sin left( frac{pi}{10} right) sin left( frac{ | vedclass

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

Question

(C) We know that $\frac{\pi}{10} = 18^\circ$ and $\frac{3\pi}{10} = 54^\circ$.Thus,the expression is $\sin 18^\circ \sin 54^\circ$.We know that $\sin

Vedclass · Accepted Answer

$1/4$

Vedclass · Answer

(C) We know that $\frac{\pi}{10} = 18^\circ$ and $\frac{3\pi}{10} = 54^\circ$.Thus,the expression is $\sin 18^\circ \sin 54^\circ$.We know that $\sin 54^\circ = \cos(90^\circ - 54^\circ) = \cos 36^\circ$.Using the standard values: $\sin 18^\circ = \frac{\sqrt{5} - 1}{4}$ and $\cos 36^\circ = \frac{\sqrt{5} + 1}{4}$.Multiplying these values: $\left( \frac{\sqrt{5} - 1}{4} \right) \left( \frac{\sqrt{5} + 1}{4} \right) = \frac{(\sqrt{5})^2 - 1^2}{16} = \frac{5 - 1}{16} = \frac{4}{16} = \frac{1}{4}$.

$\sin \left( \frac{\pi}{10} \right) \sin \left( \frac{3\pi}{10} \right) = $

Similar Questions

If $A = \cos^2 \theta + \sin^4 \theta$,then for all values of $\theta$:

Let $A$ and $B$ denote the statements:
$A: \cos \alpha + \cos \beta + \cos \gamma = 0$
$B: \sin \alpha + \sin \beta + \sin \gamma = 0$
If $\cos (\alpha - \beta) + \cos (\beta - \gamma) + \cos (\gamma - \alpha) = -\frac{3}{2}$,then:

$\cot 5^o - \tan 5^o - 2 \tan 10^o - 4 \tan 20^o - 8 \cot 40^o$ is equal to

The equation $\sec^2 \theta = \frac{4xy}{(x + y)^2}$ is only possible when

If $A + B + C = \pi$ and $\cos A = \cos B \cos C$,then $\tan B \tan C$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module