The value of cos^{3}left(frac{pi}{8}right) cd | vedclass

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

Question

(C) We know that $\cos\left(\frac{3\pi}{8}\right) = \sin\left(\frac{\pi}{2} - \frac{3\pi}{8}\right) = \sin\left(\frac{\pi}{8}\right)$ and $\sin\left(\

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) We know that $\cos\left(\frac{3\pi}{8}ight) = \sin\left(\frac{\pi}{2} - \frac{3\pi}{8}ight) = \sin\left(\frac{\pi}{8}ight)$ and $\sin\left(\frac{3\pi}{8}ight) = \cos\left(\frac{\pi}{2} - \frac{3\pi}{8}ight) = \cos\left(\frac{\pi}{8}ight)$.Substituting these into the expression:$= \cos^{3}\left(\frac{\pi}{8}ight) \cdot \sin\left(\frac{\pi}{8}ight) + \sin^{3}\left(\frac{\pi}{8}ight) \cdot \cos\left(\frac{\pi}{8}ight)$$= \sin\left(\frac{\pi}{8}ight) \cos\left(\frac{\pi}{8}ight) \left[ \cos^{2}\left(\frac{\pi}{8}ight) + \sin^{2}\left(\frac{\pi}{8}ight) ight]$Since $\sin^{2}	heta + \cos^{2}	heta = 1$,the expression simplifies to:$= \sin\left(\frac{\pi}{8}ight) \cos\left(\frac{\pi}{8}ight) \cdot 1$$= \frac{1}{2} \left( 2 \sin\left(\frac{\pi}{8}ight) \cos\left(\frac{\pi}{8}ight) ight)$$= \frac{1}{2} \sin\left(2 \cdot \frac{\pi}{8}ight) = \frac{1}{2} \sin\left(\frac{\pi}{4}ight)$$= \frac{1}{2} \cdot \frac{1}{\sqrt{2}} = \frac{1}{2\sqrt{2}}$.

The value of $\cos^{3}\left(\frac{\pi}{8}\right) \cdot \cos\left(\frac{3\pi}{8}\right) + \sin^{3}\left(\frac{\pi}{8}\right) \cdot \sin\left(\frac{3\pi}{8}\right)$ is:

Similar Questions

$2\sin^2 \beta + 4\cos(\alpha + \beta)\sin \alpha \sin \beta + \cos 2(\alpha + \beta) = $

The value of $\tan 20^\circ + 2\tan 50^\circ - \tan 70^\circ$ is equal to

If $\tan^2 \alpha \tan^2 \beta + \tan^2 \beta \tan^2 \gamma + \tan^2 \gamma \tan^2 \alpha + 2 \tan^2 \alpha \tan^2 \beta \tan^2 \gamma = 1$,then the value of $\sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma$ is

If $\operatorname{cosec} \theta + \cot \theta = p,$ then $\cos \theta =$

The value of $\frac{\tan 40^{\circ}+\tan 20^{\circ}}{1-\cot 70^{\circ} \cot 50^{\circ}}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module