sin 36^circ sin 72^circ sin 108^circ sin 144^ | vedclass

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

Question

(D) We know that $\sin 108^\circ = \sin(180^\circ - 72^\circ) = \sin 72^\circ$ and $\sin 144^\circ = \sin(180^\circ - 36^\circ) = \sin 36^\circ$.So,th

Vedclass · Accepted Answer

$5/16$

Vedclass · Answer

(D) We know that $\sin 108^\circ = \sin(180^\circ - 72^\circ) = \sin 72^\circ$ and $\sin 144^\circ = \sin(180^\circ - 36^\circ) = \sin 36^\circ$.So,the expression becomes $\sin^2 36^\circ \sin^2 72^\circ$.Using the identity $\sin^2 	heta = \frac{1 - \cos 2	heta}{2}$,we have:$\sin^2 36^\circ \sin^2 72^\circ = \left(\frac{1 - \cos 72^\circ}{2}ight) \left(\frac{1 - \cos 144^\circ}{2}ight)$$= \frac{1}{4} (1 - \cos 72^\circ)(1 - \cos 144^\circ)$$= \frac{1}{4} (1 - \sin 18^\circ)(1 + \cos 36^\circ)$Substituting the values $\sin 18^\circ = \frac{\sqrt{5}-1}{4}$ and $\cos 36^\circ = \frac{\sqrt{5}+1}{4}$:$= \frac{1}{4} \left(1 - \frac{\sqrt{5}-1}{4}ight) \left(1 + \frac{\sqrt{5}+1}{4}ight)$$= \frac{1}{4} \left(\frac{5-\sqrt{5}}{4}ight) \left(\frac{5+\sqrt{5}}{4}ight)$$= \frac{1}{4} \left(\frac{25-5}{16}ight) = \frac{1}{4} 	imes \frac{20}{16} = \frac{5}{16}$.

$\sin 36^\circ \sin 72^\circ \sin 108^\circ \sin 144^\circ = $

Similar Questions

If $x$ and $y$ are acute angles such that $\cos x + \cos y = \frac{3}{2}$ and $\sin x + \sin y = \frac{3}{4}$,then $\sin(x + y)$ equals to

$\cos ^3\left(\frac{\pi}{8}\right) \cos \left(\frac{3 \pi}{8}\right)+\sin ^3\left(\frac{\pi}{8}\right) \sin \left(\frac{3 \pi}{8}\right) = $

The exact value of $\cos^2 73^\circ + \cos^2 47^\circ + (\cos 73^\circ \cdot \cos 47^\circ)$ is

The value of $2 \sin(\frac{\pi}{8}) \sin(\frac{2\pi}{8}) \sin(\frac{3\pi}{8}) \sin(\frac{5\pi}{8}) \sin(\frac{6\pi}{8}) \sin(\frac{7\pi}{8})$ is:

Factorize the expression: $\sec ^2 x+5 \tan x+5$

Vedclass Test Series

Exam Paper Generator

Online Exam Module