Hence $\sin^4 2x = (\sin^2 2x)^2 = \left(\frac{1}{2}(1-\cos 4x)\right)^2 $ expand and use the identity $\cos^2 y = \frac{1}{2} ( 1 + \cos 2y)$ to get $\cos 8x$
Hence $\sin^4 2x = (\sin^2 2x)^2 = \left(\frac{1}{2}(1-\cos 4x)\right)^2 $ expand and use the identity $\cos^2 y = \frac{1}{2} ( 1 + \cos 2y)$ to get $\cos 8x$