The value of sin frac{pi}{16} sin frac{3pi}{1 | vedclass

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

Question

(B) Let $S = \sin \frac{\pi}{16} \sin \frac{3\pi}{16} \sin \frac{5\pi}{16} \sin \frac{7\pi}{16}$.Using $\sin 	heta = \cos(\frac{\pi}{2} - 	heta)$,we

Vedclass · Accepted Answer

$\frac{\sqrt{2}}{16}$

Vedclass · Answer

(B) Let $S = \sin \frac{\pi}{16} \sin \frac{3\pi}{16} \sin \frac{5\pi}{16} \sin \frac{7\pi}{16}$.Using $\sin 	heta = \cos(\frac{\pi}{2} - 	heta)$,we have $\sin \frac{7\pi}{16} = \cos \frac{\pi}{16}$ and $\sin \frac{5\pi}{16} = \cos \frac{3\pi}{16}$.So,$S = (\sin \frac{\pi}{16} \cos \frac{\pi}{16}) (\sin \frac{3\pi}{16} \cos \frac{3\pi}{16})$.Using $2 \sin 	heta \cos 	heta = \sin 2	heta$,we get $S = (\frac{1}{2} \sin \frac{\pi}{8}) (\frac{1}{2} \sin \frac{3\pi}{8}) = \frac{1}{4} \sin \frac{\pi}{8} \cos \frac{\pi}{8}$ (since $\sin \frac{3\pi}{8} = \cos \frac{\pi}{8}$).$S = \frac{1}{8} \sin \frac{\pi}{4} = \frac{1}{8} 	imes \frac{1}{\sqrt{2}} = \frac{1}{8\sqrt{2}} = \frac{\sqrt{2}}{16}$.

The value of $\sin \frac{\pi}{16} \sin \frac{3\pi}{16} \sin \frac{5\pi}{16} \sin \frac{7\pi}{16}$ is

Similar Questions

The value of the series $\cos 12^{\circ} + \cos 84^{\circ} + \cos 132^{\circ} + \cos 156^{\circ}$ is

If $\theta$ is eliminated from the equations $x = a \cos(\theta - \alpha)$ and $y = b \cos(\theta - \beta)$,then $\frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{2xy}{ab} \cos(\alpha - \beta)$ is equal to

$\cos ^2 76^{\circ}+\sin ^2 46^{\circ}+\sin 76^{\circ} \cos 46^{\circ} = $

If $\tan \theta_1 = k \cot \theta_2$,then $\frac{\cos (\theta_1 + \theta_2)}{\cos (\theta_1 - \theta_2)} = $

$\tan \frac{\pi}{5}+2 \tan \frac{2 \pi}{5}+4 \cot \frac{4 \pi}{5}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module