sin 12^circ sin 48^circ sin 54^circ = | vedclass

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

Question

(C) We know that $\sin 54^\circ = \cos 36^\circ = \frac{\sqrt{5}+1}{4}$.Consider the expression $E = \sin 12^\circ \sin 48^\circ \sin 54^\circ$.Using

Vedclass · Accepted Answer

$1/8$

Vedclass · Answer

(C) We know that $\sin 54^\circ = \cos 36^\circ = \frac{\sqrt{5}+1}{4}$.Consider the expression $E = \sin 12^\circ \sin 48^\circ \sin 54^\circ$.Using the formula $2 \sin A \sin B = \cos(A-B) - \cos(A+B)$:$E = \frac{1}{2} [\cos(48^\circ - 12^\circ) - \cos(48^\circ + 12^\circ)] \sin 54^\circ$$E = \frac{1}{2} [\cos 36^\circ - \cos 60^\circ] \sin 54^\circ$Since $\sin 54^\circ = \cos 36^\circ$ and $\cos 60^\circ = 1/2$:$E = \frac{1}{2} [\cos 36^\circ - 1/2] \cos 36^\circ$$E = \frac{1}{2} [\cos^2 36^\circ - \frac{1}{2} \cos 36^\circ]$Substituting $\cos 36^\circ = \frac{\sqrt{5}+1}{4}$:$E = \frac{1}{2} [(\frac{\sqrt{5}+1}{4})^2 - \frac{1}{2} (\frac{\sqrt{5}+1}{4})]$$E = \frac{1}{2} [\frac{5+1+2\sqrt{5}}{16} - \frac{\sqrt{5}+1}{8}]$$E = \frac{1}{2} [\frac{6+2\sqrt{5}}{16} - \frac{2\sqrt{5}+2}{16}]$$E = \frac{1}{2} [\frac{4}{16}] = \frac{1}{8}$.

$\sin 12^\circ \sin 48^\circ \sin 54^\circ = $

Similar Questions

If $\tan(\pi \sin \theta) = \cot(\pi \cos \theta)$,then $\left| \cot \left( \theta - \frac{\pi}{4} \right) \right|$ is -

If $\tan A - \tan B = x$ and $\cot A - \cot B = y$,then $\cot (A - B) =$

If $f_n(x) = \frac{1}{2n} [\sin^{2n} x + \cos^{2n} x]$,then $f_1(x) + f_2(x) - f_3(x) =$

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

The value of $\sin^2 5^{\circ} + \sin^2 10^{\circ} + \sin^2 15^{\circ} + \ldots + \sin^2 85^{\circ} + \sin^2 90^{\circ} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module