Question

If $\alpha$ and $\beta$ are angles in the first quadrant, and $\tan \alpha =\frac{1}{7},\sin \beta =\frac{1}{\sqrt{10}},$ then using the formula $\sin (A+B)=\sin A\cos B+\cos A\sin B,$ find the value of $(\alpha +2\beta )$.

• A. $0{}^{\circ }$
• B. $45{}^{\circ }$
• C. $60{}^{\circ }$
• D. $90{}^{\circ }$

Answer 1

The correct option is B $45{}^{\circ }$

we know that, $\sin 2\beta =\frac{2\tan \beta }{1+{\tan }^2\beta }\cos 2\beta =\frac{1-{\tan }^2\beta }{1+{\tan }^2\beta }$

also we know that, $1+{\tan }^2\alpha =se{c}^2\alpha \Rightarrow 1+(\frac{1}{7}{)}^2=se{c}^2\alpha \Rightarrow \frac{50}{49}\Rightarrow {\cos }^2\alpha =\frac{49}{50}$

$\Rightarrow \cos \alpha =\frac{7}{5\sqrt{2}}\sin \alpha =\sqrt{1-\frac{49}{50}}=\frac{1}{5\sqrt{2}}$

also, we know that, $\sin \beta =\frac{1}{\sqrt{10}}\Rightarrow \cos \beta =\sqrt{1-\frac{1}{10}}=\frac{3}{\sqrt{10}}$

$\therefore \tan \beta =\frac{\sin \beta }{\cos \beta }=\frac{}{}\frac{1}{3}$

Now, $\sin (\alpha +2\beta )=\sin \alpha \cos 2\beta +\cos \alpha \sin 2\beta =\frac{1}{5\sqrt{2}}\times \frac{1-{\tan }^2\beta }{1+{\tan }^2\beta }+\frac{7}{5\sqrt{2}}\times \frac{2\tan \beta }{1+{\tan }^2\beta }=\frac{1}{5\sqrt{2}}[\frac{1-\frac{1}{9}}{1+\frac{1}{9}}+7\times \frac{\frac{2}{3}}{1+\frac{1}{9}}]=\frac{1}{5\sqrt{2}}[\frac{8}{10}+7\times \frac{6}{10}]=\frac{1}{5\sqrt{2}}\left[\frac{50}{10}\right]=\frac{1}{5\sqrt{2}}\times 5=\frac{1}{\sqrt{2}}\Rightarrow \sin (\alpha +2\beta )=\frac{1}{\sqrt{2}}\Rightarrow (\alpha +2\beta )={45}^o$