The maximum value of 3 cos theta + 5 sin left | vedclass

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Question

(A) Let $f(	heta) = 3 \cos 	heta + 5 \sin \left( 	heta - \frac{\pi}{6} ight)$.Using the identity $\sin(A - B) = \sin A \cos B - \cos A \sin B$:$f

Vedclass · Accepted Answer

$\sqrt{19}$

Vedclass · Answer

(A) Let $f(	heta) = 3 \cos 	heta + 5 \sin \left( 	heta - \frac{\pi}{6} ight)$.Using the identity $\sin(A - B) = \sin A \cos B - \cos A \sin B$:$f(	heta) = 3 \cos 	heta + 5 \left( \sin 	heta \cos \frac{\pi}{6} - \cos 	heta \sin \frac{\pi}{6} ight)$$f(	heta) = 3 \cos 	heta + 5 \left( \frac{\sqrt{3}}{2} \sin 	heta - \frac{1}{2} \cos 	heta ight)$$f(	heta) = \frac{5\sqrt{3}}{2} \sin 	heta + \left( 3 - \frac{5}{2} ight) \cos 	heta$$f(	heta) = \frac{5\sqrt{3}}{2} \sin 	heta + \frac{1}{2} \cos 	heta$.The expression is of the form $a \sin 	heta + b \cos 	heta$,which has a maximum value of $\sqrt{a^2 + b^2}$.Maximum value $= \sqrt{\left( \frac{5\sqrt{3}}{2} ight)^2 + \left( \frac{1}{2} ight)^2}$$= \sqrt{\frac{25 	imes 3}{4} + \frac{1}{4}} = \sqrt{\frac{75 + 1}{4}} = \sqrt{\frac{76}{4}} = \sqrt{19}$.

The maximum value of $3 \cos \theta + 5 \sin \left( \theta - \frac{\pi}{6} \right)$ for any real value of $\theta$ is

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