## Trigonometry (11th Edition) Clone

$$\sin4x=4\sin x\cos x\cos2x$$ The equation is an identity, as proved below.
$$\sin4x=4\sin x\cos x\cos2x$$ The right side would be examined first, for it is more complex. $$X=4\sin x\cos x\cos 2x$$ $$X=2\times(2\sin x\cos x)\times\cos2x$$ - From Double-Angle Identity: $2\sin x\cos x=\sin2x$ So we replace $2\sin x\cos x$ with $\sin 2x$. $$X=2\sin2x\cos2x$$ - Again, we take advantage of the identity $2\sin x\cos x=\sin 2x$, but this time $$2\sin 2x\cos2x=\sin4x$$ Therefore, $$X=\sin4x$$ That means $$\sin4x=4\sin x\cos x\cos2x$$ The equation is an identity as a result.