Prove that


Prove that

$\sin \left(x-\frac{\pi}{6}\right)+\cos \left(x-\frac{\pi}{3}\right)=\sqrt{3} \sin x$



In this question the following formulas will be used:

$\sin (A-B)=\sin A \cos B-\cos A \sin B$

$\cos (A-B)=\cos A \cos B+\sin A \sin B$

$=\sin x \cos \frac{\pi}{6}-\cos x \sin \frac{\pi}{6}+\cos x \cos \frac{\pi}{3}+\sin x \sin \frac{\pi}{3}$

$=\sin x \times \frac{\sqrt{3}}{2}-\cos x \times \frac{1}{2}+\cos x \times \frac{1}{2}+\sin x \times \frac{\sqrt{3}}{2}$

$=\sin x \times \frac{\sqrt{3}}{2}+\sin x \times \frac{\sqrt{3}}{2}$

$=\left(\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}\right) \sin x$

$=\sqrt{3} \sin x$


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