I need to solve question number 19 and 20 Prove the identity

I need to solve question number 19 and 20

Prove the identity cot 2x = 1 - tan^2 x/2 tan x Prove the identity sin 3x + sin 7x/cos 3x - cos 7x = cot 2x Use the formulas for lowering powers to rewrite cos^4 x sin^4 x in terms of the first power of cosine Find all solutions of the equation 81 sin^2 theta - 1 = 0 Solve the equation cos 5 theta - cos 7 theta = 0 on the interval [0, 2 pi)

Solution

19) (sin 3x + sin 7x)/(cos 3x – cos 7x) = cot 2x

In order to prove this, we need to following two sum-to-product formulas:
(a) sin(A) + sin(B) = 2sin[(A + B)/2]cos[(A - B)/2]
(b) cos(A) - cos(B) = -2sin[(A + B)/2]sin[(A - B)/2].

So, we have:
LHS = [sin(3x) + sin(7x)]/[cos(3x) - cos(7x)]
= 2sin[(3x + 7x)/2]cos[(3x - 7x)/2]/{-2sin[(3x + 7x)/2]sin[(3x - 7x)/2]}, from (a) and (b)
= 2sin(5x)cos(-2x)/[-2sin(5x)sin(-2x)]
= -cos(-2x)/sin(-2x), by canceling 2sin(5x)
= -cos(2x)/[-sin(2x)], since cosine is even and sine is odd
= cos(2x)/cos(2x), since the negatives cancel
= cot(2x), by the definition of cotangent
= RHS.

20) cos^4(x) = (cos^2(x))^2
= [cos(2x) + 1]^2 / 4
= [cos^2(2x) + 2cos(2x) + 1] / 4
= [(cos(4x) + 1)/2 + 2cos(2x) + 1] / 4
= [cos(4x) + 1 + 4cos(2x) + 2] / 8
= [cos(4x) + 4cos(2x) + 3] / 8

sin^4(x) = (sin^2(x))^2
= [1-cos^2(x)]^2
= [1 - (cos(2x) + 1)/2]^2
= [2 - cos(2x) - 1]^2 / 4
= [1 - cos(2x)]^2 / 4
= [1 - 2cos(2x) + cos^2(2x)] / 4
= [1 - 2cos(2x) + (cos(4x) + 1)/2] / 4
= [2 - 4cos(2x) + cos(4x) + 1] / 8
= [3 - 4cos(2x) + cos(4x)] / 8


Multiply these two expressions to get
cos^4(x) sin^4(x)
= [cos(4x) + 4cos(2x) + 3] / 8 * [3 - 4cos(2x) + cos(4x)] / 8
= ...expand and simplify...
= [cos^2(4x) + 6cos(4x) - 16cos^2(2x) + 9] / 64
= [(cos(8x) + 1)/2 + 6cos(4x) - 16(cos(4x) + 1)/2 + 9]/64
= [cos(8x) + 1 + 12cos(4x) - 16cos(4x) - 16 + 18] / 128
= [cos(8x) - 4cos(4x) + 3] / 128