Question

I would really appreciate help with the following, along with work shown so I can understand how these were solved.

Answer 1

First, you have to know the double angle formula:

sin 2A  =  2 sin A cos A

cos 2A  =  cos² A − sin² A  =  2 cos² A − 1  =  1 − 2 sin² A

tan 2A = 2 tan A / (1 − tan² A)

Next, find the other leg: 17^2-15^=64, √64 is 8, so the unknown leg length is 8.

sinα=15/17, cosα=8/17, tanα=15/8

sinβ=8/17, cosβ15/17, tanβ=8/15

Plug these values in the above formula, you'll have your answers.

For the half angles, know the half-angle formula:

sin(B/2) = ± √(1 − cos B)/2

cos(B/2) = ± √(1 + cos B)/2

tan(B/2)  =  (1 − cos B) / sin B  =  sin B / (1 + cos B)

refer to this website for details:

https://brownmath.com/twt/double.htm

For the power reduction formula, please search for it. I tried to copy and paste here, but it wouldn't paste. 

for number 4, factor the quadratic equation into (cscx-8)(cscx+1)=0

cscx=8 or cscx=-1

You may not have the csc button on your calculator. Recall that csc is 1/sin, so sinx=1/8 or sinx=-1, so x is either 7.18 or 270 degrees