The period of sin theta - sqrt{3} cos theta i | vedclass

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Question

(D) We can rewrite the expression as: $\sin 	heta - \sqrt{3} \cos 	heta = 2 \left( \frac{1}{2} \sin 	heta - \frac{\sqrt{3}}{2} \cos 	heta ight)$

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$2\pi$

Vedclass · Answer

(D) We can rewrite the expression as: $\sin 	heta - \sqrt{3} \cos 	heta = 2 \left( \frac{1}{2} \sin 	heta - \frac{\sqrt{3}}{2} \cos 	heta ight)$.Using the identity $\sin(A - B) = \sin A \cos B - \cos A \sin B$,we get: $2 \sin \left( 	heta - \frac{\pi}{3} ight)$.The period of $\sin(k	heta)$ is $\frac{2\pi}{|k|}$.Here,$k = 1$,so the period is $\frac{2\pi}{1} = 2\pi$.

The period of $\sin \theta - \sqrt{3} \cos \theta$ is

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