The maximum value of cos^2 left( frac{pi}{3} | vedclass

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

Question

(C) Given expression: $f(x) = \cos^2 \left( \frac{\pi}{3} - x \right) - \cos^2 \left( \frac{\pi}{3} + x \right)$.Using the identity $\cos^2 A - \cos^2

Vedclass · Accepted Answer

$\frac{\sqrt{3}}{2}$

Vedclass · Answer

(C) Given expression: $f(x) = \cos^2 \left( \frac{\pi}{3} - x ight) - \cos^2 \left( \frac{\pi}{3} + x ight)$.Using the identity $\cos^2 A - \cos^2 B = \sin(B - A) \sin(B + A)$:Let $A = \frac{\pi}{3} - x$ and $B = \frac{\pi}{3} + x$.Then $B - A = (\frac{\pi}{3} + x) - (\frac{\pi}{3} - x) = 2x$.And $B + A = (\frac{\pi}{3} + x) + (\frac{\pi}{3} - x) = \frac{2\pi}{3}$.Thus,$f(x) = \sin(2x) \sin\left( \frac{2\pi}{3} ight)$.Since $\sin\left( \frac{2\pi}{3} ight) = \frac{\sqrt{3}}{2}$,we have $f(x) = \frac{\sqrt{3}}{2} \sin(2x)$.The maximum value of $\sin(2x)$ is $1$.Therefore,the maximum value of the expression is $\frac{\sqrt{3}}{2} 	imes 1 = \frac{\sqrt{3}}{2}$.

The maximum value of $\cos^2 \left( \frac{\pi}{3} - x \right) - \cos^2 \left( \frac{\pi}{3} + x \right)$ is

Similar Questions

If the sides of a right-angled triangle are $\{cos2\alpha + cos2\beta + 2cos(\alpha + \beta )\}$ and $\{sin2\alpha + sin2\beta + 2sin(\alpha + \beta )\}$,then the length of the hypotenuse is:

If $\frac{3\pi}{4} < \alpha < \pi,$ then $\sqrt{\csc^2 \alpha + 2\cot \alpha}$ is equal to

If $\pi < \alpha < \frac{3\pi}{2}$,then $\sqrt{\frac{1 - \cos \alpha}{1 + \cos \alpha}} + \sqrt{\frac{1 + \cos \alpha}{1 - \cos \alpha}} = $

The length of an arc which subtends an angle $18^{\circ}$ at the centre of the circle of radius $6 \text{ cm}$ is (in $\text{cm}$)

$A$ ladder having a length of $16\,m$ is leaning against a wall at an angle of $60^{\circ}$ with the ground. Find the distance between the wall and the foot of the ladder (in $m$).

Vedclass Test Series

Exam Paper Generator

Online Exam Module