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

$A$ circular wire of diameter $10\,cm$ is cut and placed along the circumference of a circle of diameter $1\,m$. The angle subtended by the wire at the centre of the circle is equal to

The value of $\frac{\cot 54^\circ}{\tan 36^\circ} + \frac{\tan 20^\circ}{\cot 70^\circ}$ is

If $\sin \theta = -\frac{7}{25}$ and $\theta$ is in the third quadrant,then $\frac{7 \cot \theta - 24 \tan \theta}{7 \cot \theta + 24 \tan \theta} = $

If $x=a(\sin \theta+\cos \theta)$ and $y=b(\sin \theta-\cos \theta)$,then the value of $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}$ is

The simplest value of $\tan 1^{\circ} \tan 2^{\circ} \tan 3^{\circ} \ldots \tan 89^{\circ}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module