$$\sin 4θ=4\cos^3 θ \sin θ - 4\cos θ \sin^3θ.$$
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.
On
You've already verified that $\sin 4x=4\sin x\cos x\cos 2x$. To finish, use $\cos 2x = \cos^2 x - \sin^2 x$.
On
$$\begin{align}\sin(n+1)x&=\sin nx\cos x+\cos nx\sin x\\ \sin(n-1)x&=\sin nx\cos x-\cos nx\sin x\\ \hline\sin(n+1)x+\sin(n-1)x&=2\sin nx\cos x\end{align}$$ Now that we know that $$\sin(n+1)x=2\sin nx\cos x-\sin(n-1)x$$ We can go forward to $$\begin{align}\sin 0x&=0\\ \sin1x&=\sin x\\ \sin2x&=2\sin x\cos x-\sin0x=2\sin x\cos x\\ \sin 3x&=2\sin 2x\cos x-\sin1x=4\sin x\cos^2x-\sin x\\ \sin4x&=2\sin3x\cos x-\sin2x=8\sin x\cos^3x-4\sin x\cos x\\ &=8\sin x\cos^3x-4\sin x\cos x(\cos^2x+\sin^2x)\\ &=4\sin x\cos^3x-4\sin^3x\cos x\end{align}$$
On
Note that:
${\sin n \theta = \dbinom{n}{1}\cos^{n-1}\theta\sin \theta- \dbinom{n}{3}\cos^{n-3}\theta \sin^3 \theta + \dbinom n 5\cos^{n-5}\theta\sin ^{5}\theta...}\\= \color{blue}{\displaystyle\sum_{r=0, 2r+1\le n}(-1)^r\dbinom{n}{2r+1}\cos^{n-2r-1}\theta \sin^{2r+1}\theta} $
For proof, see this.
Therefore,
$\sin 4\theta = \dbinom{4}{1}\cos^3\theta \sin \theta- \dbinom{4}{3}\cos\theta \sin^3\theta = 4\cos^3 \theta \sin\theta - 4 \cos\theta \sin^3\theta$
On
$$e^{ix} = i\sin{x}+\cos{x}$$ $$e^{4ix} = (i\sin{x}+\cos{x})^4$$ $$i\sin{4x}+\cos{4x} = \sin^4{x}-4i\sin^3x\cos{x}-6\sin^2x\cos^2{x}+4i\sin x\cos^3{x} + \cos^4x$$ As all coefficients are real, we can say the imaginary parts of each side are equal: $$i\sin{4x} = -4i\sin^3x\cos{x}+4i\sin x\cos^3{x}$$ $$\sin{4x} = 4\sin x\cos^3{x}-4\sin^3x\cos{x}$$
$$4\cos^3\theta\sin\theta-4\sin^3\theta\cos\theta=4\sin\theta\cos\theta(\cos^2\theta-\sin^2\theta)=2\sin2\theta\cos2\theta=\sin4\theta.$$