How to rewrite $\sin^4 \theta$ in terms of $\cos \theta, \cos 2\theta,\cos3\theta,\cos4\theta$? I need help with writing $\sin^4 \theta$ in terms of $\cos \theta, \cos 2\theta,\cos3\theta, \cos4\theta$.
My attempts so far has been unsuccessful and I constantly get developments that are way to cumbersome and not elegant at all. What is the best way to approach this problem?
I know that the answer should be:
$\sin^4 \theta =\frac{3}{8}-\frac{1}{2}\cos2\theta+\frac{1}{8}\cos4\theta$
Please explain how to do this.
Thank you!

$$\begin{align}\sin^4 \theta &= (\sin^2\theta)^2\\ &= \left(\frac12-\frac12\cos(2\theta)\right)^2\\ &= \frac14 \left(1 - \cos(2\theta)\right)^2\\ &= \frac14\left(1 - 2 \cos(2\theta) + \cos^2(2 \theta)\right)\\ &= \frac14\left(1 - 2 \cos(2 \theta) + \frac12(\cos (4\theta) + 1)\right)\\ &= \frac14\left(\frac32 - 2\cos(2\theta) + \frac12\cos(4 \theta)\right)\\ &= \frac38 - \frac12\cos(2\theta) + \frac18\cos(4\theta).\end{align}$$
