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!

Write
$$\sin^4{\theta} = \left ( \frac{e^{i \theta} - e^{-i \theta}}{2 i} \right )^4$$
and use the binomial theorem.
$$\begin{align}\left ( \frac{e^{i \theta} - e^{-i \theta}}{2 i} \right )^4 &= \frac{1}{16} (e^{i 4 \theta} - 4 e^{i 2 \theta} + 6 - 4 e^{-i 2 \theta} + e^{-i 4 \theta}) \\ &= \frac{1}{8} (\cos{4 \theta} - 4 \cos{2 \theta} + 3)\end{align}$$
Item of curiosity: the Chebyshev polynomials are defined such that 
$$T_n(\cos{\theta}) = \cos{n \theta}$$
