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. 

Using $\sin 2\theta=2\sin\theta\cos\theta$ and $\sin^2\theta=(1-\cos2\theta)/2$ we get
$$\cos^2x ~\sin^3x=(\cos x\sin x)^2\sin x=\left(\frac{\sin 2x}{2}\right)^2\sin x=\frac{1}{4}\frac{1-\cos 4x}{2}\sin x \tag{$\circ$}$$
With the sum rule we have $\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\sin\beta\cos\alpha.$ Therefore we have the identity
$$\frac{\sin(\alpha+\beta)-\sin(\alpha-\beta)}{2}=\sin\beta\cos\alpha.$$
Apply with $\alpha=4x$ and $\beta=x$ to $(\circ)$.
