Expressing sin(4x) and sin(−2x+π/6) via sine and cosine of x

Problem

Express the following quantities using only $\sin x$ and $\cos x$: a) $\sin(4x)$; b) $\sin\!\left(-2x+\dfrac{\pi}{6}\right)$. c) Use the double-angle formulas for sine and cosine to prove the double-angle formula for the tangent $\;\tan(2\alpha)=\dfrac{2\tan\alpha}{1-\tan^{2}\alpha}$.

Solution

a) $\sin 4x=2\sin 2x\cos 2x=2(2\sin x\cos x)(\cos^{2}x-\sin^{2}x)=4\sin x\cos x(\cos^{2}x-\sin^{2}x)$. b) By the addition formula: $\sin\!\left(-2x+\tfrac{\pi}{6}\right)=\sin(-2x)\cos\tfrac{\pi}{6}+\cos(-2x)\sin\tfrac{\pi}{6}=-\tfrac{\sqrt3}{2}\sin 2x+\tfrac12\cos 2x=-\sqrt3\,\sin x\cos x+\tfrac12\big(\cos^{2}x-\sin^{2}x\big).$ c) $\tan 2\alpha=\dfrac{\sin 2\alpha}{\cos 2\alpha}=\dfrac{2\sin\alpha\cos\alpha}{\cos^{2}\alpha-\sin^{2}\alpha}$; dividing numerator and denominator by $\cos^{2}\alpha$: $\tan 2\alpha=\frac{2\dfrac{\sin\alpha}{\cos\alpha}}{1-\dfrac{\sin^{2}\alpha}{\cos^{2}\alpha}}=\frac{2\tan\alpha}{1-\tan^{2}\alpha}.$ $\boxed{\,\sin 4x=4\sin x\cos x(\cos^{2}x-\sin^{2}x)\,;\quad \sin\!\left(-2x+\tfrac{\pi}{6}\right)=-\sqrt3\sin x\cos x+\tfrac12(\cos^{2}x-\sin^{2}x)\,}$