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 $\cos^6 \alpha + \sin^6 \alpha + K \sin^2 2\alpha = 1,$ then $K =$

$\tan 15^\circ = $

$2\cos x - \cos 3x - \cos 5x = $

Maximum value of $5 \cos \theta + 3 \cos \left( \theta + \frac{\pi}{3} \right) - 1$ is

$\tan \left( \frac{\pi }{4} + \theta \right) - \tan \left( \frac{\pi }{4} - \theta \right) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module