Expanding $4\sin^3(x)$ using the Complex Exponential

Question

How would I go about proving: $$ 4 \sin^3(x)=3\sin(x)-\sin(3x) $$ Using the complex exponential, i.e. $$ e^{ix}=\cos(x)+i\sin(x) $$

Answer 1

$$(\cos x+i\ \sin x)^n=\cos nx+i\ \sin nx....(1)$$

Since we want $\sin^3x$ we will expand $(\cos x+i\ \sin x)^3$

$(\cos x+i\ \sin x)^3=\cos^3x-3\cos x\sin^2x+i\ (3\cos^2x\sin c-\sin^3x)$

From $(1)$ we have $$\cos3x+i\ \sin3x=\cos^3x-3\cos x\sin^2x+i(3\cos^2x\sin x-\sin^3 x)$$

Now equate the imaginary parts and we get

$$\sin3x=3\cos^2x\sin x-\sin^3x$$ $$\sin3x=3(1-\sin^2x)\sin x-\sin^3x$$ $$\sin3x=3\sin x-3\sin^2x\sin x-\sin^3 x$$ $$\sin3x=3\sin x-4\sin^3x$$ Therefore, $$4\sin^3x=3\sin x-\sin3x$$