If k = sin frac{pi}{18} cdot sin frac{5pi}{18 | vedclass

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

Question

(B) Given $k = \sin \frac{\pi}{18} \sin \frac{5\pi}{18} \sin \frac{7\pi}{18}$.Using the identity $\sin 	heta = \cos \left( \frac{\pi}{2} - 	heta i

Vedclass · Accepted Answer

$\frac{1}{8}$

Vedclass · Answer

(B) Given $k = \sin \frac{\pi}{18} \sin \frac{5\pi}{18} \sin \frac{7\pi}{18}$.Using the identity $\sin 	heta = \cos \left( \frac{\pi}{2} - 	heta ight)$:$k = \cos \left( \frac{\pi}{2} - \frac{\pi}{18} ight) \cos \left( \frac{\pi}{2} - \frac{5\pi}{18} ight) \cos \left( \frac{\pi}{2} - \frac{7\pi}{18} ight)$$k = \cos \frac{8\pi}{18} \cos \frac{4\pi}{18} \cos \frac{2\pi}{18} = \cos \frac{4\pi}{9} \cos \frac{2\pi}{9} \cos \frac{\pi}{9}$.Using the formula $\cos 	heta \cos 2	heta \cos 4	heta = \frac{\sin 8	heta}{8 \sin 	heta}$ where $	heta = \frac{\pi}{9}$:$k = \frac{\sin \left( 8 \cdot \frac{\pi}{9} ight)}{8 \sin \frac{\pi}{9}} = \frac{\sin \frac{8\pi}{9}}{8 \sin \frac{\pi}{9}}$.Since $\sin \frac{8\pi}{9} = \sin \left( \pi - \frac{\pi}{9} ight) = \sin \frac{\pi}{9}$:$k = \frac{\sin \frac{\pi}{9}}{8 \sin \frac{\pi}{9}} = \frac{1}{8}$.

If $k = \sin \frac{\pi}{18} \cdot \sin \frac{5\pi}{18} \cdot \sin \frac{7\pi}{18}$,then the numerical value of $k$ is

Similar Questions

If $\alpha, \beta, \gamma \in \left( 0, \frac{\pi}{2} \right)$,then $\frac{\sin(\alpha + \beta + \gamma)}{\sin \alpha + \sin \beta + \sin \gamma}$ is

Let $A_0 A_1 A_2 A_3 A_4 A_5$ be a regular hexagon inscribed in a circle of unit radius. Then the product of the lengths of the line segments $A_0 A_1$,$A_0 A_2$,and $A_0 A_4$ is

$\cos ^{2} 5^{\circ}+\cos ^{2} 10^{\circ}+\cos ^{2} 15^{\circ}+\cdots+\cos ^{2} 90^{\circ}=$

$\sin 75^\circ = $

If $\tan \beta = \cos \theta \tan \alpha ,$ then ${\tan ^2}\frac{\theta }{2} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module