## Trigonometry (11th Edition) Clone

$$\frac{2(\sin x-\sin^3x)}{\cos x}=\sin 2x$$ The process is explained step by step below.
$$\frac{2(\sin x-\sin^3x)}{\cos x}=\sin 2x$$ 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.