Question

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

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

Answer 1

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

$\tan \alpha =\frac{1}{7}$, $\sin \beta =\frac{1}{\sqrt{10}}$

From these we get,

$\sin \alpha =\frac{1}{\sqrt{50}},\cos \alpha =\frac{7}{\sqrt{50}}$, $\cos \beta =\frac{3}{\sqrt{10}}$

Using formula,

$\sin (A+B)=\sin A\cos B+\cos A\sin B$

$\sin (\alpha +2\beta )=\sin \alpha \cos 2\beta +\cos \alpha \sin 2\beta$

$=\sin \alpha (2{\cos }^2\beta -1)+\cos \alpha .2.\sin \beta \cos \beta$

$=\frac{1}{\sqrt{50}}(2.\frac{9}{10}-1)+\frac{7}{\sqrt{50}}.2.\frac{3}{\sqrt{10}}.\frac{1}{\sqrt{10}}$

$=\frac{1}{\sqrt{2}}$

$\alpha +2\beta ={\sin }^{-1}\frac{1}{\sqrt{2}}={45}^0$