<s>[INST] Find constants $a$, $b$, $c$, $d$ and $e$ such that $\cos4x=a\sin^4x+b\sin^3x+c\sin^2x+d\sin x+e$ for all angles $x$ Basically, write $\cos4x$ as a polynomial in $\sin x$.
I've tried the double angles theorem and $\cos2x=\cos^2x-\sin^2x$. I'm still having trouble right now though.
Please help! Thanks!
 [/INST] If you prefer we can use complex numbers (yes!).
Let $z=\cos x+i\sin x$. Using de Moivre's theorem:
$$\begin{align} z^4&=\cos 4x+i\sin 4x=(\cos x+i\sin x)^4\\&=\cos^4x+4\cos x(i\sin x)+6\cos^2x(i\sin x)^2+4\cos x(i\sin x)^3+(i\sin x)^4\end{align}$$
Since we're only concerned with the real part of this expression, $\cos 4x$, we see that
$$\cos4x=\cos^4x-6\cos^2x\sin^2x+\sin^4x$$
and now use $\cos^2x=1-\sin^2x$ repeatedly.
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