EQUATIONS

Example #1:
sin2x=sinx [0,2pi]
2sinxcosx=sinx
2sinxcosx-sinx=0
sinx(2cosx-1)=0
Solve by setting terms equal to 0.
sinx=0 2cosx-1=0
x=0, 2pi, pi cosx=1/2
x=pi/3, 5pi/3

Example #2:
sin2x+cos3x=0 [0, 2pi]
2sinxcosx+cos(2x+x)=0
2sinxcosx+cos2xcosx-sin2xsinx=0
2sinxcosx+cos2xcosx-2sinxcosxsinx=0
2sinxcosx+(1-2sin²x)cosx-2sin²xcosx=0
2sinxcosx+cosx-2sin²xcosx-2sin²xcosx=0
2sinxcosx+cosx-4sin²xcosx=0
cosx(2sinx+1-4sin²x)=0
cosx=0

x=0, pi/2, 3pi/2

-4sin²x+2sinx+1=0
Use quadratic formula to get answer for sinx=
Use inverse sign to find remaining 4 x values
5.97, 3.45, .942, 2.19

Example #3: (Proof)
sin4x=2sin2xcos2x
sin2x=2sinxcosx
=sin2(2x)
sin4x=sin4x
QED

Homework: pg 510 #35, 36 #51-57 Odd
Pg 518 # 3-33 multiples of 3

More on Solving Trig Equations

Homework: page 502 #9-31 odd

Sorry this post is SO late, I forgot about it over Thanksgiving and unfortunately, I am just now posting. SO SORRY! Also, sorry the sizes are so large, I did not know how to adjust that easily :P

Using Law of Sines and Law of Cosines

When do you use Law of Sines and when do you use Law of Cosines?

Well,

Law of Sines

more

AAS

SSA (ambiguous)

Law of Cosines

more sides given

SAS

SSS

Ambiguous Case: SSA

There are five possible combinations of triangles when you are given two sides and an angle.



In Case 1, you can see that side b >a

In Case 2, b forms a single right angle triangle with c. Here b equals h, h being the height of the triangle, and yields a single right angle triangle.

In Case 3, a>b>h, forming two triangles. Side b is too long to form a single right angle triangle, but yet is also too short to swing out farther than side a which would result in only one triangle. Instead it forms to triangles, one acute triangle and one obtuse triangle.

In Case 4, side b is equal side a, resulting in a single isosceles triangle. Being an isosceles triangle, angle A and angle B are also equal. Side b cannot be placed anywhere else or it would not form a triangle.

In Case 5, side b is > side a. It forms one triangle only, with side b stretching out opposite of a. It cannot be on the other side of a because then it would not form a triangle.

Ok here is how we use the Law of Sines to solve a triangle.

Lets say we are given:
a = 21
b=20

Lets start by solving for
(sinA)/a = (sinB)/b

So we plug in the numbers that we have:

(sin 33)/21 = (sinB)/20
20(0.545)/21=sinB
0.519=sinB
B=31.268°

We have found
180-33-31.268=115.732°

We then can find c with the Law of Cosines:

c(c)=a(a)+b(b)-2abcosC
c(c)=441+400-2(21)(20)cos115.732
c(c)=1205.696
c=34.723

Sum and Difference Identities

sin (x + π) ≠ sin x + sin π Cannot distribute Cosine of a difference (proof) cos (u-v) = cos u cos v + sin u sin v

Identities

Sine of a Sum

sin (u+v) = sin u cos v + cos u sin v

Sine of a Difference

sin (u-v) = sin u cos v - cos u sin v

Cosine of a Sum

cos (u+v) = cos u cos v - sin u sin v

Cosine of a Difference

cos (u-v) = cos u cos v + sin u sin v

Tangent of a Sum

tan (u+v) = (tan u + tan v)/(1-tan u tan v)

Tangent of a Difference

tan (u-v) = (tan u - tan v)/(1+ tan u tan v)

When to use it:

Example:

cos(15) = cos (45-30)

Simplifying Expressions
Example: x +1 + x -3
X2-4x+4 x -2
= x +1 + x -3
x2-4x+4 x -2

1) Factor the denominator and then find a common denominator.
= x+1 + (x-3) (x-2)
(x-2)(x-2) (x-2) (x-2)
2) Combine the fractions
= x+1+ (x-3)(x-2)
(x-2)(x-2)
3) Simplify
= 2x-2
x-2



Example: cosx – sinx
1-sinx cosx
= cosx – sinx
1-sinx cosx
1) Find a common denominator for both fractions
= cosx (cosx) – sinx (1-sinx)
(1-sinx)(cosx) cosx (1-sinx)
2) Combine the fractions
= cos2x – sinx(1-sinx)
(1-sinx)(cosx)
3) Simplify
= cos2x – sinx+sin2x
(1-sinx)(cosx)
4) Use the identity sin2x+cos2x = 1 in the numerator.
= 1-sinx
(1-sinx)(cosx)
5) Simplify
= 1
cosx
6) Use the reciprocal identity
= secx



Factoring
Example: 1 + cosx - sin2x
= 1 + cosx - sin2x
1) Use the identity sin2x + cos2x = 1 to substitute sin2x for (1-cos2x).
= 1 + cosx – (1 - cos2x)
2) Distribute the negative sign into the parenthesis.
= 1 + cosx – 1 + cos2x
3) Simplify
= cosx + cos2x
= cosx(1 + cosx)









Example: sec2x + tanx - 3
= sec2x + tanx - 3
1) Use the identity 1+tan2x = sec2x
= 1 + tan2x + tanx – 3
2) Simplify
= tan2x + tanx – 2
3) Factor
= (tanx + 2)(tanx – 1)




You can check if two expressions are equivalent by using your graphing calculator. Graph the two expressions, but change one of the expressions to the bouncing ball.

Homework: pg 487 # 1-7 odd, 15-43 odd

Solving Trig Equations - 11.12.07

Data/Trig Functions

sine curve regression


make sure your calculator is in radian mode

calculator:

1) stat button-enter data
2) 2nd/stat plot
3) move to dot plot-hit enter
4) set window
5) view graph


sine curve function

after creating the sine curve...

1) stat
2) move to calc-SinReg
3) enter: L1, L2, (commas are important)
4) vars-move to y-vars
5) function
6) enter Y1
7) hit enter

you should see something like:

y=
a=
b=
c=
d=

the data will also be a y= function in Y1

to check other types of curves for fit:

calculator:

1) stat-calc
2) choose a regression-CubReg, QuartReg, etc.-enter
3) enter: L1, L2,
4) vars
5) y-vars-function
6) enter Y2-hit enter
7) view graph



Quiz-Tuesday, November 6

Topics:

1) graphing sine/ cosine by hand
2) identifying graphs of tan/cot/sec/csc
3) unit circle
4) inverse trig
5) application