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

$\cos^2 \left( \frac{\pi}{4} - \beta \right) - \sin^2 \left( \alpha - \frac{\pi}{4} \right) = $

The maximum value of $(7 \cos\theta + 24 \sin\theta) \times (7 \sin\theta - 24 \cos\theta)$ for every $\theta \in R$.

The value of $\frac{\cos 2 \theta}{1-\sin 2 \theta}=$

For $A = 133^\circ$,$2\cos \frac{A}{2}$ is equal to

$\sec {50^o} + \tan {50^o}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module