The value of sin frac{5 pi}{12} sin frac{pi}{ | vedclass

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

Question

(D) Given expression is $\sin \frac{5 \pi}{12} \sin \frac{\pi}{12}$.We use the trigonometric identity $2 \sin A \sin B = \cos(A - B) - \cos(A + B)$.Mu

Vedclass · Accepted Answer

$\frac{1}{4}$

Vedclass · Answer

(D) Given expression is $\sin \frac{5 \pi}{12} \sin \frac{\pi}{12}$.We use the trigonometric identity $2 \sin A \sin B = \cos(A - B) - \cos(A + B)$.Multiplying and dividing by $2$,we get:$\frac{1}{2} [2 \sin \frac{5 \pi}{12} \sin \frac{\pi}{12}]$$= \frac{1}{2} [\cos(\frac{5 \pi}{12} - \frac{\pi}{12}) - \cos(\frac{5 \pi}{12} + \frac{\pi}{12})]$$= \frac{1}{2} [\cos(\frac{4 \pi}{12}) - \cos(\frac{6 \pi}{12})]$$= \frac{1}{2} [\cos(\frac{\pi}{3}) - \cos(\frac{\pi}{2})]$Since $\cos(\frac{\pi}{3}) = \frac{1}{2}$ and $\cos(\frac{\pi}{2}) = 0$,$= \frac{1}{2} [\frac{1}{2} - 0] = \frac{1}{4}$.

The value of $\sin \frac{5 \pi}{12} \sin \frac{\pi}{12}$ is

Similar Questions

$\frac{\tan 52^{\circ} - \tan 38^{\circ}}{\tan 14^{\circ}} = $

If $\sin A = \frac{4}{5}$ and $\cos B = -\frac{12}{13}$,where $A$ and $B$ lie in the first and third quadrant respectively,then $\cos(A + B) = $

If $\sin (A+B) \sin (A-B)+\cos (A+B) \cos (A-B)=\frac{1}{2}$ and $0 < B < \frac{\pi}{2}$,then $B=$

If $(1+\tan \alpha)(1+\tan 4 \alpha)=2$ and $\alpha \in \left(0, \frac{\pi}{16}\right)$,then $\alpha$ is equal to

$\sin \left(\frac{\pi}{3}+x\right)-\cos \left(\frac{\pi}{6}+x\right) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module