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. 

We know $e^{iy}=\cos y+i\sin y$
So, $\cos 5x+i\sin 5x=e^{i5x}=(e^{ix})^5=(\cos x+i \sin x)^5$
Using binomial theorem & equating imaginary parts,
$sin 5x = sin^5x − 10\cdot sin^3x cos^2x + 5\cdot sin x cos^4x$
$=sin^5x − 10\cdot sin^3x(1 - sin^2x) + 5\cdot sin x (1 - sin^2x)^2$
So, $sin 5x= 16\cdot\sin^5x − 20\cdot\sin^3x + 5\cdot\sin x$
We know, $\sin3x=3\sin x -4sin^3x $
$\sin^3x\cos^2x=\sin^3x(1-\sin^2x)=\sin^3x-\sin^5x$
Let $A\sin5x+B\sin3x+C\sin x=\sin^3x-\sin^5x$
Or, $A(16\cdot\sin^5x − 20\cdot\sin^3x + 5\cdot\sin x) + B(3\sin x -4sin^3x)+C\sin x$
$=\sin^3x-\sin^5x$
Comparing the coefficients of different powers of sinx,
5th power=>16A=-1=>A=$-\frac{1}{16}$
3rd power=>-20A-4B=1=>B=$\frac{1}{16}$
1st power=>5A+3B+C=0=>C=-(5A+3B)=$\frac{1}{8}$
