If tan alpha = frac{1}{7} and sin beta = frac | vedclass

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Question

(A) Given $\sin \beta = \frac{1}{\sqrt{10}}$. Since $0 < \beta < \frac{\pi}{2}$,we have $\cos \beta = \sqrt{1 - \sin^2 \beta} = \sqrt{1 - \frac{1}{10}

Vedclass · Accepted Answer

$\frac{\pi}{4} - \alpha$

Vedclass · Answer

(A) Given $\sin \beta = \frac{1}{\sqrt{10}}$. Since $0 Thus,$	an \beta = \frac{\sin \beta}{\cos \beta} = \frac{1/\sqrt{10}}{3/\sqrt{10}} = \frac{1}{3}$.Now,calculate $	an 2\beta = \frac{2 	an \beta}{1 - 	an^2 \beta} = \frac{2(1/3)}{1 - (1/3)^2} = \frac{2/3}{1 - 1/9} = \frac{2/3}{8/9} = \frac{2}{3} 	imes \frac{9}{8} = \frac{3}{4}$.Next,consider $	an(\alpha + 2\beta) = \frac{	an \alpha + 	an 2\beta}{1 - 	an \alpha 	an 2\beta} = \frac{1/7 + 3/4}{1 - (1/7)(3/4)} = \frac{(4+21)/28}{(28-3)/28} = \frac{25}{25} = 1$.Since $0  0$).Thus,$0 Therefore,$2\beta = \frac{\pi}{4} - \alpha$.

If $\tan \alpha = \frac{1}{7}$ and $\sin \beta = \frac{1}{\sqrt{10}}$ where $0 < \alpha, \beta < \frac{\pi}{2}$,then $2\beta$ is equal to:

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