I've been working on this problem on and off for a couple hours and have not been able to find out how to go about it even after looking through the other questions and some google searching.
Prove that the given equations are identities $ \sin^4 \theta = \frac{3-4\cos2\theta+\cos4\theta}{8} $
So I started off trying this
$\sin^4 = (\sin^2\theta)^2=(1-\cos^2\theta)^2$
$(1-\cos^2\theta)^2 = \left[1-\left(\frac 12\cos2\theta + 1\right)\right]^2$
But that hasn't led me to anywhere. Trying it from the other side hasn't yielded much results for me either.
$\frac{(3-4\cos2\theta+\cos4\theta)}{8}= \frac18(3-4[2\cos^2\theta-1]+\cos4\theta)$
$\frac18(3-4[2\cos^2\theta-1]+\cos4\theta) = \frac18(3-4[2(1-\sin^2\theta)-1]+\cos4\theta)$
From there I seem to have just hit a loop with $\sin^2$ and $\cos^2$ . I have A LOT of trouble with these types of problems in this section. I would greatly appreciate it if someone could help me out.