<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] No, just take $x=0,\pi/4,\pi/2$, and you'll get three equations on $A,B,C$:
\begin{align}
0&=A+B+C\\ \ \\
\frac14&=A+0-C\\ \\
1&=A-B+C
\end{align}
Solving, we get $A=3/8$, $B=-1/2$, $C=1/8$. 
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