Question

If $0<\beta <\alpha <\frac{\pi }{4},\cos (\alpha +\beta )=\frac{3}{5}$ and $\cos (\alpha -\beta )=\frac{4}{5}$ then $\sin 2\alpha$ is equal to

• A. $1$
• B. $0$
• C. $2$
• D. none of these

Answer 1

The correct option is A $1$

$\cos (\alpha +\beta )=\frac{3}{5}\Rightarrow \sin (\alpha +\beta )=\frac{4}{5}$

$\cos (\alpha -\beta )=\frac{4}{5}\Rightarrow \sin (\alpha -\beta )=\frac{3}{5}$

$\frac{\sin (\alpha +\beta )+\sin (\alpha -\beta )}{\cos (\alpha +\beta )+\cos (\alpha -\beta )}=\frac{\frac{4}{5}+\frac{3}{5}}{\frac{3}{5}+\frac{4}{5}}$

$\Rightarrow \frac{2\sin \alpha \cos \beta }{2\cos \alpha \cos \beta }=1\Rightarrow \tan \alpha =1$

$\sin 2\alpha =\frac{2\tan \alpha }{1+{\tan }^2\alpha }=\frac{2}{1+1}=1$