The value of x for the maximum value of sqrt{ | vedclass

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

Question

(A) Let $f(x) = \sqrt{3} \cos x + \sin x$.We can rewrite this expression by multiplying and dividing by $2$:$f(x) = 2 \left( \frac{\sqrt{3}}{2} \cos x

Vedclass · Accepted Answer

$30$

Vedclass · Answer

(A) Let $f(x) = \sqrt{3} \cos x + \sin x$.We can rewrite this expression by multiplying and dividing by $2$:$f(x) = 2 \left( \frac{\sqrt{3}}{2} \cos x + \frac{1}{2} \sin x \right)$.Using the identity $\sin(A + B) = \sin A \cos B + \cos A \sin B$,where $\sin 60^\circ = \frac{\sqrt{3}}{2}$ and $\cos 60^\circ = \frac{1}{2}$,we get:$f(x) = 2 \sin(x + 60^\circ)$.The maximum value of the sine function is $1$,which occurs when the angle is $90^\circ$.Therefore,for the maximum value of $f(x)$:$x + 60^\circ = 90^\circ$$x = 90^\circ - 60^\circ = 30^\circ$.

The value of $x$ for the maximum value of $\sqrt{3} \cos x + \sin x$ is .....$^o$

Similar Questions

$\tan 9^\circ - \tan 27^\circ - \tan 63^\circ + \tan 81^\circ = $

$\tan \theta \sin \left( \frac{\pi }{2} + \theta \right) \cos \left( \frac{\pi }{2} - \theta \right) = $

If $\frac{\sin(x + y)}{\sin(x - y)} = \frac{a + b}{a - b},$ then $\frac{\tan x}{\tan y}$ is equal to

$ABCD$ is a rectangle where $AC$ is a diagonal. The value of $(\tan^2 \angle CAD + 1) \sin^2 \angle BAC$ is

If $\tan (A - B) = 1$ and $\sec (A + B) = \frac{2}{\sqrt{3}}$,then the smallest positive value of $B$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module