sin^4 frac{pi}{8} + sin^4 frac{3pi}{8} + sin^ | vedclass

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

Question

(C) Given expression: $S = \sin^4 \frac{\pi}{8} + \sin^4 \frac{3\pi}{8} + \sin^4 \frac{5\pi}{8} + \sin^4 \frac{7\pi}{8}$.Since $\sin(\pi - 	heta) = \

Vedclass · Accepted Answer

$\frac{3}{2}$

Vedclass · Answer

(C) Given expression: $S = \sin^4 \frac{\pi}{8} + \sin^4 \frac{3\pi}{8} + \sin^4 \frac{5\pi}{8} + \sin^4 \frac{7\pi}{8}$.Since $\sin(\pi - 	heta) = \sin 	heta$,we have $\sin \frac{7\pi}{8} = \sin \frac{\pi}{8}$ and $\sin \frac{5\pi}{8} = \sin \frac{3\pi}{8}$.Thus,$S = 2 \left( \sin^4 \frac{\pi}{8} + \sin^4 \frac{3\pi}{8} ight)$.Using the identity $2 \sin^2 	heta = 1 - \cos 2	heta$,we have $\sin^2 	heta = \frac{1 - \cos 2	heta}{2}$.So,$\sin^4 	heta = \left( \frac{1 - \cos 2	heta}{2} ight)^2 = \frac{1}{4} (1 - 2 \cos 2	heta + \cos^2 2	heta)$.$S = 2 \left[ \frac{1}{4} (1 - 2 \cos \frac{\pi}{4} + \cos^2 \frac{\pi}{4}) + \frac{1}{4} (1 - 2 \cos \frac{3\pi}{4} + \cos^2 \frac{3\pi}{4}) ight]$.$S = \frac{1}{2} \left[ (1 - 2(\frac{1}{\sqrt{2}}) + \frac{1}{2}) + (1 - 2(-\frac{1}{\sqrt{2}}) + \frac{1}{2}) ight]$.$S = \frac{1}{2} \left[ (1 - \sqrt{2} + 0.5) + (1 + \sqrt{2} + 0.5) ight] = \frac{1}{2} [1.5 + 1.5] = \frac{1}{2} (3) = \frac{3}{2}$.

$\sin^4 \frac{\pi}{8} + \sin^4 \frac{3\pi}{8} + \sin^4 \frac{5\pi}{8} + \sin^4 \frac{7\pi}{8} = $

Similar Questions

If $\tan A = \frac{1}{2}$ and $\tan B = \frac{1}{3},$ the value of $A + B$ is

The value of $\sin \frac{\pi }{14} \sin \frac{3\pi }{14} \sin \frac{5\pi }{14} \sin \frac{7\pi }{14} \sin \frac{9\pi }{14} \sin \frac{11\pi }{14} \sin \frac{13\pi }{14}$ is equal to

$\sin^6 \theta + \cos^6 \theta + 3 \sin^2 \theta \cos^2 \theta = $

Let $A, B$ and $C$ be the angles of a plane triangle and $\tan \frac{A}{2} = \frac{1}{3}, \tan \frac{B}{2} = \frac{2}{3}$. Then $\tan \frac{C}{2}$ is equal to

If $\tan \alpha = n \tan \beta$ and $\sin \alpha = m \sin \beta,$ then $\frac{m^{2}-1}{n^{2}-1} =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module