<s>[INST] Expressing $\sin^4x-\sin^6x$ in another way I am slightly confused on how one would subtract $\sin^4x-\sin^6x$.
I know that $\sin^2x=(1/2)(1-\cos2x)$,
so $\sin^4x$ would logically be $[(1/2)(1-\cos2x)]^2=(1/4)(1-2\cos(2x)+\cos^2(2x)$
However the value of $\sin^6x$ eludes me. Would it be $(1-\cos2x)^3$? I did that and got $1-\cos2x-2\cos2x-2\cos^2(2x)+\cos^2(2x)-\cos^3(2x)$
I am not sure if that is correct. 
 [/INST] Rather than messing about with trig identities, it may be simpler to use complex exponentials.  If $z = e^{ix} = \cos(x) + i \sin(x)$, so $\sin(x) = (z - z^{-1})/(2i)$ and $\cos(x) = (z + z^{-1})/2$,
$$\eqalign{\sin^4 x - \sin^6 x &= \frac{(z - z^{-1})^4}{16} + \frac{(z - z^{-1})^6}{64}\cr
&= \frac{z^4 - 4 z^2 + 6 - 4 z^{-2} + z^{-4}}{16} \\
&\qquad+\frac{z^6 - 6 z^4 + 15 z^2 - 20 + 15 z^{-2} - 6 z^{-4} + z^{-6}}{64}\cr
&= \frac{z^6 + z^{-6}}{64} - \frac{z^4 + z^{-4}}{32} - \frac{z^2 + z^{-2}}{64} + \frac{1}{16}\cr
&= \frac{\cos(6x)}{32} - \frac{\cos(4x)}{16}  - \frac{\cos(2x)}{32} + \frac{1}{16}\cr}$$
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