## Precalculus (6th Edition) Blitzer

The expression on the left- side is $\sin \left( x+\frac{\pi }{6} \right)-\cos \left( x+\frac{\pi }{3} \right)$, which can be simplified by using the identity $\sin \left( C+D \right)$: $\sin \left( C+D \right)=\sin C\text{cos }D+\cos C\sin D$ Also, we use the identity $\cos \left( x+y \right)$: $\cos \left( x+y \right)=\cos x\cos y-\sin x\sin y$ Now, applying both of the identities, the expression can be simplified as follows: \begin{align} & \sin \left( x+\frac{\pi }{6} \right)-\cos \left( x+\frac{\pi }{3} \right)=\left( \sin x\cos \frac{\pi }{6}+\cos x\sin \frac{\pi }{6} \right)-\left( \cos x\cos \frac{\pi }{3}-\sin x\sin \frac{\pi }{3} \right) \\ & =\left( \sin x.\frac{\sqrt{3}}{2}+\cos x.\frac{1}{2} \right)-\left( \cos x.\frac{1}{2}-\sin x.\frac{\sqrt{3}}{2} \right) \\ & =\sin x.\frac{\sqrt{3}}{2}+\cos x.\frac{1}{2}-\cos x.\frac{1}{2}+\sin x.\frac{\sqrt{3}}{2} \end{align} The expression can be further simplified as $\sin x.\frac{\sqrt{3}}{2}+\cos x.\frac{1}{2}-\cos x.\frac{1}{2}+\sin x.\frac{\sqrt{3}}{2}=\sqrt{3}\sin x$ Thus, the expression on the left-side is equal to the expression on the right-side. $\sin \left( x+\frac{\pi }{6} \right)-\cos \left( x+\frac{\pi }{3} \right)=\sqrt{3}\sin x$ Thus, it is proved that $\sin \left( x+\frac{\pi }{6} \right)-\cos \left( x+\frac{\pi }{3} \right)=\sqrt{3}\sin x$.