sin 20^circ sin 40^circ sin 60^circ sin 80^ci | vedclass

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

Question

(C) We know that $\sin 60^\circ = \frac{\sqrt{3}}{2}$.Expression $= \frac{\sqrt{3}}{2} \sin 20^\circ \sin 40^\circ \sin 80^\circ$.Using the identity $

Vedclass · Accepted Answer

$3/16$

Vedclass · Answer

(C) We know that $\sin 60^\circ = \frac{\sqrt{3}}{2}$.Expression $= \frac{\sqrt{3}}{2} \sin 20^\circ \sin 40^\circ \sin 80^\circ$.Using the identity $\sin 	heta \sin(60^\circ - 	heta) \sin(60^\circ + 	heta) = \frac{1}{4} \sin 3	heta$:Here,$	heta = 20^\circ$,so $\sin 20^\circ \sin 40^\circ \sin 80^\circ = \sin 20^\circ \sin(60^\circ - 20^\circ) \sin(60^\circ + 20^\circ) = \frac{1}{4} \sin(3 	imes 20^\circ) = \frac{1}{4} \sin 60^\circ = \frac{1}{4} 	imes \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{8}$.Therefore,the total expression $= \frac{\sqrt{3}}{2} 	imes \frac{\sqrt{3}}{8} = \frac{3}{16}$.

$\sin 20^\circ \sin 40^\circ \sin 60^\circ \sin 80^\circ = $

Similar Questions

If $\text{cosec} \theta = \frac{p + q}{p - q}$,then $\cot \left( \frac{\pi}{4} + \frac{\theta}{2} \right) = $

If $\frac{\tan 3A}{\tan A} = a$,then $\frac{\sin 3A}{\sin A}$ is equal to

If $\sin \alpha = \frac{336}{625}$ and $450^\circ < \alpha < 540^\circ$,then $\sin \left( \frac{\alpha}{4} \right) = $

If $33\theta = \pi$, then the value of $\cos \theta \cos 2\theta \cos 4\theta \cos 8\theta \cos 16\theta$ is

$\frac{\sin \theta + \sin 2\theta}{1 + \cos \theta + \cos 2\theta} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module