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.
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