cotΘ
Pythagorean Relationships:
sin²Θ+cos²Θ=1
1+tan²Θ=sec²Θ
1+cot²Θ=csc²Θ
Cofunction Relationships:
sinΘ=cos(90°-Θ) and cosΘ=sin(90°-Θ)
tanΘ=cot(90°-Θ) and cotΘ=tan(90°-Θ)
secΘ=csc(90°-Θ) and cscΘ=sec(90°-Θ)
When you get an equation you have to first check to see if there are any identities you can use, if not you go to algebra, after that you go back to your identities and finish the problem. This is really easy you just need to memorize the relationships.
One thing I had trouble with this week and could use some help on however was writing the equation for shifts.
Reciprocal Relationships:
cscΘ=1/sinΘ
secΘ=1/cosΘ
cotΘ=1/tanΘ
Relationships with Negatives:
sin -Θ= -sinΘ and cos -Θ= -cosΘ
csc -Θ= -cscΘ and sec -Θ= -secΘ
tan -Θ= -tanΘ and cot -Θ= -cotΘ
Pythagorean Relationships:
sin²Θ+cos²Θ=1
1+tan²Θ=sec²Θ
1+cot²Θ=csc²Θ
Cofunction Relationships:
sinΘ=cos(90°-Θ) and cosΘ=sin(90°-Θ)
tanΘ=cot(90°-Θ) and cotΘ=tan(90°-Θ)
secΘ=csc(90°-Θ) and cscΘ=sec(90°-Θ)
When you get an equation you have to first check to see if there are any identities you can use, if not you go to algebra, after that you go back to your identities and finish the problem. This is really easy you just need to memorize the relationships.
One thing I had trouble with this week and could use some help on however was writing the equation for shifts.
Alicia's Reflection #16
Alright well this week we learned identities and equations. At first it was really hard for me until I learned my trig identities. Here are the steps to solving these equations:
1. check identities
2. algebra
3. identities
These are the identities to memorize:
Reciprocal Relationships:
cscΘ=1/sinΘ
secΘ=1/cosΘ
cotΘ=1/tanΘ
Relationships with Negatives:
sin -Θ= -sinΘ and cos -Θ= -cosΘ
csc -Θ= -cscΘ and sec -Θ= -secΘ
tan -Θ= -tanΘ and cot -Θ= -cotΘ
Pythagorean Relationships:
sin²Θ+cos²Θ=1
1+tan²Θ=sec²Θ
1+cot²Θ=csc²Θ
Cofunction Relationships:
sinΘ=cos(90°-Θ) and cosΘ=sin(90°-Θ)
tanΘ=cot(90°-Θ) and cotΘ=tan(90°-Θ)
secΘ=csc(90°-Θ) and cscΘ=sec(90°-Θ)
***Cautions***
You cannot divide trig functions to cancel them
You can move everything to one side and factor out a trig function
You can divide by a trig function to create a new one
Helpful information:
I - II : make negative then add 180
I- III : add 180
I - IV : make negative then add 360
Sin- positive in 1 and 2... negative in 3 and 4
Cos- positive in 1 and 4. ..negative in 2 and 3
Tan- positive in 1 and 3... negative in 2 and 4
I could use some help with proving equations in section 8-4 if anyone has any helpful hints!!
Goodluck on the Chapter 8 Test tomorrow :)
1. check identities
2. algebra
3. identities
These are the identities to memorize:
Reciprocal Relationships:
cscΘ=1/sinΘ
secΘ=1/cosΘ
cotΘ=1/tanΘ
Relationships with Negatives:
sin -Θ= -sinΘ and cos -Θ= -cosΘ
csc -Θ= -cscΘ and sec -Θ= -secΘ
tan -Θ= -tanΘ and cot -Θ= -cotΘ
Pythagorean Relationships:
sin²Θ+cos²Θ=1
1+tan²Θ=sec²Θ
1+cot²Θ=csc²Θ
Cofunction Relationships:
sinΘ=cos(90°-Θ) and cosΘ=sin(90°-Θ)
tanΘ=cot(90°-Θ) and cotΘ=tan(90°-Θ)
secΘ=csc(90°-Θ) and cscΘ=sec(90°-Θ)
***Cautions***
You cannot divide trig functions to cancel them
You can move everything to one side and factor out a trig function
You can divide by a trig function to create a new one
Helpful information:
I - II : make negative then add 180
I- III : add 180
I - IV : make negative then add 360
Sin- positive in 1 and 2... negative in 3 and 4
Cos- positive in 1 and 4. ..negative in 2 and 3
Tan- positive in 1 and 3... negative in 2 and 4
I could use some help with proving equations in section 8-4 if anyone has any helpful hints!!
Goodluck on the Chapter 8 Test tomorrow :)
This week wasn't that difficult and we even had a lucky break by having the test pushed back until monday. The week consisted of learning the section of chapter eight that focuses on using the Trig identites
TRIG IDENTITIES
Reciprocal Relationships
cscΘ=1/sinΘ
secΘ=1/cosΘ
cotΘ=1/tanΘ
Relationships with Negatives
sin -Θ= -sinΘ and cos -Θ= -cosΘ
csc -Θ= -cscΘ and sec -Θ= -secΘ
tan -Θ= -tanΘ and cot -Θ= -cotΘ
Pythagorean Relationsihps
sin²Θ+cos²Θ=1
1+tan²Θ=sec²Θ
1+cot²Θ=csc²Θ
Cofunction Relationships
sinΘ=cos(90°-Θ) and cosΘ=sin(90°-Θ)
tanΘ=cot(90°-Θ) and cotΘ=tan(90°-Θ)
secΘ=csc(90°-Θ) and cscΘ=sec(90°-Θ)
This section isnt a difficult section to comprehend it just takes alot of memorization.
What i dont feel like i understand is the section involving quadrent shifts to find the answer
if anyone could give me basic easy peasy to remember and follow steps that would be great
Thanks!
TRIG IDENTITIES
Reciprocal Relationships
cscΘ=1/sinΘ
secΘ=1/cosΘ
cotΘ=1/tanΘ
Relationships with Negatives
sin -Θ= -sinΘ and cos -Θ= -cosΘ
csc -Θ= -cscΘ and sec -Θ= -secΘ
tan -Θ= -tanΘ and cot -Θ= -cotΘ
Pythagorean Relationsihps
sin²Θ+cos²Θ=1
1+tan²Θ=sec²Θ
1+cot²Θ=csc²Θ
Cofunction Relationships
sinΘ=cos(90°-Θ) and cosΘ=sin(90°-Θ)
tanΘ=cot(90°-Θ) and cotΘ=tan(90°-Θ)
secΘ=csc(90°-Θ) and cscΘ=sec(90°-Θ)
This section isnt a difficult section to comprehend it just takes alot of memorization.
What i dont feel like i understand is the section involving quadrent shifts to find the answer
if anyone could give me basic easy peasy to remember and follow steps that would be great
Thanks!
Amy's Reflection #16
This week we learn about Idenities and Equations...
Steps for solving:
1. check Idenities
2.Use Algebra: combining, factoring, multiplacation, fraction, and sandwiching.
3. then go back to idenities
Example:
Simplify SecX-SinX TanX
use and identity for secX and tanX
1/cosX-sinX(sinX/cosX)
distribute sinx into ( sinx/cosx)
1-sin^2X/cosX
use the iden. and solve for cos^2X with 1-sin^2X
Cos^2 X/ CosX= Cosx
And here are some things you're gonna need to know for the test...
Reciprocal Relationships:
csc (theta)=1/sin (theta)
sec (theta) =1/cos (theta)
cot (theta) =1/tan (theta)
Relationships with Negatives:
sin (-theta) = -sin (theta) and cos (-theta) = -cos (theta)
csc (-theta) = -csc (theta) and sec (-theta)= -sec (theta)
tan (-theta)= -tan (theta) and cot (-theta)= -cot (theta)
Pythagorean Relationship:
sin² (theta) + cos² (theta) =1
1+tan² (theta) = sec² (theta)
1+cot² (theta) = csc² (theta)
Cofunction Relationships
sin (theta) = cos(90°- theta) and cos (theta) = sin(90°-theta)
tan (theta) = cot(90°-theta) and cot (theta) =tan(90°-theta)
sec (theta) = csc(90°-theta) and csc (theta)=sec(90°-theta)
Cautions:
**you can't divide trig functions to cancel them
Example:
2 cos (theta) / cos (theta) = cos² (theta)/ cos (theta)
**but you can move everything to one side & factor out a trig function
**or divide by a trig function to create a new one:
Example:
sinxtanx = 3sinx
sinxtanx - 3sinx = 0
sinx (tanx - 3) = 0
sinx = 0
x = sin^-1 (0)
x = 0, pie, 2pie
tanx -3 = 0
tanx = 3
x = tan^-1 (3)
x= 71.565 pie/180, 257.565 pie/180
ok so what i dont get is how to do problems like #10 on the chapter test..so if anyone would be kind enough to help me out with that that would be awesome...AND DONT FORGET WE HAVE A TEST TOMORROW...GOOD LUCK!!!
Steps for solving:
1. check Idenities
2.Use Algebra: combining, factoring, multiplacation, fraction, and sandwiching.
3. then go back to idenities
Example:
Simplify SecX-SinX TanX
use and identity for secX and tanX
1/cosX-sinX(sinX/cosX)
distribute sinx into ( sinx/cosx)
1-sin^2X/cosX
use the iden. and solve for cos^2X with 1-sin^2X
Cos^2 X/ CosX= Cosx
And here are some things you're gonna need to know for the test...
Reciprocal Relationships:
csc (theta)=1/sin (theta)
sec (theta) =1/cos (theta)
cot (theta) =1/tan (theta)
Relationships with Negatives:
sin (-theta) = -sin (theta) and cos (-theta) = -cos (theta)
csc (-theta) = -csc (theta) and sec (-theta)= -sec (theta)
tan (-theta)= -tan (theta) and cot (-theta)= -cot (theta)
Pythagorean Relationship:
sin² (theta) + cos² (theta) =1
1+tan² (theta) = sec² (theta)
1+cot² (theta) = csc² (theta)
Cofunction Relationships
sin (theta) = cos(90°- theta) and cos (theta) = sin(90°-theta)
tan (theta) = cot(90°-theta) and cot (theta) =tan(90°-theta)
sec (theta) = csc(90°-theta) and csc (theta)=sec(90°-theta)
Cautions:
**you can't divide trig functions to cancel them
Example:
2 cos (theta) / cos (theta) = cos² (theta)/ cos (theta)
**but you can move everything to one side & factor out a trig function
**or divide by a trig function to create a new one:
Example:
sinxtanx = 3sinx
sinxtanx - 3sinx = 0
sinx (tanx - 3) = 0
sinx = 0
x = sin^-1 (0)
x = 0, pie, 2pie
tanx -3 = 0
tanx = 3
x = tan^-1 (3)
x= 71.565 pie/180, 257.565 pie/180
ok so what i dont get is how to do problems like #10 on the chapter test..so if anyone would be kind enough to help me out with that that would be awesome...AND DONT FORGET WE HAVE A TEST TOMORROW...GOOD LUCK!!!
Stephanie's Reflection
y=Asin(Bx-h)+C
amplitude is hight
b is period p=2π/b
h is horizontal shift
c is vertical shift
EG: y=2sin(3x+π)-4
up 2, down 2, total height +4
p=2π/3
Phase Shift = -π (-1) moving to the left
=1/cosx - sinx/1 (sinx/cos)
=1/cosx - sinx²/cos
=1-sinx²/cosx
sin²+cox²=1
cos²1-sin²
=cos²x/cosx
-cosx
Proving:
EG: cotA(1+tanA)/tanA=csc²A
cotA(sec²A)/tanA
((cosA/sinA)(1/cos²A))/(sin/cos)
(cosA/sinAcosA)/(sinA/cosA)
cos²A/sin²AcosA
=1/sin²A
=csc²A
Factoring:
EG: 2sin²Θ-1=0
2sin²Θ=1
sin²Θ=1/2
sinΘ=+/-√1/2
Θ=sin-¹(+/-√1/2)
Θ=30°, 150°, 210°, 330°
-30+180=150
150+180=210
-210+360=330
Reciprocal Relationships
amplitude is hight
b is period p=2π/b
h is horizontal shift
c is vertical shift
EG: y=2sin(3x+π)-4
up 2, down 2, total height +4
p=2π/3
- 0 +1/6π
- π/6 +1/6
- π/3 +1/6
- π/2 +1/6
- 2π/3
Phase Shift = -π (-1) moving to the left
- 0 -1 = -π
- π/6 -1 = -5π/6
- π/3 -1 = -2π/3
- π/2 -1 = -π/2
- 2π/3 -1 = -π/3
- check identities
- algebra
- identities
=1/cosx - sinx/1 (sinx/cos)
=1/cosx - sinx²/cos
=1-sinx²/cosx
sin²+cox²=1
cos²1-sin²
=cos²x/cosx
-cosx
Proving:
EG: cotA(1+tanA)/tanA=csc²A
cotA(sec²A)/tanA
((cosA/sinA)(1/cos²A))/(sin/cos)
(cosA/sinAcosA)/(sinA/cosA)
cos²A/sin²AcosA
=1/sin²A
=csc²A
Factoring:
EG: 2sin²Θ-1=0
2sin²Θ=1
sin²Θ=1/2
sinΘ=+/-√1/2
Θ=sin-¹(+/-√1/2)
Θ=30°, 150°, 210°, 330°
-30+180=150
150+180=210
-210+360=330
Reciprocal Relationships
- cscΘ=1/sinΘ
- secΘ=1/cosΘ
- cotΘ=1/tanΘ
- sin -Θ= -sinΘ and cos -Θ= -cosΘ
- csc -Θ= -cscΘ and sec -Θ= -secΘ
- tan -Θ= -tanΘ and cot -Θ= -cotΘ
- sin²Θ+cos²Θ=1
- 1+tan²Θ=sec²Θ
- 1+cot²Θ=csc²Θ
- sinΘ=cos(90°-Θ) and cosΘ=sin(90°-Θ)
- tanΘ=cot(90°-Θ) and cotΘ=tan(90°-Θ)
- secΘ=csc(90°-Θ) and cscΘ=sec(90°-Θ)
Wednesday, December 2, 2009
Another Reflection to comment on
I have trouble determining if the curve is sine or cosine. The two curves look very similar except for the fact that the points start in different places.
Monday, November 30, 2009
Alicia's Comments
okay im going to comment one of my own questions.
I have trouble with determining when to use the law of sines and law of cosines. If you have this same problem, maybe this explanation will help.
Basically, only use the law of sines when you have a non-right triangle that has a pair. If it's a right triangle, then you would use SOHCAHTOA. You also use law of sines to find the other angles when doing law of cosines. Use law of cosines for non right triangles that have no pairs.
I hoped this helped more than just myself :)
I have trouble with determining when to use the law of sines and law of cosines. If you have this same problem, maybe this explanation will help.
Basically, only use the law of sines when you have a non-right triangle that has a pair. If it's a right triangle, then you would use SOHCAHTOA. You also use law of sines to find the other angles when doing law of cosines. Use law of cosines for non right triangles that have no pairs.
I hoped this helped more than just myself :)
Alicia's Reflection #15
okay so im going to go over some older concepts from the beginning of trigonometry since our midterm is accumulative.
Unit Circle:
90 degs. = (0,1) pi/2
180 degs. = (-1,0) pi2
70 degs. = (0,-1) 3pi/2
360 degs. = (1,0) 2pi
sin=y/r
cos=x/r
tan=y/x
cot=x/y
sec=r/x
csc=r/y
SOHCAHTOA:
S = sin
O = opposite angle
H = hypotenuse
(sin = opposite/hypotenuse)
C = cos
A = adjacent angle
H = hypotenuse
(cos = adjacent/hypotenuse)
T = tan
O = opposite angle
A = adjacent angle
(tan = opposite/adjacent)
*the hypotenuse is opposite the right angle.
*A= 1/2 bh*
*To find the area of a non right triangle use this formula:
*A= 1/2 (leg)(leg)SIN(angle b/w)
*When you have a non right triangle that has pairs, use the law of sines:
Sin A/a = Sin B/b= Sin C/c
*All you are doing is setting up a proportion.
**Remember to solve for an angle, you have to take the inverse.
*To solve a triangle with no angles, use the Law of Cosines:
(opp leg)^2= (adj leg)^2 + (other adj leg)^2 -2(adj leg)(adj leg) Cos(angle b/w)
*use an angle to orient yourself like SOHCAHTOA!!
Hope yall had a good thanksgiving holidays!!
Unit Circle:
90 degs. = (0,1) pi/2
180 degs. = (-1,0) pi2
70 degs. = (0,-1) 3pi/2
360 degs. = (1,0) 2pi
sin=y/r
cos=x/r
tan=y/x
cot=x/y
sec=r/x
csc=r/y
SOHCAHTOA:
S = sin
O = opposite angle
H = hypotenuse
(sin = opposite/hypotenuse)
C = cos
A = adjacent angle
H = hypotenuse
(cos = adjacent/hypotenuse)
T = tan
O = opposite angle
A = adjacent angle
(tan = opposite/adjacent)
*the hypotenuse is opposite the right angle.
*A= 1/2 bh*
*To find the area of a non right triangle use this formula:
*A= 1/2 (leg)(leg)SIN(angle b/w)
*When you have a non right triangle that has pairs, use the law of sines:
Sin A/a = Sin B/b= Sin C/c
*All you are doing is setting up a proportion.
**Remember to solve for an angle, you have to take the inverse.
*To solve a triangle with no angles, use the Law of Cosines:
(opp leg)^2= (adj leg)^2 + (other adj leg)^2 -2(adj leg)(adj leg) Cos(angle b/w)
*use an angle to orient yourself like SOHCAHTOA!!
Hope yall had a good thanksgiving holidays!!
Taylor's Reflection (( 22 November))
i missed this blog. OOPS! i totally forgot. i guess i was looking forward too much to the holiday. but ill make up the missed blog now.
i think i understand all there is that goes along with the graphing we learned last week
if i just make sure i remember and follow the steps one by one carefully it really isnt a hard concept to grasp.
i do have a little blip of confusion when it comes to distinguishing wether there is a horizontal shift or a vertical shift.
i think i just need an explanation of what to look for to decipher which shift is necessary
any help though would be wonderful!
thanks
i think i understand all there is that goes along with the graphing we learned last week
if i just make sure i remember and follow the steps one by one carefully it really isnt a hard concept to grasp.
i do have a little blip of confusion when it comes to distinguishing wether there is a horizontal shift or a vertical shift.
i think i just need an explanation of what to look for to decipher which shift is necessary
any help though would be wonderful!
thanks
Sunday, November 29, 2009
Taylor's thanksgiving week reflection
So basically i remember everything we have learned so far in the latest weeks
Area of a non-right triangle
A=1/2(leg)(leg)sin(angle between)
Area of Right Triangles
A=1/2bh
SOHCAHTOA
sinΘ=opposite leg/hypotenuse
cosΘ=adjacent leg/hypotenuse
tanΘ=opposite leg/adjacent leg
Law of Sines
sinA/a - sinB/b = sinC/c
(used when you know pairs or opposites in a non-right triangle)
Law of Cosines
(opposite leg)²=(adjacent leg)² + (other leg)² - 2(adjacent leg)(adjacent leg)cos°
Area of Inscribed Shapes
A=nr²sinΘcosΘ
All im worried about is completing the nine study guides in time for the exams although they will help me get a super good grade.
i predict these upcoming weeks to be the three most killer week of the school year.
everything is due
then its exams
my motivation will be the holidays just three weeks away
as soon as i get the study guides ill be sure to post all my questions
Area of a non-right triangle
A=1/2(leg)(leg)sin(angle between)
Area of Right Triangles
A=1/2bh
SOHCAHTOA
sinΘ=opposite leg/hypotenuse
cosΘ=adjacent leg/hypotenuse
tanΘ=opposite leg/adjacent leg
Law of Sines
sinA/a - sinB/b = sinC/c
(used when you know pairs or opposites in a non-right triangle)
Law of Cosines
(opposite leg)²=(adjacent leg)² + (other leg)² - 2(adjacent leg)(adjacent leg)cos°
Area of Inscribed Shapes
A=nr²sinΘcosΘ
All im worried about is completing the nine study guides in time for the exams although they will help me get a super good grade.
i predict these upcoming weeks to be the three most killer week of the school year.
everything is due
then its exams
my motivation will be the holidays just three weeks away
as soon as i get the study guides ill be sure to post all my questions
Stephen's Reflection from last week
Ok so i didnt think about school all this week so i forgot about the whole blog and comment thing so ima come do this right now to make it up. Im going to talk about something that gives me problems and is from chapter 1....graphing parabolas. Its not so much that its hard it just that i forget how to find x intercept, y intercept, vertex and axis of symmetry. so ill talk about this as best as i can.
There are 7 steps to this: 1. finding what way it opens, 2. how many intercepts, 3. finding the x intercepts, 4. finding y intercepts, 5. finding the axis of symmetry, 6. finding the vertex, and 7. graphing the parabola. to find the opening, you have to look infront of the x^2...if the number is positive then it opens up, if negative it opens down.
To find how many intercepts you plug in the formula b^2-4ac.
I kinda dont understand how to find x intercepts but i think you do completing the square .
when finding the y intercepts, you plug 0 in for x and solve for y.
to find the axis of symmetry you plug in the formula x=-b/2a
for the vertex you plug in your answer of the axis in for x and solve
then you just plug your answers in on a graph and you should get a parabola, if you dont know how to do this then you can use ur calculator
As said before im a little fuzzy on how to do these steps so if someone would be able to help or correct what i wrote if wrong then i would be grateful.
There are 7 steps to this: 1. finding what way it opens, 2. how many intercepts, 3. finding the x intercepts, 4. finding y intercepts, 5. finding the axis of symmetry, 6. finding the vertex, and 7. graphing the parabola. to find the opening, you have to look infront of the x^2...if the number is positive then it opens up, if negative it opens down.
To find how many intercepts you plug in the formula b^2-4ac.
I kinda dont understand how to find x intercepts but i think you do completing the square .
when finding the y intercepts, you plug 0 in for x and solve for y.
to find the axis of symmetry you plug in the formula x=-b/2a
for the vertex you plug in your answer of the axis in for x and solve
then you just plug your answers in on a graph and you should get a parabola, if you dont know how to do this then you can use ur calculator
As said before im a little fuzzy on how to do these steps so if someone would be able to help or correct what i wrote if wrong then i would be grateful.
Stephen's Reflection
Ok so guess what...school tomorrow :( but oh well. I have been swamped with basically soccer with the practices and games and the tourny so I barely remember anything and im not ready to go back to school and learn more. But i do remember law of sines so i guess i will talk about that.
Law of sines works for non right triangles and it only works if you have a angle/side pair. The formula for law of sines is sinA/a=sinB/b=sinC/c. Big letters are the angles and little letters are the sides in triangle ABC. If you have angle A, angle B and side a then you will look for the side opposite of angle B. you will cross multiply to where ur problem will look like: bsinA=asinB. Then you divide by sin A and solve completely using ur calculator. Then you found your side b.
For what i need help with is remembering law of cosines so that way that bridge thing will go easier and im looking at my notes and im lost for some reason so i would be thankful if someone helped me.
Law of sines works for non right triangles and it only works if you have a angle/side pair. The formula for law of sines is sinA/a=sinB/b=sinC/c. Big letters are the angles and little letters are the sides in triangle ABC. If you have angle A, angle B and side a then you will look for the side opposite of angle B. you will cross multiply to where ur problem will look like: bsinA=asinB. Then you divide by sin A and solve completely using ur calculator. Then you found your side b.
For what i need help with is remembering law of cosines so that way that bridge thing will go easier and im looking at my notes and im lost for some reason so i would be thankful if someone helped me.
Stephanie's Reflection
Area of a non-right triangle
A=1/2(leg)(leg)sin(angle between)
EG: The area of Triangle PQR is 15. If p is 5 and q is 10, find all possible measures of R.
A=15
P=5
Q=10
15=1/2(5)(10)sinR
15=25sinR
sinR=3/5
R=sin-1(3/5)
R≈36.870°,143.130°
Right Triangles
Law of Sines
sinA/a - sinB/b = sinC/c
Law of Cosines
(opposite leg)²=(adjacent leg)² + (other leg)² - 2(adjacent leg)(adjacent leg)cos°
EG: x, 5, 6 angle = 35°
x²=5²+6²-2(5)(6)cos35°
x=√(5²+6²-2(5)(6)cos35°)
x≈3.443
Area of Inscribed Shapes
A=nr²sinΘcosΘ
A=1/2(leg)(leg)sin(angle between)
EG: The area of Triangle PQR is 15. If p is 5 and q is 10, find all possible measures of R.
A=15
P=5
Q=10
15=1/2(5)(10)sinR
15=25sinR
sinR=3/5
R=sin-1(3/5)
R≈36.870°,143.130°
Right Triangles
- A=1/2bh
- SOHCAHTOA
- sinΘ=opposite leg/hypotenuse
- cosΘ=adjacent leg/hypotenuse
- tanΘ=opposite leg/adjacent leg
Law of Sines
sinA/a - sinB/b = sinC/c
- used when you know pairs in a non-right triangle
- you are setting up proportions
Law of Cosines
(opposite leg)²=(adjacent leg)² + (other leg)² - 2(adjacent leg)(adjacent leg)cos°
EG: x, 5, 6 angle = 35°
x²=5²+6²-2(5)(6)cos35°
x=√(5²+6²-2(5)(6)cos35°)
x≈3.443
Area of Inscribed Shapes
A=nr²sinΘcosΘ
Reflection #....um....Terrio's Reflection
Yeah so this week sucked! I did not have any time to do anything fun. The highlight of my week was going to Alicia's huge house and spending two full days to work on our bridge, but we finished it =]. The worst part of the week was the soccer tournament. Not only did I get in a fight with two guys on one team, but I could not walk after the first two games...and we played four. Well back to math.
So right now the only thing I really remember is the area of a non right triangle.
Formula:1/2(leg)(leg)sin(angle b/w)
So if you had 1/2(3)(6)sin(52) your answer would be: 7.100
Now for what I don't remember. I could really use some help remembering is the Law of Sines, and the Law of Cosines. Happy Thanksgiving...well late Thanksgiving =]
So right now the only thing I really remember is the area of a non right triangle.
Formula:1/2(leg)(leg)sin(angle b/w)
So if you had 1/2(3)(6)sin(52) your answer would be: 7.100
Now for what I don't remember. I could really use some help remembering is the Law of Sines, and the Law of Cosines. Happy Thanksgiving...well late Thanksgiving =]
Reflection 15
Even though we did not have school this week you are to reflect back on an older concept. This is in preparation for your mid-term. As we talked about in class, pick anything from the year and write your reflection based on that.
Saturday, November 28, 2009
Amy's Reflection #15
well obviously, we didn't learn anything new this week...so here's are some of the trig functions we learned so far...
pythagorean theorem: c^2 = a^2 + b^2 (use to find the hypotenuse of a right triangle)
area of a right triangle = 1/2 bh
SOHCAHTOA (used to solve right triangles):
Sin θ= opposite leg/hypotenuse
Cos θ= adjacent leg/hypotenuse
Tan θ= opposite leg/adjacent leg
area of a non right triangle = 1/2 (leg)(leg)SIN(angle b/w)
Example: non-right triangle: HIJ (left to right)H = 65 degrees, j = 2, i = 6. Find the area.
A = (1/2)(2)(6)sin(65)
A = 5.438
area of an isoceles triangle = bh
Law of Sines(used to non-right triangles):
Sin A/a = Sin B/b= Sin C/c
Example: you have a triangle with the sides 4 and 5 & you also have an angle of 30 degrees.
A = 1/2 (4) (5) Sin 30 degrees
A = 10 Sin 30 degrees which is aproximately = 5
Law of Cosines (used when you can't use Law of Sines):
(opposite leg)^2 = (adjacent leg)^2 + (other adjacent leg)^2 - 2(adjacent leg) (adjacent leg) cos (angle between)
Example: you have a triangle with the sides of 5, 6, and 7. find the angle between 5 and 6.
7^2=6^2+5^2-2(5)(6)
cos a7^2-6^2-5^2= 2(5)(6)
cos acos a= 7^2-6^2-5^2 / -2(6)(5)
a= cos-1 ((7^2-^6^2-5^2)/(-2(5)(6))
a= 78.463 degrees
what i still don't understand is how you solve a whole trigangle using law of cosines..if someone could help me with that that would be great...!!!
pythagorean theorem: c^2 = a^2 + b^2 (use to find the hypotenuse of a right triangle)
area of a right triangle = 1/2 bh
SOHCAHTOA (used to solve right triangles):
Sin θ= opposite leg/hypotenuse
Cos θ= adjacent leg/hypotenuse
Tan θ= opposite leg/adjacent leg
area of a non right triangle = 1/2 (leg)(leg)SIN(angle b/w)
Example: non-right triangle: HIJ (left to right)H = 65 degrees, j = 2, i = 6. Find the area.
A = (1/2)(2)(6)sin(65)
A = 5.438
area of an isoceles triangle = bh
Law of Sines(used to non-right triangles):
Sin A/a = Sin B/b= Sin C/c
Example: you have a triangle with the sides 4 and 5 & you also have an angle of 30 degrees.
A = 1/2 (4) (5) Sin 30 degrees
A = 10 Sin 30 degrees which is aproximately = 5
Law of Cosines (used when you can't use Law of Sines):
(opposite leg)^2 = (adjacent leg)^2 + (other adjacent leg)^2 - 2(adjacent leg) (adjacent leg) cos (angle between)
Example: you have a triangle with the sides of 5, 6, and 7. find the angle between 5 and 6.
7^2=6^2+5^2-2(5)(6)
cos a7^2-6^2-5^2= 2(5)(6)
cos acos a= 7^2-6^2-5^2 / -2(6)(5)
a= cos-1 ((7^2-^6^2-5^2)/(-2(5)(6))
a= 78.463 degrees
what i still don't understand is how you solve a whole trigangle using law of cosines..if someone could help me with that that would be great...!!!
Wednesday, November 25, 2009
Amy's Comments
ok since nobody asked an questions im gonna just give two examples...
Example #1: Find the angle of inclination...
2x+5y=15
m=-2/5
tan=-2/5
alpha=tan^-1(-2/5)
Example #2: For a triangle with C=36 degrees a=5 b=6, find side c....
c^2=5^2+6^2-2(5)(6)
cos36c=sqrt(25+36-2(5)(6)cos36)
c= 3.53
i hope this might help someone...& Happy Thanksgiving Everyone!!!
Example #1: Find the angle of inclination...
2x+5y=15
m=-2/5
tan=-2/5
alpha=tan^-1(-2/5)
Example #2: For a triangle with C=36 degrees a=5 b=6, find side c....
c^2=5^2+6^2-2(5)(6)
cos36c=sqrt(25+36-2(5)(6)cos36)
c= 3.53
i hope this might help someone...& Happy Thanksgiving Everyone!!!
Sunday, November 22, 2009
Alicia's Reflection #14
Okay, this week we started chapter 8 Trigonometry equations. Here are some formulas to remember:
Solving for a line:
m=tan alpha
Solving for a conic:
tan 2alpha= B/A-C
If A=C then alpha = pi/4 (always)
m=slope
alpha=angle of inclination
Example:
Find the angle of inclination of
x^2-2xy+3y^2=1
tan 2alpha=B/A-C
A=1 B=-2 C=3 (tan is +ve in 1&3)
tan 2alpha= -2/1-3=1
tan 2alpha=1
2alpha=tan^-1(1)
2alpha=45,225 degrees
2alpha=45/2, 225/2
alpha= pi/4
Happy Thanksgiving!
Solving for a line:
m=tan alpha
Solving for a conic:
tan 2alpha= B/A-C
If A=C then alpha = pi/4 (always)
m=slope
alpha=angle of inclination
Example:
Find the angle of inclination of
x^2-2xy+3y^2=1
tan 2alpha=B/A-C
A=1 B=-2 C=3 (tan is +ve in 1&3)
tan 2alpha= -2/1-3=1
tan 2alpha=1
2alpha=tan^-1(1)
2alpha=45,225 degrees
2alpha=45/2, 225/2
alpha= pi/4
Happy Thanksgiving!
Amy's Rlection #14
ok this week we started chapter 8. Here's an explanation for 8.1:
1. For any line : m = tan (alpha)
**m = slope , (alpha) = angle of inclination
2. For a conic: tan 2 (alpha) = B/A-C
**if A=C then pie/4 (always)
**A = coefficient of x^2, B = coefficient of xy, C = coefficient of y^2
Examples:
1. Find the angle of inclination of x^2 - 2xy + 3y^2 = 1.
tan 2 (alpha) = B/A-C
A = 1 , B = -2 , C = 3
tan 2 (alpha) = -2/1 -3 = 1
tan 2 (alpha) = 1
2A = tan^-1 (1)
**if the # you're taking the inverse of is positive then find where it's negative and of course if it's negative then find where it's positive
**tan is negative in quad. #1 & 3
2 (alpha) = 45 , 225
alpha = 45/2 , 225/2
alpha = 22.5 , 112.5
2. x^2 + y^2 - 3xy + 4x - sqrt. x = 1
alpha = 1 (because A = 1 & C = 1 so A = C)
Helpful Hints:
quad #1: solve inverse
quad. #2 : make negative and add 180
quad. #3 : just add 180
quad. #4 : make negative and add 360
can someone tell me how you find the x-points and the phase shift. i know you need add something but what..i dont know..someone please help :(
1. For any line : m = tan (alpha)
**m = slope , (alpha) = angle of inclination
2. For a conic: tan 2 (alpha) = B/A-C
**if A=C then pie/4 (always)
**A = coefficient of x^2, B = coefficient of xy, C = coefficient of y^2
Examples:
1. Find the angle of inclination of x^2 - 2xy + 3y^2 = 1.
tan 2 (alpha) = B/A-C
A = 1 , B = -2 , C = 3
tan 2 (alpha) = -2/1 -3 = 1
tan 2 (alpha) = 1
2A = tan^-1 (1)
**if the # you're taking the inverse of is positive then find where it's negative and of course if it's negative then find where it's positive
**tan is negative in quad. #1 & 3
2 (alpha) = 45 , 225
alpha = 45/2 , 225/2
alpha = 22.5 , 112.5
2. x^2 + y^2 - 3xy + 4x - sqrt. x = 1
alpha = 1 (because A = 1 & C = 1 so A = C)
Helpful Hints:
quad #1: solve inverse
quad. #2 : make negative and add 180
quad. #3 : just add 180
quad. #4 : make negative and add 360
can someone tell me how you find the x-points and the phase shift. i know you need add something but what..i dont know..someone please help :(
Stephanie's Reflection
SOLVING TRIG EQUATIONS
for any line m = tan alpha
m = slope , alpha = angle of inclination
For a conic: tan 2 alpha = B/A-C
if A=C then pi/4
A = coefficient of x^2, B = coefficient of xy, C = coefficient of y^2
Eg: find the value of x between 0 and 2pi for which sinx=6.
xsin-1=6
x=36.870, 143.130
2cos theta+9=7
theta=cos-17-9//3
theta=48.190
theta=131.810, 228.190
2tan theta+1=0
theta=tan -1 -1/2
theta=26.465
theta=153.435
153.435 times pi/180
theta=.853 pi
for any line m = tan alpha
m = slope , alpha = angle of inclination
For a conic: tan 2 alpha = B/A-C
if A=C then pi/4
A = coefficient of x^2, B = coefficient of xy, C = coefficient of y^2
Eg: find the value of x between 0 and 2pi for which sinx=6.
xsin-1=6
x=36.870, 143.130
2cos theta+9=7
theta=cos-17-9//3
theta=48.190
theta=131.810, 228.190
2tan theta+1=0
theta=tan -1 -1/2
theta=26.465
theta=153.435
153.435 times pi/180
theta=.853 pi
Monday, November 16, 2009
Stephen's Reflection
Ok so this week was basically a review for the test on Tuesday. I think that everyone basically has it down pack with how to do SOHCAHTOA, law of sines, and law of cosines. But what might confuse some people is when SOHCAHTOA, law of sines, and law of cosines is actually used. To start, SOHCAHTOA is used when you have a right triangle. This is the only formula that you use on a right triangle. Then for law of sines, there must be a pair. By a pair i mean that an angle has to have an opposite side and visa versa. If there are two of those then law of sines is used. And last for law of cosines. Law of cosines is used when you have a two legs and an angle inbetween. For law of cosines you would usually try to find all angles and sides.
There is one thing im a little confused about. I get a little confused when it askes me to find the are of a parallelogram. I know that you have to cut the parallelogram into two triangles and find the triangle's angles and sides if not listed but what im confused about is when you actually cut an angle because then you cant just cut it in half and so i get confused about how to find that angle of that specific triangle.
There is one thing im a little confused about. I get a little confused when it askes me to find the are of a parallelogram. I know that you have to cut the parallelogram into two triangles and find the triangle's angles and sides if not listed but what im confused about is when you actually cut an angle because then you cant just cut it in half and so i get confused about how to find that angle of that specific triangle.
Taylor #13
This week we mostly reviewed the stuff we had previously learned in the chapter. We also learned about the law of cosines. I feel like I understand it. I think the only trouble I am having with it is sometimes when I type it into the calculator I get an error over and over again.
The only thing to using law of cosines is memorizing the formula
FORMULA FOR LAW OF COSINES
(opposite leg)²=(adjacent leg)² + (other leg)² - 2(adjacent leg)(adjacent leg)cos°
And typing it into the calculator also has some part of answering the equation
When typing in an inverse you must put parenthesis around the both the nominator and the denominator
I know things like this are what is confusing most people
As far as what I don’t understand there really isn’t much
Basically what gets me tripped up is when we get those weird problems on homework that look and sound nothing like anything we have learned but there really isn’t much anyone else can do to to help me with this its more of a matter of working them out as best as I can.
The other thing is I get caught up on the finer details of problems. In other words, things like buttons used when typing into calculator are what get me
So if anyone has any tricks to remember things that are little details any little advice could help
THANKS!
The only thing to using law of cosines is memorizing the formula
FORMULA FOR LAW OF COSINES
(opposite leg)²=(adjacent leg)² + (other leg)² - 2(adjacent leg)(adjacent leg)cos°
And typing it into the calculator also has some part of answering the equation
When typing in an inverse you must put parenthesis around the both the nominator and the denominator
I know things like this are what is confusing most people
As far as what I don’t understand there really isn’t much
Basically what gets me tripped up is when we get those weird problems on homework that look and sound nothing like anything we have learned but there really isn’t much anyone else can do to to help me with this its more of a matter of working them out as best as I can.
The other thing is I get caught up on the finer details of problems. In other words, things like buttons used when typing into calculator are what get me
So if anyone has any tricks to remember things that are little details any little advice could help
THANKS!
Taylor #13
This week we mostly reviewed the stuff we had previously learned in the chapter. We also learned about the law of cosines. I feel like I understand it. I think the only trouble I am having with it is sometimes when I type it into the calculator I get an error over and over again.
The only thing to using law of cosines is memorizing the formula
FORMULA FOR LAW OF COSINES
(opposite leg)²=(adjacent leg)² + (other leg)² - 2(adjacent leg)(adjacent leg)cos°
And typing it into the calculator also has some part of answering the equation
When typing in an inverse you must put parenthesis around the both the nominator and the denominator
I know things like this are what is confusing most people
As far as what I don’t understand there really isn’t much
Basically what gets me tripped up is when we get those weird problems on homework that look and sound nothing like anything we have learned but there really isn’t much anyone else can do to to help me with this its more of a matter of working them out as best as I can.
The other thing is I get caught up on the finer details of problems. In other words, things like buttons used when typing into calculator are what get me
So if anyone has any tricks to remember things that are little details any little advice could help
THANKS!
The only thing to using law of cosines is memorizing the formula
FORMULA FOR LAW OF COSINES
(opposite leg)²=(adjacent leg)² + (other leg)² - 2(adjacent leg)(adjacent leg)cos°
And typing it into the calculator also has some part of answering the equation
When typing in an inverse you must put parenthesis around the both the nominator and the denominator
I know things like this are what is confusing most people
As far as what I don’t understand there really isn’t much
Basically what gets me tripped up is when we get those weird problems on homework that look and sound nothing like anything we have learned but there really isn’t much anyone else can do to to help me with this its more of a matter of working them out as best as I can.
The other thing is I get caught up on the finer details of problems. In other words, things like buttons used when typing into calculator are what get me
So if anyone has any tricks to remember things that are little details any little advice could help
THANKS!
Taylor #13
This week we mostly reviewed the stuff we had previously learned in the chapter. We also learned about the law of cosines. I feel like I understand it. I think the only trouble I am having with it is sometimes when I type it into the calculator I get an error over and over again.
The only thing to using law of cosines is memorizing the formula
FORMULA FOR LAW OF COSINES
(opposite leg)²=(adjacent leg)² + (other leg)² - 2(adjacent leg)(adjacent leg)cos°
And typing it into the calculator also has some part of answering the equation
When typing in an inverse you must put parenthesis around the both the nominator and the denominator
I know things like this are what is confusing most people
As far as what I don’t understand there really isn’t much
Basically what gets me tripped up is when we get those weird problems on homework that look and sound nothing like anything we have learned but there really isn’t much anyone else can do to to help me with this its more of a matter of working them out as best as I can.
The other thing is I get caught up on the finer details of problems. In other words, things like buttons used when typing into calculator are what get me
So if anyone has any tricks to remember things that are little details any little advice could help
THANKS!
The only thing to using law of cosines is memorizing the formula
FORMULA FOR LAW OF COSINES
(opposite leg)²=(adjacent leg)² + (other leg)² - 2(adjacent leg)(adjacent leg)cos°
And typing it into the calculator also has some part of answering the equation
When typing in an inverse you must put parenthesis around the both the nominator and the denominator
I know things like this are what is confusing most people
As far as what I don’t understand there really isn’t much
Basically what gets me tripped up is when we get those weird problems on homework that look and sound nothing like anything we have learned but there really isn’t much anyone else can do to to help me with this its more of a matter of working them out as best as I can.
The other thing is I get caught up on the finer details of problems. In other words, things like buttons used when typing into calculator are what get me
So if anyone has any tricks to remember things that are little details any little advice could help
THANKS!
Sunday, November 15, 2009
Devin's Reflection #
This week we took another deadly step further into trigonometry. We learned more ways to solve for angle of a triangle, and also how to solve for the sides of a triangle. We also learned some formulas and ways to solve for the area of a triangle. To find the area of a right triangle,
1. Be sure that the triangle is right
2. Solve for the sides of the triangle
3. Use the formula : bh/2
To find the measures of a right triangle, use SOH CAH TOA
Sine= opp side/hyp
Cosine= adj side/hyp
Tangent= opp side/adj side
To find the measures of non-right triangles, you can use multiple techniques.
If the triangle has a pair (the measure of an angle and corresponding side), use The Law of Sine
sinA/a=sinB/b=sinC/c
If the triangles has no pairs, then use
The Law of Cosine
c^2=a^2+b^2-2(a)(b)cos(theta)
To find the area of a non-right triangle use the formula:
1/2(leg)(leg)sin(theta)
1. Be sure that the triangle is right
2. Solve for the sides of the triangle
3. Use the formula : bh/2
To find the measures of a right triangle, use SOH CAH TOA
Sine= opp side/hyp
Cosine= adj side/hyp
Tangent= opp side/adj side
To find the measures of non-right triangles, you can use multiple techniques.
If the triangle has a pair (the measure of an angle and corresponding side), use The Law of Sine
sinA/a=sinB/b=sinC/c
If the triangles has no pairs, then use
The Law of Cosine
c^2=a^2+b^2-2(a)(b)cos(theta)
To find the area of a non-right triangle use the formula:
1/2(leg)(leg)sin(theta)
Alicia's Reflection #13
heyy, so this week was pretty much a review for our chapter 9 test on tuesday. We had a take home quiz for chapter 9 that was pretty easy once we learned the law of cosines. I do really good with those kind of questions but the minute i see a word problem its like i completely forget how to solve them. i need to figure out a way to know what method to use based upon the information that the word problem gives. if anyone has any tricks to know how to solve the word problems, it would help me out alot for the test on tuesday. Also, if anyone can help me with # 8 and #10 on the chapter test it would help me out greatly.
Review of chapter 9 Right triangles:
*the hypotenuse is opposite the right angle.
*A= 1/2 bh
*SOHCAHTOA---stands for....
Sin θ= opposite leg/hypotenuse
Cos θ= adjacent leg/hypotenuse
Tan θ= opposite leg/adjacent leg
*To find the area of a non right triangle use this formula:
*A= 1/2 (leg)(leg)SIN(angle b/w)
*When you have a non right triangle that has pairs, use the law of sines:
Sin A/a = Sin B/b= Sin C/c
*All you are doing is setting up a proportion.
**Remember to solve for an angle, you have to take the inverse.
*To solve a triangle with no angles, use the Law of Cosines:
(opp leg)^2= (adj leg)^2 + (other adj leg)^2 -2(adj leg)(adj leg) Cos(angle b/w)
*use an angle to orient yourself like SOHCAHTOA!!
Goodluck to everyone on the test!!
Review of chapter 9 Right triangles:
*the hypotenuse is opposite the right angle.
*A= 1/2 bh
*SOHCAHTOA---stands for....
Sin θ= opposite leg/hypotenuse
Cos θ= adjacent leg/hypotenuse
Tan θ= opposite leg/adjacent leg
*To find the area of a non right triangle use this formula:
*A= 1/2 (leg)(leg)SIN(angle b/w)
*When you have a non right triangle that has pairs, use the law of sines:
Sin A/a = Sin B/b= Sin C/c
*All you are doing is setting up a proportion.
**Remember to solve for an angle, you have to take the inverse.
*To solve a triangle with no angles, use the Law of Cosines:
(opp leg)^2= (adj leg)^2 + (other adj leg)^2 -2(adj leg)(adj leg) Cos(angle b/w)
*use an angle to orient yourself like SOHCAHTOA!!
Goodluck to everyone on the test!!
Stephanie's Reflection
Area of a non-right triangle
A=1/2(leg)(leg)sin(angle between)
EG: The area of Triangle PQR is 15. If p is 5 and q is 10, find all possible measures of R.
A=15
P=5
Q=10
15=1/2(5)(10)sinR
15=25sinR
sinR=3/5
R=sin-1(3/5)
R≈36.870°,143.130°
Right Triangles
Law of Sines
sinA/a - sinB/b = sinC/c
Law of Cosines
(opposite leg)²=(adjacent leg)² + (other leg)² - 2(adjacent leg)(adjacent leg)cos°
EG: x, 5, 6 angle = 35°
x²=5²+6²-2(5)(6)cos35°
x=√(5²+6²-2(5)(6)cos35°)
x≈3.443
Area of Inscribed Shapes
A=nr²sinΘcosΘ
I don't understand how you know when to use Law of Sines or Law of Cosines. I especially don't understand how to plug them in (for some reason). Also, getting the Area of Inscribed Shapes eludes me.
A=1/2(leg)(leg)sin(angle between)
EG: The area of Triangle PQR is 15. If p is 5 and q is 10, find all possible measures of R.
A=15
P=5
Q=10
15=1/2(5)(10)sinR
15=25sinR
sinR=3/5
R=sin-1(3/5)
R≈36.870°,143.130°
Right Triangles
- A=1/2bh
- SOHCAHTOA
- sinΘ=opposite leg/hypotenuse
- cosΘ=adjacent leg/hypotenuse
- tanΘ=opposite leg/adjacent leg
Law of Sines
sinA/a - sinB/b = sinC/c
- used when you know pairs in a non-right triangle
- you are setting up proportions
Law of Cosines
(opposite leg)²=(adjacent leg)² + (other leg)² - 2(adjacent leg)(adjacent leg)cos°
EG: x, 5, 6 angle = 35°
x²=5²+6²-2(5)(6)cos35°
x=√(5²+6²-2(5)(6)cos35°)
x≈3.443
Area of Inscribed Shapes
A=nr²sinΘcosΘ
I don't understand how you know when to use Law of Sines or Law of Cosines. I especially don't understand how to plug them in (for some reason). Also, getting the Area of Inscribed Shapes eludes me.
Amy's Reflection #13
ok this week went by pretty quick. i guess it's because we had monday off. well anyway, we didn't really learn much this quick. the new formula we did learn this week was the Law of Cosines.
Law of Cosines:
(opp. side) = (adj. side)^2 + (other adj. side)^2 - 2(adj. side)(other adj. side) cos (angle between)
*you use the law of cosines when you can't use law of sines
* in other words, when you have: a non-right triangle, an angle measure with an unknown opposite side, and two side lengths on each side of that angle measure or when you have all three sides and no angle
Example1 (when given an angle & looking for the 3rd side):
you have a triangle ABC, angle C is 32 degrees, a = 5, and b = 6. Find c.
first you plug the values into the formula..
x^2 = 5^2 + 6^2 - 2(5)(6) cos (32 degrees)
x = sqrt of ((6^2 + 5^2 - 2(5)(6) cos 32 degrees))
x = 3.181
Example2 (when given the sides & looking for the angle in b/w):
you have a triangle with the sides of 5, 6, and 7. Find the angle in between 5 and 6.
7^2=6^2+5^2-2(5)(6)cos a
7^2-6^2-5^2= 2(5)(6)cos a
cos a= 7^2-6^2-5^2/-2(6)(5)
a= cos-1 ((7^2-^6^2-5^2)/(-2(5)(6))
a= 78.463 degrees
*when you plug it in your calc. make sure you put two (( in the beginning and two )) in the end because if you don't you won't get the right answer...
ok now for what i dont understand....you know problems we get this weird shape and we have to make two triangles out of it..what do you after that? i also don't understand some of these word problem we get for homework like #s 17, 18, &19 on page 355..if anyone could help me with these problems that would be awesome..
Law of Cosines:
(opp. side) = (adj. side)^2 + (other adj. side)^2 - 2(adj. side)(other adj. side) cos (angle between)
*you use the law of cosines when you can't use law of sines
* in other words, when you have: a non-right triangle, an angle measure with an unknown opposite side, and two side lengths on each side of that angle measure or when you have all three sides and no angle
Example1 (when given an angle & looking for the 3rd side):
you have a triangle ABC, angle C is 32 degrees, a = 5, and b = 6. Find c.
first you plug the values into the formula..
x^2 = 5^2 + 6^2 - 2(5)(6) cos (32 degrees)
x = sqrt of ((6^2 + 5^2 - 2(5)(6) cos 32 degrees))
x = 3.181
Example2 (when given the sides & looking for the angle in b/w):
you have a triangle with the sides of 5, 6, and 7. Find the angle in between 5 and 6.
7^2=6^2+5^2-2(5)(6)cos a
7^2-6^2-5^2= 2(5)(6)cos a
cos a= 7^2-6^2-5^2/-2(6)(5)
a= cos-1 ((7^2-^6^2-5^2)/(-2(5)(6))
a= 78.463 degrees
*when you plug it in your calc. make sure you put two (( in the beginning and two )) in the end because if you don't you won't get the right answer...
ok now for what i dont understand....you know problems we get this weird shape and we have to make two triangles out of it..what do you after that? i also don't understand some of these word problem we get for homework like #s 17, 18, &19 on page 355..if anyone could help me with these problems that would be awesome..
AHHHHHH! I'm falling so far behind in this class now. As soon as we learned the Law of Cosines and I saw how much work it is I quit feeling like doing homework and I didn't feel like doing that whole quiz. I'm too lazy for this class. At least I know how to do the Law of Cosines I guess.
Law of Cosines:
(opposite side)^2=(adjacent leg)^2 + (other adjacent leg)^2 -2(adj leg)(other adj leg)cos(angle between the legs)
So all you really do is plug in the numbers from the triangle and then solve for whatever you need to solve for haha. When you plug in the numbers you just square root the whole thing then solve however you're suppose to be solving it.
OK. I need help with the Law of Sines. I have a feeling I was sleeping in class or something when we learned this because I do not remember a single thing about it and I can't understand my notes either.
Law of Cosines:
(opposite side)^2=(adjacent leg)^2 + (other adjacent leg)^2 -2(adj leg)(other adj leg)cos(angle between the legs)
So all you really do is plug in the numbers from the triangle and then solve for whatever you need to solve for haha. When you plug in the numbers you just square root the whole thing then solve however you're suppose to be solving it.
OK. I need help with the Law of Sines. I have a feeling I was sleeping in class or something when we learned this because I do not remember a single thing about it and I can't understand my notes either.
Wednesday, November 11, 2009
Amy's Comments
1.) C = 90 degrees, b = 7, c = 12
angle A
cos A = 7/12
A = cos^-1(7/12)
= 54.315 degrees
angle B
sin B = 7/12
B = sin^-1(7/12)
= 35.685 degrees
length of a
tan 54.315 = a/7
(7) tan 54.315 = a/7 (7)
a = 7 tan 54.315
a = 9.747
area
A = 1/2 bh= 1/2 (7) (9.747)= 34.115
2.) two sides of a triangle have the lengths 7cm & 4cm. the angle between the sides measures 73 degrees. Find the area of the triangle.
1/2 (7) (4) sin (73) = 13.388 cm^2
angle A
cos A = 7/12
A = cos^-1(7/12)
= 54.315 degrees
angle B
sin B = 7/12
B = sin^-1(7/12)
= 35.685 degrees
length of a
tan 54.315 = a/7
(7) tan 54.315 = a/7 (7)
a = 7 tan 54.315
a = 9.747
area
A = 1/2 bh= 1/2 (7) (9.747)= 34.115
2.) two sides of a triangle have the lengths 7cm & 4cm. the angle between the sides measures 73 degrees. Find the area of the triangle.
1/2 (7) (4) sin (73) = 13.388 cm^2
Monday, November 9, 2009
Stephen's Reflection
Ok so this week we are in Chapter 9 and i think its pretty easy but at times i think i understand something but then i get confused. But anyway we did stuff SOHCAHTOA, law of sines, and area of non right triangles. I think that the easiest thing for me was the law of sines because i understand that completely.
The law of sines is:
sinA/a=sinB/b=sinC/c
This is for any triangle and it basically says angle A over side a=angle B over side b=angle C over side c. If you have a triangle ABC and it gives u the angle for A and B and gives you side b only then you would want to find side a which is directly across from angle a. so you do sinA divided by side a = sinB divided by side b and then you cross multiply and divide and your answer will be a=bsinA/sinB.
I understand basically all everything. Theres not much i dont understand but i do get a little confused on what formula to use out of all that we were given like on the ch. 9 test stuff but other than that im good.
The law of sines is:
sinA/a=sinB/b=sinC/c
This is for any triangle and it basically says angle A over side a=angle B over side b=angle C over side c. If you have a triangle ABC and it gives u the angle for A and B and gives you side b only then you would want to find side a which is directly across from angle a. so you do sinA divided by side a = sinB divided by side b and then you cross multiply and divide and your answer will be a=bsinA/sinB.
I understand basically all everything. Theres not much i dont understand but i do get a little confused on what formula to use out of all that we were given like on the ch. 9 test stuff but other than that im good.
TERRIO REFLECTION #12
OK so I did have a little more trouble this week than last. The beginning of the week was pretty easy but as we went on it got harder. On thing I really picked up was SOHCAHTOA. This makes it so much easier to remember the sin, cos, tan formulas. So here's what SOHCAHTOA means.
SOHCAHTOA:
SOH: sin=opposite leg/hypotenuse
CAH: cos=adjacent leg/hypotenuse
TOA: tan=opposite leg/adjacent leg
You use this stuff for right triangles. You use this to either find one leg or the hypotenuse, or you can use it to find angle measures. If you have the hypotenuse and one angle measure other than the right angle you would use either sin or cos to find one of the other legs, depending on which leg you want to find. If you have both the legs and the hypotenuse you can use SOHCAHTOA to find the angle measures too.
I did have a good bit of trouble finding the area of a non right triangle.
SOHCAHTOA:
SOH: sin=opposite leg/hypotenuse
CAH: cos=adjacent leg/hypotenuse
TOA: tan=opposite leg/adjacent leg
You use this stuff for right triangles. You use this to either find one leg or the hypotenuse, or you can use it to find angle measures. If you have the hypotenuse and one angle measure other than the right angle you would use either sin or cos to find one of the other legs, depending on which leg you want to find. If you have both the legs and the hypotenuse you can use SOHCAHTOA to find the angle measures too.
I did have a good bit of trouble finding the area of a non right triangle.
Alicia's Reflection #12
Okay so this week we had a quiz on 9-1 which was all word problems. I did not do well on it at all. I guess word problems are just not my thing. I could use some help with problems like on our homework worksheet (ch. 9 quiz). I do not understand how to set up the ones that do not give you any angles. If anyone can help me with this it would help me out tremendously. Lets review some trigonometry.
--The hypotenuse is opposite the right angle.
--A= 1/2 bh
SOHCAHTOA---stands for....
Sin θ= opposite leg/hypotenuse
Cos θ= adjacent leg/hypotenuse
Tan θ= opposite leg/adjacent leg
--To find the area of a non right triangle use this formula:
--A= 1/2 (leg)(leg)SIN(angle b/w)
--When you have a non right triangle that has pairs, use the law of sines.
Sin A/a = Sin B/b= Sin C/c
All you are doing is setting up a proportion.
**Remember to solve for an angle, you have to take the inverse.
--The hypotenuse is opposite the right angle.
--A= 1/2 bh
SOHCAHTOA---stands for....
Sin θ= opposite leg/hypotenuse
Cos θ= adjacent leg/hypotenuse
Tan θ= opposite leg/adjacent leg
--To find the area of a non right triangle use this formula:
--A= 1/2 (leg)(leg)SIN(angle b/w)
--When you have a non right triangle that has pairs, use the law of sines.
Sin A/a = Sin B/b= Sin C/c
All you are doing is setting up a proportion.
**Remember to solve for an angle, you have to take the inverse.
Sunday, November 8, 2009
Trigonometry Pt. 2
If you are given a right triangle like shown above, you can find the missing values (side a and angle B) with a little bit of simple trigonometry and geometry.Using SOH CAH TOA you should be able infer that sin(30.7) = (a/c)
Since c is 10.5, 10.5sin(30.7) = a
So a≈5.3607
You've solved for the sides
Now, subtract the two angles from 180° to find your final angle.
B=59.3°
Taylor Reflection #12
Everything we learned this week was very simple formulas to memorize and plugging in to solve.
I also took the chapter seven test finally
I NEED SOMEONE TO SHOW ME HOW TO SOLVE WHAT WE WILL NEED TO SOLVE WITH OUT THE CALCULATOR if that makes sense,
i dont know if that makes sense but in other words... i need someone to show me how to work the problems that i work now with a calculator that i wont have a calculator for later because everthing from chapter seven i use a calculator for..
I ALSO COULD USE HELP ON APPLYING THE UNIT CIRCLE.
ive learned it and i kind of know how to apply it
but i don't know all of the times when it is applied..
What i do know is how to clear up any possible confusion for plugging into the formulas of this past week ::
to solve a right triangle you use SOHCAHTOA
The hypotenuse is opposite of the right angle.
The adjacent leg is the side that is attached to the "given angle (not right angle)"
The opposite leg is the side that is across (or opposite) the "given angle (not right angle)"
SOHCAHTOA is really a anagram to remember the way to work out a right angle triangle problem using sine cosine and tangent
SOH: sin(theta)=opposite leg/hypotenuse
CAH: cos(theta)=adjacent leg/hypotenuse
TOA: tan(theta)=opposite leg/adjacent leg
solving for a right triangle is just a matter of drawing and plugging in
#2 Laws of Sines: sin A/a = sin B/b = sin C/c
*used when pairs are known in non-right angles
*set up a proportion
Must use a "pair" for sin A over a
a "pair" is given after you take the given information from a problem and you look for an angle given and a number to represent the leg directly across from that angle. Sin B/b is solved by simple algerbra and Sin C/ c is what you"ll get to have the measurements for the rest of the triangle.
#3 Area of a non-right triangle = 1/2 (leg) (leg) sin (angle b/w)
This can possibly be the most simple formula to remember and use
legs are numbers for either the hypotenuse, the adjacent, or the opposite lines in the drawn problem. any "leg" will work for the problem.
I also took the chapter seven test finally
I NEED SOMEONE TO SHOW ME HOW TO SOLVE WHAT WE WILL NEED TO SOLVE WITH OUT THE CALCULATOR if that makes sense,
i dont know if that makes sense but in other words... i need someone to show me how to work the problems that i work now with a calculator that i wont have a calculator for later because everthing from chapter seven i use a calculator for..
I ALSO COULD USE HELP ON APPLYING THE UNIT CIRCLE.
ive learned it and i kind of know how to apply it
but i don't know all of the times when it is applied..
What i do know is how to clear up any possible confusion for plugging into the formulas of this past week ::
to solve a right triangle you use SOHCAHTOA
The hypotenuse is opposite of the right angle.
The adjacent leg is the side that is attached to the "given angle (not right angle)"
The opposite leg is the side that is across (or opposite) the "given angle (not right angle)"
SOHCAHTOA is really a anagram to remember the way to work out a right angle triangle problem using sine cosine and tangent
SOH: sin(theta)=opposite leg/hypotenuse
CAH: cos(theta)=adjacent leg/hypotenuse
TOA: tan(theta)=opposite leg/adjacent leg
solving for a right triangle is just a matter of drawing and plugging in
#2 Laws of Sines: sin A/a = sin B/b = sin C/c
*used when pairs are known in non-right angles
*set up a proportion
Must use a "pair" for sin A over a
a "pair" is given after you take the given information from a problem and you look for an angle given and a number to represent the leg directly across from that angle. Sin B/b is solved by simple algerbra and Sin C/ c is what you"ll get to have the measurements for the rest of the triangle.
#3 Area of a non-right triangle = 1/2 (leg) (leg) sin (angle b/w)
This can possibly be the most simple formula to remember and use
legs are numbers for either the hypotenuse, the adjacent, or the opposite lines in the drawn problem. any "leg" will work for the problem.
Amy's Reflection #12!
here are some things i understand:
to solve a right triangle you use SOHCAHTOA
Area of a right angle = 1/2 bh
Laws of Sines: sin A/a = sin B/b = sin C/c
*used when pairs are known in non-right angles
*set up a proportion
Area of a non-right triangle = 1/2 (leg) (leg) sin (angle b/w)
Examples:
1. B = 90 degrees C = 50 degrees b = 10
Angle A
180 - 90 - 50 = 40
Side a
cos 50 = a/10
a = 10 cos 50
a = 6.428
Side c
sin 50 = c/10
c = 10 sin 50
c = 7.660
area = 1/2 (6.428) (7.660) = 24.620
2. two sides of a triangle have the lengths 7cm & 4cm. the angle between the sides measures 73 degrees. Find the area of the triangle.
1/2 (7) (4) sin (73) = 13.388 cm^2
Hints (word problems):
if it mentions anything about anything that has height or something on the ground that means you're gonna draw a right angle and solve...and the other problems you're gonna draw an isosceles triangle and remember to draw down the middle to make it two separate triangles...
(i hope that makes sense....)
the worksheets are due tuesday, huh?
to solve a right triangle you use SOHCAHTOA
Area of a right angle = 1/2 bh
Laws of Sines: sin A/a = sin B/b = sin C/c
*used when pairs are known in non-right angles
*set up a proportion
Area of a non-right triangle = 1/2 (leg) (leg) sin (angle b/w)
Examples:
1. B = 90 degrees C = 50 degrees b = 10
Angle A
180 - 90 - 50 = 40
Side a
cos 50 = a/10
a = 10 cos 50
a = 6.428
Side c
sin 50 = c/10
c = 10 sin 50
c = 7.660
area = 1/2 (6.428) (7.660) = 24.620
2. two sides of a triangle have the lengths 7cm & 4cm. the angle between the sides measures 73 degrees. Find the area of the triangle.
1/2 (7) (4) sin (73) = 13.388 cm^2
Hints (word problems):
if it mentions anything about anything that has height or something on the ground that means you're gonna draw a right angle and solve...and the other problems you're gonna draw an isosceles triangle and remember to draw down the middle to make it two separate triangles...
(i hope that makes sense....)
the worksheets are due tuesday, huh?
Stephanie's Reflection
AREA OF A NON-RIGHT TRIANGLE
A=1/2(leg)(leg)sin(angle between)
EX: 2 sides of a triangle have legs 7cm and 4cm. the angle between the sides measures 73 degrees. find the area of the triangle
1/2(7)(4)sin(73)=13.388com^2
the area of triangle PQR is 15. if p is 5 and q is 10, find all possible measures of R
15=1/2(5)(10)sinR
15=25sinR
sinR=3/5
R=sin^-1(3/5)
R=36.870, 143.130
RIGHT TRIANGLES
cos theta=adjacent/hypotenuse
tan theta=opposite/adjacent
LAW OF SINES
sinA/a sinB/b sinC/c
sin115/123X=sin theta/16
123sin theta=12sin115
sin theta=16sin115/123
theta=sin^-1 16sin115/123
=6.771
sin30/4=sin15/x
xsin30=4sin15
x=4sin15/sin30
x=4sin15/.5
x=2.071
A civil engineer wants to determine the distances from points A and B to an inaccessible point C. From direct measurements, AB=25cm, CA=110 and angle B=20. Find AC and BC.
sin50/25=sin20/b
bsin50=25sin20
b=25sin20/sin50
b=11.162m
sin50/25=sin110/a
asin50=25sin110
a=25sin110/sin50
a=30.667
A=1/2(leg)(leg)sin(angle between)
EX: 2 sides of a triangle have legs 7cm and 4cm. the angle between the sides measures 73 degrees. find the area of the triangle
1/2(7)(4)sin(73)=13.388com^2
the area of triangle PQR is 15. if p is 5 and q is 10, find all possible measures of R
15=1/2(5)(10)sinR
15=25sinR
sinR=3/5
R=sin^-1(3/5)
R=36.870, 143.130
RIGHT TRIANGLES
- hypotenuse opposite angle
- a=1/2bh
- SOHCAHTOA
cos theta=adjacent/hypotenuse
tan theta=opposite/adjacent
LAW OF SINES
sinA/a sinB/b sinC/c
- used when you know pairs in a non-right triangle
- you are setting up proportions
sin115/123X=sin theta/16
123sin theta=12sin115
sin theta=16sin115/123
theta=sin^-1 16sin115/123
=6.771
sin30/4=sin15/x
xsin30=4sin15
x=4sin15/sin30
x=4sin15/.5
x=2.071
A civil engineer wants to determine the distances from points A and B to an inaccessible point C. From direct measurements, AB=25cm, CA=110 and angle B=20. Find AC and BC.
sin50/25=sin20/b
bsin50=25sin20
b=25sin20/sin50
b=11.162m
sin50/25=sin110/a
asin50=25sin110
a=25sin110/sin50
a=30.667
Monday, November 2, 2009
Stephen's Reflection
Ok so this week we finished up with trig and did the test on thursday. Hopefully everyone did well on it and i also found that it was kinda easy. Maybe because i actually studied for this one. Anyway after the test we started Chapter 9 with SOHCAHTOA. This is fairly easy when you learn what the sides of a right triangle is.
The hypotenuse is opposite of the right angle.
The adjacent leg is the side that is opposite of the hypotenuse.
The opposite leg is the side that is opposite the top angle.
What SOHCAHTOA really means is:
SOH: sin(theta)=opposite leg/hypotenuse
CAH: cos(theta)=adjacent leg/hypotenuse
TOA: tan(theta)=opposite leg/adjacent leg
When given a right triangle, all you have to do is plug it in. If it only gives u a hypotenuse and an opposite leg, you would use sin(theta). If it gives u a hypotenuse and adjacent leg, you will use cos(theta) and if it only give you the opp leg and adj leg, you will use tan(theta).
That is basically it for SOHCAHTOA. Theres not much that i dont understand because i find trig pretty easy and this is fairly easy as well.
The hypotenuse is opposite of the right angle.
The adjacent leg is the side that is opposite of the hypotenuse.
The opposite leg is the side that is opposite the top angle.
What SOHCAHTOA really means is:
SOH: sin(theta)=opposite leg/hypotenuse
CAH: cos(theta)=adjacent leg/hypotenuse
TOA: tan(theta)=opposite leg/adjacent leg
When given a right triangle, all you have to do is plug it in. If it only gives u a hypotenuse and an opposite leg, you would use sin(theta). If it gives u a hypotenuse and adjacent leg, you will use cos(theta) and if it only give you the opp leg and adj leg, you will use tan(theta).
That is basically it for SOHCAHTOA. Theres not much that i dont understand because i find trig pretty easy and this is fairly easy as well.
Sunday, November 1, 2009
Taylor Reflection #11
This week was okay i guess
im still trying very hard to catch up on what i missed
i am facing the dreded chapter seven test this week
and im very nervous because although i got to review with the class i didnt get to experience any quizes of this chapter or hear and learn the information first hand
i think i might just be okay
i have learned and memorized all the formulas and rules
however im still trying to practice the application of the rules and formulas
and whats funny is i completely understand chapter nine
i guess it will all come down to doing all the problems the book has to offer and hoping for the best
what i definitely understand is the trig chart and i also made up a quick way to remember it
what i also know is the methods
such as what the variables for the four word problem formulas stand for
R- radius
((distance will mean radius))
S- arc length
Theta- angle
(( apparent size will mean Theta)
K- area of a sector
Take the word problems slowly
take the time necessary to discover what is given in the problem and work step by step from there
just tell your self there is a way to solve this problema and it cant be too hard
what i think i need help on will really be the key way for recognizing which problem will be worked with which method.
any tricks to help me decipher when i take the test would be awesome
Thanks!
im still trying very hard to catch up on what i missed
i am facing the dreded chapter seven test this week
and im very nervous because although i got to review with the class i didnt get to experience any quizes of this chapter or hear and learn the information first hand
i think i might just be okay
i have learned and memorized all the formulas and rules
however im still trying to practice the application of the rules and formulas
and whats funny is i completely understand chapter nine
i guess it will all come down to doing all the problems the book has to offer and hoping for the best
what i definitely understand is the trig chart and i also made up a quick way to remember it
what i also know is the methods
such as what the variables for the four word problem formulas stand for
R- radius
((distance will mean radius))
S- arc length
Theta- angle
(( apparent size will mean Theta)
K- area of a sector
Take the word problems slowly
take the time necessary to discover what is given in the problem and work step by step from there
just tell your self there is a way to solve this problema and it cant be too hard
what i think i need help on will really be the key way for recognizing which problem will be worked with which method.
any tricks to help me decipher when i take the test would be awesome
Thanks!
Stephanie's Reflection
Unit Circle
90 pi/2
180 pi
270 3pi/2
360 2pi
y=3
x=4
sin theta = y/r
cos theta = x/r
tan theta = y/x
csc theta = r/x
sec theta = x/y
cot theta = x/y
r=sqrtx^2+y^2
180 pi
270 3pi/2
360 2pi
sin pos in one and two and neg in three and four
cos pos in one and four and neg in two and three
tan pos in one and three and neg in two and four
cot pos in one and neg in two three and four
sec pos in one and neg in two three and four
csc pos in one and two and neg in three and four
sec = 1/cos
cos = 1/sin
EG: 3^2+3^2=5
sqrt25=h
r=5y=3
x=4
sin theta=3/5
cos theta=4/5
tan theta=3/4
csc theta=5/3
sec theta=5/4
cot theta=4/3
find the reference angle for sin 210 degrees
sin 210-__-___sin_30___
210-180=30
sin 210=-1/2
find the reference angle for cos 315 degrees
cos 215=__+___cos__45___
cos 315=+/-sqrt2/2
315-180=45
tan 695=__-__tan__25__
tan 695=-466
695-360=335
c=2pir
2pir=1/2 2piR
r=1/2R
cos theta-r/R
=1/2
theta=60
find the reference angle for sin 210 degrees
sin 210-__-___sin_30___
210-180=30
sin 210=-1/2
find the reference angle for cos 315 degrees
cos 215=__+___cos__45___
cos 315=+/-sqrt2/2
315-180=45
tan 695=__-__tan__25__
tan 695=-466
695-360=335
c=2pir
2pir=1/2 2piR
r=1/2R
cos theta-r/R
=1/2
theta=60
Alicia's Reflection #11
Okay so this week we had our chapter 7 text on trigonometry. I hope everyone did good on the test. I thought I did alright except for the 3 word problems on latitude and earth and all that so I may need some help with those types of word problems. We also started chapter 9 this week on Right triangles.
Notes:
*the hypotenuse is opposite the right angle.
*A= 1/2 bh
*SOHCAHTOA---stands for....
Sin θ= opposite leg/hypotenuse
Cos θ= adjacent leg/hypotenuse
Tan θ= opposite leg/adjacent leg
Example: a right triangle has an angle of 28 degrees, an opposite leg of 40 degrees, and an adjacent leg b.
Tan θ= opposite leg/ adjacent leg
Tan 28 degrees=40/b
b Tan 28 degrees=4
b Tan
Notes:
*the hypotenuse is opposite the right angle.
*A= 1/2 bh
*SOHCAHTOA---stands for....
Sin θ= opposite leg/hypotenuse
Cos θ= adjacent leg/hypotenuse
Tan θ= opposite leg/adjacent leg
Example: a right triangle has an angle of 28 degrees, an opposite leg of 40 degrees, and an adjacent leg b.
Tan θ= opposite leg/ adjacent leg
Tan 28 degrees=40/b
b Tan 28 degrees=4
b Tan
Yeah so I'm diggin' this trigonometry stuff. I think I really am picking up this stuff a lot better than the stuff we were learning the first quarter. One thing this week that I learned was Right Triangles. When I saw this I was surprised that I already knew how to do it. I learned it in Geometry last year. I even remembered SOHCAHTOA
Notes:
hypotenuse opposite right angle
A=1/2bh
SOHCAHTOA
A right triangle has an angle of 32 degrees. The adjacent leg is 5. Find the hypotenuse.
Answer=cos32=5/x
xcos32=5
x=5/cos5 or 5/5.896
The only thing that I'm still having trouble with is the word problems with the arc length and the planets and stuff.
Notes:
hypotenuse opposite right angle
A=1/2bh
SOHCAHTOA
A right triangle has an angle of 32 degrees. The adjacent leg is 5. Find the hypotenuse.
Answer=cos32=5/x
xcos32=5
x=5/cos5 or 5/5.896
The only thing that I'm still having trouble with is the word problems with the arc length and the planets and stuff.
Alaina's Reflection
This week was pretty easy, I think. This week, we learned to use trig functions to find the measurements of the sides and angles of an isosceles triangles.
What happened to chapter 8, by the way?
What happened to chapter 8, by the way?
- drop down from the top of the triangle to create the altitude and two right triangles.
- divide the base in half to give you the base measurement of both right triangles.
- to find you altitude, your newly found leg, you would use the Pythagorean Theorem.
- to find the angle measures, you would use trig inverses. (sin-1, cos-1, and tan-1)
As far as things that I don't understand, there isn't much. Actually, I can't think of anything at the moment that I don't understand that has to do with Trig. The test on Thursday was pretty easy, mainly because I studied my butt off and had it retaught to me about four times. So I understand pretty much all that I've learned so far concerning Trig.
An Introduction to Triangle Trigonometry
Okay, here we are, basic triangle trigonometry. Here are a few things that will come in handy:
Trigonometric Properties:
SOH CAH TOA
sine=((opposite)/(hypotenuse))
cosine=((adjacent)/(hypotenuse))
tangent=((opposite)/(adjacent))
And the less commonly used: CHO SHA CAO
cosecant=((hypotenuse)/(opposite))
secant=((hypotenuse)/(adjacent))
cotangent=((adjacent)/(opposite))
~ Use the functions above when solving for a side
~ Use the inverses of the functions above when solving for an angle.
Naming Parts of a Triangle:
A triangle’s sides are labeled as lowercase letters.
A triangle’s edges are labeled as either capital or greek letters (alpha, beta, gamma, delta, phi, psy, omega, theta, etc).
Geometric Properties:
The sum of the angles of a triangle add up to 180°. In other words, if you know 2 angles of a triangle you can subtract those from 180 to find the third angle!
The sum of the angles of any polygon with 4 or more sides add up to 360°.
Hope this helps!
Amy's Reflection #11
ok ths week started chapter 9..i wonder if we're gonna go back and do chapter 8??..anyway im not really good at solving right angles so im not..dont want solve a problem wrong and confuse someone...but here are some things do understand:
Coversion:
Degrees to Radians is: degrees*pi/180
Radians to Degree is: radians*180/pi
Unit Circle:
90 degs. = (0,1) pi/2
180 degs. = (-1,0) pi
270 degs. = (0,-1) 3pi/2
360 degs. = (1,0) 2pi
sin=y/r
cos=x/r
tan=y/x
cot=x/y
sec=r/x
csc=r/y
SOHCAHTOA:
S = sin
O = opposite angle
H = hypotenuse
(sin = opposite/hypotenuse)
C = cos
A = adjacent angle
H = hypotenuse
(cos = adjacent/hypotenuse)
T = tan
O = opposite angle
A = adjacent angle
(tan = opposite/adjacent)
and i still need with the inverse stuff..if someone could help me with that..i'll really appreciate it..
Coversion:
Degrees to Radians is: degrees*pi/180
Radians to Degree is: radians*180/pi
Unit Circle:
90 degs. = (0,1) pi/2
180 degs. = (-1,0) pi
270 degs. = (0,-1) 3pi/2
360 degs. = (1,0) 2pi
sin=y/r
cos=x/r
tan=y/x
cot=x/y
sec=r/x
csc=r/y
SOHCAHTOA:
S = sin
O = opposite angle
H = hypotenuse
(sin = opposite/hypotenuse)
C = cos
A = adjacent angle
H = hypotenuse
(cos = adjacent/hypotenuse)
T = tan
O = opposite angle
A = adjacent angle
(tan = opposite/adjacent)
and i still need with the inverse stuff..if someone could help me with that..i'll really appreciate it..
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