The value of sin^2 frac{pi}{8} + sin^2 frac{3 | vedclass

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

Question

(B) We know that $\sin(\pi - 	heta) = \sin 	heta$.Therefore,$\sin \frac{7\pi}{8} = \sin(\pi - \frac{\pi}{8}) = \sin \frac{\pi}{8}$ and $\sin \frac{5

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) We know that $\sin(\pi - 	heta) = \sin 	heta$.Therefore,$\sin \frac{7\pi}{8} = \sin(\pi - \frac{\pi}{8}) = \sin \frac{\pi}{8}$ and $\sin \frac{5\pi}{8} = \sin(\pi - \frac{3\pi}{8}) = \sin \frac{3\pi}{8}$.The given expression becomes: $2 \sin^2 \frac{\pi}{8} + 2 \sin^2 \frac{3\pi}{8} = 2 [\sin^2 \frac{\pi}{8} + \sin^2 \frac{3\pi}{8}]$.Since $\sin \frac{3\pi}{8} = \sin(\frac{\pi}{2} - \frac{\pi}{8}) = \cos \frac{\pi}{8}$,we have $\sin^2 \frac{3\pi}{8} = \cos^2 \frac{\pi}{8}$.Substituting this into the expression: $2 [\sin^2 \frac{\pi}{8} + \cos^2 \frac{\pi}{8}] = 2(1) = 2$.

The value of $\sin^2 \frac{\pi}{8} + \sin^2 \frac{3\pi}{8} + \sin^2 \frac{5\pi}{8} + \sin^2 \frac{7\pi}{8}$ is

Similar Questions

If the maximum value of $y = \frac{7 + 6 \tan x - \tan^2 x}{1 + \tan^2 x}$ is $\lambda$,then the value of $\log_{\sqrt{2}}(\lambda)$ is:

If $\tan (A + B) = p$ and $\tan (A - B) = q,$ then the value of $\tan 2A$ in terms of $p$ and $q$ is

If $3 \sin \theta + 5 \cos \theta = 5$,then the value of $5 \sin \theta - 3 \cos \theta$ will be

If $\sin A = \sin B$ and $\cos A = \cos B,$ then

If $\tan x + \tan \left( \frac{\pi }{3} + x \right) + \tan \left( \frac{2\pi }{3} + x \right) = 3,$ then

Vedclass Test Series

Exam Paper Generator

Online Exam Module