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=\frac{1-\cos(2x)}2`

We have

`2\sin^4(2x)=2(\sin^2(2x))^2=2\left(\frac{1-\cos(4x)}2\right)^2=\frac12(1-\cos(4x))^2`

Expand the binomial:

`2\sin^4(2x)=\frac12(1-2\cos(4x)+\cos^2(4x))`

Using the identity in the previous question,

`\cos^2x=\frac{1+\cos(2x)}2`

we get

`\cos^2(4x)=\frac{1+\cos(8x)}2`

So,

`2\sin^4(2x)=\frac12\left(1-2\cos(4x)+\frac{1+\cos(8x)}2\right)`

`2\sin^4(2x)=\frac14\left(3-4\cos(4x)+\cos(8x)\right)`