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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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

Similar Questions

If $2\cos\theta + \sin\theta = 1$,then the value of $4\cos\theta + 3\sin\theta$ is equal to

The value of $\sin \theta + \cos \theta$ will be greatest when

If $ABCD$ is a cyclic quadrilateral,then the value of $\cos A - \cos B + \cos C - \cos D = $

If $A$ lies in the second quadrant and $B$ lies in the third quadrant and $\cos A = -\frac{\sqrt{3}}{2}$ and $\sin B = -\frac{3}{5}$,then find the value of $\frac{2 \tan B + \sqrt{3} \tan A}{\cot^2 A + \cos B}$.

If $\sin A + \sin^2 A = 1$,then the value of $\cos^2 A + \cos^4 A$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module