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 use the identity $\sin 	heta \sin(60^\circ - 	heta) \sin(60^\circ + 	heta) = \frac{1}{4} \sin 3	heta$.Given expression: $E = \sin 20^\circ

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) We use the identity $\sin 	heta \sin(60^\circ - 	heta) \sin(60^\circ + 	heta) = \frac{1}{4} \sin 3	heta$.Given expression: $E = \sin 20^\circ \sin 40^\circ \sin 60^\circ \sin 80^\circ$.Substitute $\sin 60^\circ = \frac{\sqrt{3}}{2}$:$E = \frac{\sqrt{3}}{2} (\sin 20^\circ \sin 40^\circ \sin 80^\circ)$.Using the identity with $	heta = 20^\circ$:$\sin 20^\circ \sin(60^\circ - 20^\circ) \sin(60^\circ + 20^\circ) = \sin 20^\circ \sin 40^\circ \sin 80^\circ = \frac{1}{4} \sin(3 	imes 20^\circ) = \frac{1}{4} \sin 60^\circ$.Since $\sin 60^\circ = \frac{\sqrt{3}}{2}$,we have:$\sin 20^\circ \sin 40^\circ \sin 80^\circ = \frac{1}{4} 	imes \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{8}$.Now substitute this back into the expression for $E$:$E = \frac{\sqrt{3}}{2} 	imes \frac{\sqrt{3}}{8} = \frac{3}{16}$.

$\sin 20^\circ \sin 40^\circ \sin 60^\circ \sin 80^\circ = $ (in $/16$)

Similar Questions

The value of $\cos^2 10^\circ - \cos 10^\circ \cos 50^\circ + \cos^2 50^\circ$ is

From an aeroplane flying vertically over a straight horizontal road,the angles of depression of two consecutive milestones on opposite sides of the aeroplane are observed to be $30^{\circ}$ and $60^{\circ}$. Find the height of the aeroplane above the road in miles.

$\tan 3A - \tan 2A - \tan A = $

$2\sin A{\cos ^3}A - 2{\sin ^3}A\cos A = $

If $a \tan \theta = b$,then $a \cos 2\theta + b \sin 2\theta = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module