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!
 A: 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 
$$T_n(\cos{\theta}) = \cos{n \theta}$$
A: $$\begin{align}\sin^4 \theta &= (\sin^2\theta)^2\\ &= \left(\frac12-\frac12\cos(2\theta)\right)^2\\ &= \frac14 \left(1 - \cos(2\theta)\right)^2\\ &= \frac14\left(1 - 2 \cos(2\theta) + \cos^2(2 \theta)\right)\\ &= \frac14\left(1 - 2 \cos(2 \theta) + \frac12(\cos (4\theta) + 1)\right)\\ &= \frac14\left(\frac32 - 2\cos(2\theta) + \frac12\cos(4 \theta)\right)\\ &= \frac38 - \frac12\cos(2\theta) + \frac18\cos(4\theta).\end{align}$$
A: 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.
A: By repeatedly applying the formulas
\begin{eqnarray*}
\sin(x)\sin(y)&=&{1\over2}\left[\cos(x-y)-\cos(x+y)\right]\\[5pt]
\sin(x)\cos(y)&=&{1\over2}\left[\sin(x-y)+\sin(x+y)\right]
\end{eqnarray*}
you will see how to write odd powers of sine as a linear combination of sines,
 and even powers of sine as a linear combination of cosines.
