Trigonometry 7th Edition

4 $\cos(2\times\frac{\pi}{6}+\frac{\pi}{3})$ = $4\cos(\frac{6\pi+6\pi}{18})$ = $4\cos(\frac{2\pi}{3})$ = $4\cos(\pi-\frac{\pi}{3})$ Using the identity $\cos(\pi-x)= -\cos x$, we get $4\cos(\pi-\frac{\pi}{3})$ = $4\times(- \cos \frac{\pi}{3})$ = $4\times(-\frac{1}{2})$ = -2