The value of sin left(frac{5 pi}{24}right) cd | vedclass

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

Question

(A) Using the formula $2 \sin A \cos B = \sin(A+B) + \sin(A-B)$,we have: $\sin \left(\frac{5 \pi}{24}\right) \cos \left(\frac{\pi}{24}\right) = \frac{

Vedclass · Accepted Answer

$\frac{1+\sqrt{2}}{4}$

Vedclass · Answer

(A) Using the formula $2 \sin A \cos B = \sin(A+B) + \sin(A-B)$,we have: $\sin \left(\frac{5 \pi}{24}\right) \cos \left(\frac{\pi}{24}\right) = \frac{1}{2} \left[ \sin \left(\frac{5 \pi}{24} + \frac{\pi}{24}\right) + \sin \left(\frac{5 \pi}{24} - \frac{\pi}{24}\right) \right]$ $= \frac{1}{2} \left[ \sin \left(\frac{6 \pi}{24}\right) + \sin \left(\frac{4 \pi}{24}\right) \right]$ $= \frac{1}{2} \left[ \sin \left(\frac{\pi}{4}\right) + \sin \left(\frac{\pi}{6}\right) \right]$ $= \frac{1}{2} \left[ \frac{1}{\sqrt{2}} + \frac{1}{2} \right] = \frac{1}{2} \left[ \frac{\sqrt{2}+1}{2} \right] = \frac{\sqrt{2}+1}{4}$

The value of $\sin \left(\frac{5 \pi}{24}\right) \cdot \cos \left(\frac{\pi}{24}\right)$ is

Similar Questions

$\cos ^4 \frac{\pi}{24} - \sin ^4 \frac{\pi}{24} = $

$\cos 20^{\circ} + \cos 30^{\circ} + \cos 40^{\circ} = $

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

If $A+B=\frac{\pi}{4}$,then $\frac{\cos B-\sin B}{\cos B+\sin B}=$

If $\sin \alpha = 1/\sqrt{5}$ and $\sin \beta = 3/5$,then $\beta - \alpha$ lies in the interval

Vedclass Test Series

Exam Paper Generator

Online Exam Module