If theta+phi=frac{2 pi}{3} and cos theta=frac | vedclass

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

Question

(D) Given that $	heta+\phi=\frac{2 \pi}{3}$ and $\cos 	heta=\frac{\sqrt{3}}{2}.$Since $\cos 	heta = \frac{\sqrt{3}}{2},$ we know that $	heta = \fr

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(D) Given that $	heta+\phi=\frac{2 \pi}{3}$ and $\cos 	heta=\frac{\sqrt{3}}{2}.$Since $\cos 	heta = \frac{\sqrt{3}}{2},$ we know that $	heta = \frac{\pi}{6}$ (or $30^{\circ}$).Substituting the value of $	heta$ into the first equation:$\frac{\pi}{6} + \phi = \frac{2 \pi}{3}$$\phi = \frac{2 \pi}{3} - \frac{\pi}{6}$$\phi = \frac{4 \pi - \pi}{6} = \frac{3 \pi}{6} = \frac{\pi}{2}$Now,we need to find $\sin \phi$:$\sin \phi = \sin \left(\frac{\pi}{2}ight) = 1.$

If $\theta+\phi=\frac{2 \pi}{3}$ and $\cos \theta=\frac{\sqrt{3}}{2},$ what is the value of $\sin \phi ?$

Similar Questions

Two towers $A$ and $B$ have heights $45 \, m$ and $15 \, m$ respectively. The angle of elevation from the bottom of tower $B$ to the top of tower $A$ is $60^{\circ}$. If the angle of elevation from the bottom of tower $A$ to the top of tower $B$ is $\theta$,then the value of $\sin \theta$ is:

If $\sqrt{2} \tan 2 \theta = \sqrt{6}$ and $0^{\circ} < \theta < 45^{\circ},$ then the value of $\sin \theta + \sqrt{3} \cos \theta - 2 \tan^{2} \theta$ is

The expression $\frac{\tan A}{1 - \cot A} + \frac{\cot A}{1 - \tan A}$ can be written as:

The value of $\sec ^{2} \theta - \frac{\sin ^{2} \theta - 2 \sin ^{4} \theta}{2 \cos ^{4} \theta - \cos ^{2} \theta}$ is

$\sin^6 \theta + \cos^6 \theta + 3 \sin^2 \theta \cos^2 \theta = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module