<s>[INST] If $\sin^4(x)=A+B\cos(2x)+C\cos(4x)$, then find $A$, $B$, and $C$. 
If $\sin^4(x)=A+B\cos(2x)+C\cos(4x)$, then find $A, B$, and $C$.

Would I use half-angle identities for this problem?
 [/INST] By the double-angle formula for $\cos{2x}$,
$$ \cos{2x} = \cos^2{x} - \sin^2{x} = 2\cos^2{x}-1 = 1-2\sin^2{x}. $$
Rearranging,
$$ \sin^2{x} = \frac{1}{2}(1-\cos{2x}). $$
Squaring both sides an expanding,
$$ \sin^4{x} = \frac{1}{4}(1-2\cos{2x}+\cos^2{2x}), $$
and then using the double-angle formula $\cos{4x} = 2\cos^2{x}-1 $ gives
$$ \sin^4{x} = \frac{1}{4}-\frac{1}{2}\cos{2x}+\frac{1}{4}\frac{1}{2}(1+\cos{4x}) = \frac{3}{8} - \frac{1}{2}\cos{2x}+\frac{1}{8}\cos{4x}. $$

Alternatively,
$$ \sin^4{x} = \frac{(e^{ix}-e^{-ix})^4}{2^4i^4} = \frac{1}{16}(e^{4ix} - 4e^{2ix} + 6 - 4e^{-2ix} + e^{-4ix} ) = \frac{1}{8}(\cos{4x}-4\cos{2x}+3). $$
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