Answer
$$\frac{2(\sin x-\sin^3x)}{\cos x}=\sin 2x$$
The process is explained step by step below.
Work Step by Step
$$\frac{2(\sin x-\sin^3x)}{\cos x}=\sin 2x$$
Let's see to the left side first.
$$X=\frac{2(\sin x-\sin^3x)}{\cos x}$$
$$X=\frac{2\sin x(1-\sin^2x)}{\cos x}$$
- As to $(1-\sin^2x)$, recall that $\cos^2x=1-\sin^2x$, meaning that
$$X=\frac{2\sin x\cos^2x}{\cos x}$$
$$X=2\sin x\cos x$$
- Now recall that $2\sin x\cos x=\sin 2x$, as stated in double-angle identity for sine.
Therefore,
$$X=\sin2x$$
That means $$\frac{2(\sin x-\sin^3x)}{\cos x}=\sin 2x$$
The verification is complete. The equation is an identity.
