Verifying $\sin 4θ=4\cos^3 θ \sin θ - 4\cos θ \sin^3θ$ $$\sin 4θ=4\cos^3 θ \sin θ - 4\cos θ \sin^3θ.$$
Ηere is what I have so far
$$\sin 4θ = 2\sin 2θ \cos 2θ = 4\sin θ \cos θ \cos 2θ.$$
Not sure if this is the correct path I should take to solve this problem. I have been stuck hard for about an hour now.

$$\begin{align}\sin(n+1)x&=\sin nx\cos x+\cos nx\sin x\\
\sin(n-1)x&=\sin nx\cos x-\cos nx\sin x\\
\hline\sin(n+1)x+\sin(n-1)x&=2\sin nx\cos x\end{align}$$
Now that we know that
$$\sin(n+1)x=2\sin nx\cos x-\sin(n-1)x$$
We can go forward to
$$\begin{align}\sin 0x&=0\\
\sin1x&=\sin x\\
\sin2x&=2\sin x\cos x-\sin0x=2\sin x\cos x\\
\sin 3x&=2\sin 2x\cos x-\sin1x=4\sin x\cos^2x-\sin x\\
\sin4x&=2\sin3x\cos x-\sin2x=8\sin x\cos^3x-4\sin x\cos x\\
&=8\sin x\cos^3x-4\sin x\cos x(\cos^2x+\sin^2x)\\
&=4\sin x\cos^3x-4\sin^3x\cos x\end{align}$$
