Question:
Prove that
$\sin \left(x-\frac{\pi}{6}\right)+\cos \left(x-\frac{\pi}{3}\right)=\sqrt{3} \sin x$
Solution:
In this question the following formulas will be used:
$\sin (A-B)=\sin A \cos B-\cos A \sin B$
$\cos (A-B)=\cos A \cos B+\sin A \sin B$
$=\sin x \cos \frac{\pi}{6}-\cos x \sin \frac{\pi}{6}+\cos x \cos \frac{\pi}{3}+\sin x \sin \frac{\pi}{3}$
$=\sin x \times \frac{\sqrt{3}}{2}-\cos x \times \frac{1}{2}+\cos x \times \frac{1}{2}+\sin x \times \frac{\sqrt{3}}{2}$
$=\sin x \times \frac{\sqrt{3}}{2}+\sin x \times \frac{\sqrt{3}}{2}$
$=\left(\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}\right) \sin x$
$=\sqrt{3} \sin x$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.