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.
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