Express $8\sin^2(2x) -4$ as a single sine or cosine

Question

My (incorrect) work so far:

$8\sin^2(2x)-4$

= $-4[1-\sin^2(2x)-1]$

= $-4[\cos(2x)]$

= $-4[2\cos^2(x)-1]$

= $-8\cos^2(x)+4$

The answer on the answer key : $-4\cos4(x)$. I'd appreciate any help.

Answer 1

$$8\sin^2(2x)-4=4(2\sin^2(2x)-1)$$ Since $-\cos(2y)=2\sin ^2(y)-1$, let $y=2x$. Then, $2\sin^2(2x)-1=-\cos(4x)$. This implies that $$8\sin^2(2x)-4=4(2\sin^2(2x)-1)=-4\cos(4x)$$