<s>[INST] Prove $\cos^2 x \,\sin^3 x=\frac{1}{16}(2 \sin x + \sin 3x - \sin 5x)$ How would I prove the following?
$$\cos^2 x \,\sin^3 x=\frac{1}{16}(2 \sin x + \sin 3x - \sin 5x)$$
I do not know how to do do the problem I do know $\sin(3x)$ can be $\sin(2x+x)$ and such yet I am not sure how to commence. 
 [/INST] $2\sin x+\sin3x-\sin5x$
$=2\sin x-2\cos4x\sin x$ applying $\sin2C-\sin2D=2\sin(C-D)\cos(C+D)$ formula
$=2\sin x(1-\cos4x)$
$=4\sin x \cdot \sin^22x$, applying $\cos2A=1-2\sin^2A$
$=4\sin x (2\sin x\cos x)^2$
$=16\sin^3x\cos^2x$

Alternatively, we know $e^{iy}=\cos y+i\sin y$ => $e^{-iy}=\cos y - i\sin y$
So, $\cos y=\frac{e^{iy}+e^{-iy}}{2}$  and $\sin y=\frac{e^{iy} - e^{-iy}}{2i}$
$\sin^3x\cos^2x$
$=(\frac{e^{ix}-e^{-ix}}{2i})^3(\frac{e^{ix} + e^{-ix}}{2})^2$
$=\frac{-1}{32i}(e^{ix}-e^{-ix})^2(e^{ix} + e^{-ix})^2(e^{ix}-e^{-ix})$
$=\frac{-1}{32i}((e^{ix}-e^{-ix})(e^{ix} + e^{-ix}))^2(e^{ix}-e^{-ix})$
$=\frac{-1}{32i}(e^{2ix}-e^{-2ix})^2(e^{ix}-e^{-ix})$
$=\frac{1}{32i}(2(e^{ix}-e^{-ix}) + (e^{3ix}-e^{-3ix}) - (e^{5ix}-e^{-5ix}))$
$=\frac{1}{32i}(2(2i\sin x) + (2i\sin3x) - (2i\sin5x))$
$=\frac{2\sin x+\sin3x-\sin5x}{16}$

Alternatively, $\sin^3x\cos^2x$
$=\frac{1}{8}(4\sin^3x\cdot2\cos^2x)$
$=\frac{1}{8}(3\sin x-\sin3x)(1+\cos2x)$  as $\sin3A=3\sin A-4\sin^3A$  and $\cos2A=2\cos^2A-1$
$=\frac{1}{8}(3\sin x-\sin3x+3\sin x\cos2x-\sin3x\cos2x)$
$=\frac{1}{16}(6\sin x-2\sin3x+6\sin x\cos2x-2\sin3x\cos2x)$
$=\frac{1}{16}(6\sin x-2\sin3x+3(\sin3x-\sin x)-(\sin5x+\sin x))$ (applying $2\sin A\cos B=\sin(A+B)+\sin(A-B))$
$=\frac{1}{16}(2\sin x+\sin3x-\sin5x)$
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