Answer
The expression on the left-side is equal to the expression on the right-side.
Work Step by Step
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 $\sin \left( x+\frac{\pi }{6} \right)-\cos \left( x+\frac{\pi }{3} \right)=\sqrt{3}\sin x$.
