Answer
$$\sin4x=4\sin x\cos x-8\sin^3x\cos x$$
The equation is verified to be an identity as below.
Work Step by Step
$$\sin4x=4\sin x\cos x-8\sin^3x\cos x$$
We would tackle from the right side.
$$X=4\sin x\cos x-8\sin^3x\cos x$$
$$X=4\sin x\cos x-(4\sin x\cos x)\times(2\sin^2x)$$
$$X=4\sin x\cos x(1-2\sin^2x)$$
$$X=2\times(2\sin x\cos x)\times(1-2\sin^2x)$$
- From Double-Angle Identities for sine and cosine:
$$2\sin x\cos x=\sin2x$$
$$1-2\sin^2x=\cos2x$$
Therefore,
$$X=2\sin2x\cos2x$$
- Finally, we can still apply Double-Angle Identity for sine ($2\sin\theta\cos\theta=\sin2\theta$) to $X$, but as $\theta=2x$. That means,
$$X=\sin(2\times2x)$$
$$X=\sin4x$$
That concludes,
$$\sin4x=4\sin x\cos x-8\sin^3x\cos x$$
2 sides are equal, and the equation is an identity.
