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

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(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

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

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(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:

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