If theta and phi are angles in the 1^{st} qua | vedclass

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

Question

(D) Given,$	an 	heta = \frac{1}{7}$ and $\sin \phi = \frac{1}{\sqrt{10}}$.Since $	heta$ is in the $1^{st}$ quadrant,$\sin 	heta = \frac{1}{\sqrt{5

Vedclass · Accepted Answer

$	heta + 2\phi = 45^\circ$

Vedclass · Answer

(D) Given,$	an 	heta = \frac{1}{7}$ and $\sin \phi = \frac{1}{\sqrt{10}}$.Since $	heta$ is in the $1^{st}$ quadrant,$\sin 	heta = \frac{1}{\sqrt{50}}$ and $\cos 	heta = \frac{7}{\sqrt{50}}$.Since $\phi$ is in the $1^{st}$ quadrant,$\cos \phi = \sqrt{1 - \sin^2 \phi} = \sqrt{1 - \frac{1}{10}} = \frac{3}{\sqrt{10}}$.Now,$\cos 2\phi = 2\cos^2 \phi - 1 = 2(\frac{9}{10}) - 1 = \frac{18}{10} - 1 = \frac{8}{10} = \frac{4}{5}$.And $\sin 2\phi = 2\sin \phi \cos \phi = 2(\frac{1}{\sqrt{10}})(\frac{3}{\sqrt{10}}) = \frac{6}{10} = \frac{3}{5}$.Now,consider $\cos(	heta + 2\phi) = \cos 	heta \cos 2\phi - \sin 	heta \sin 2\phi$.$\cos(	heta + 2\phi) = (\frac{7}{\sqrt{50}})(\frac{8}{10}) - (\frac{1}{\sqrt{50}})(\frac{6}{10}) = \frac{56 - 6}{10\sqrt{50}} = \frac{50}{10(5\sqrt{2})} = \frac{5}{5\sqrt{2}} = \frac{1}{\sqrt{2}}$.Since $	heta$ and $\phi$ are in the $1^{st}$ quadrant,$	heta + 2\phi = 45^\circ$.

If $\theta$ and $\phi$ are angles in the $1^{st}$ quadrant such that $\tan \theta = 1/7$ and $\sin \phi = 1/\sqrt{10}$. Then:

Similar Questions

If $\sin \theta = \sin 15^{\circ} + \sin 45^{\circ}$,where $0^{\circ} < \theta < 180^{\circ}$,then $\theta =$ (in $^{\circ}$)

If $\tan A = -\frac{1}{2}$ and $\tan B = -\frac{1}{3},$ then $A + B = $

If $\cot \alpha = 1$ and $\sec \beta = -\frac{5}{3}$,where $\pi < \alpha < \frac{3\pi}{2}$ and $\frac{\pi}{2} < \beta < \pi$,then the value of $\tan(\alpha + \beta)$ and the quadrant in which $\alpha + \beta$ lies,respectively,are

If $\cos 43^{\circ} + \sin 43^{\circ} = k$, then find the value of $\cos 2^{\circ}$.

$\frac{\cos 10^o + \sin 10^o}{\cos 10^o - \sin 10^o} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module