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]
