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

This is the same question as 17042785 (the question number in the site URL), with the exception of using the other identity,
[tex]\sin^2x=\dfrac{1-\cos(2x)}2[/tex]
We have
[tex]2\sin^4(2x)=2(\sin^2(2x))^2=2\left(\dfrac{1-\cos(4x)}2\right)^2=\dfrac12(1-\cos(4x))^2[/tex]
Expand the binomial:
[tex]2\sin^4(2x)=\dfrac12(1-2\cos(4x)+\cos^2(4x))[/tex]
Using the identity in the previous question,
[tex]\cos^2x=\dfrac{1+\cos(2x)}2[/tex]
we get
[tex]\cos^2(4x)=\dfrac{1+\cos(8x)}2[/tex]
So,
[tex]2\sin^4(2x)=\dfrac12\left(1-2\cos(4x)+\dfrac{1+\cos(8x)}2\right)[/tex]
[tex]2\sin^4(2x)=\dfrac14\left(3-4\cos(4x)+\cos(8x)\right)[/tex]