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!

Hint: Start by noting $\sin^4 (\theta)=\left(\sin^2(\theta)\right)\cdot\left(\sin^2(\theta)\right)$. Then use  the double angle formula derived by $\cos(2\theta)=\cos^2(\theta)-\sin^2(\theta)=1-2\sin^2 (\theta)$ so $$\sin^2(\theta)={1\over 2}\cdot\left(1-\cos(2\theta)\right)$$ and you now plug into the factors of  $\sin^4 (\theta) $, and use the double angle formula for $\cos^2(2\theta) $ to get the required answer.
