Answer
$$\sin4x=4\sin x\cos^3x-4\sin^3 x\cos x$$
Work Step by Step
$$\sin4x$$
The question asks to write $\sin4x$ in terms of a trigonometric function of $x$. We will need both Sum and Difference Identities and Double-Angle Identities to help here.
$$\sin4x=\sin(2\times2x)$$
- Double-Angle Identity for sine: $\sin2A=2\sin A\cos A$ with here $A=2x$
$$\sin4x=2\sin2x\cos2x$$
- For $\sin2x$, we apply again $\sin2A=2\sin A\cos A$ with $A=x$.
- For $\cos2x$, we apply Double-Angle Identity for cosine: $\cos2A=\cos^2A-\sin^2A$.
$$\sin4x=2(2\sin x\cos x)(\cos^2x-\sin^2x)$$
$$\sin4x=(4\sin x\cos x)(\cos^2x-\sin^2x)$$
$$\sin4x=4\sin x\cos^3x-4\sin^3 x\cos x$$
We can stop here as all trigonometric functions are functions of $x$ now.
