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!
 A: $$\begin{align}\cos4x&=1-2\sin^22x\\&=1-2(1-\cos^22x)\\&=1-2(1-(1-2\sin^2x)^2)\\&=1-2(4\sin^2x-4\sin^4x)\\&=1-8\sin^2x+8\sin^4x\end{align}$$
A: Double angle theorem once:
$\cos 4x = \cos^2 2x + \sin^2 2x$.
Double angle theorem twice:
$= (\cos^2 x -\sin^2 x)^2 + (2\sin x\cos x)^2$.
Expand chicken soup and rice:
$= \cos^4 x- 2\cos^2 x\sin^2 x + \sin^4 x +4\sin^2 x\cos^2 x$
Can you finish?
To get it entirely in terms of $\sin^k x$ replace $\cos^2 x$ with $1-\sin^2 x$ (keeping in mind $\cos^4 x = (\cos^2 x)^2$)
Can you finish?
$= (1 - \sin^2 x)^2  - 2(1-\sin^2 x) \sin^2 x + \sin^4 x +4\sin^2 x(1-\sin^2 x) $
A: 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.
