Question

Use the power-reducing formulas to rewrite the expression in terms of first powers of the cosines of multiple angles.

Answer 1

This is the same question as 17042785 (the question number in the site URL), with the exception of using the other identity,

$\sin^2x=\dfrac{1-\cos(2x)}2$

We have

$2\sin^4(2x)=2(\sin^2(2x))^2=2\left(\dfrac{1-\cos(4x)}2\right)^2=\dfrac12(1-\cos(4x))^2$

Expand the binomial:

$2\sin^4(2x)=\dfrac12(1-2\cos(4x)+\cos^2(4x))$

Using the identity in the previous question,

$\cos^2x=\dfrac{1+\cos(2x)}2$

we get

$\cos^2(4x)=\dfrac{1+\cos(8x)}2$

So,

$2\sin^4(2x)=\dfrac12\left(1-2\cos(4x)+\dfrac{1+\cos(8x)}2\right)$

$2\sin^4(2x)=\dfrac14\left(3-4\cos(4x)+\cos(8x)\right)$