Question

If $0<\alpha ,\beta <\frac{\pi }{4},\cos (\alpha +\beta )=\frac{4}{5},\sin (\alpha -\beta )=\frac{5}{13}$, then $\tan 2\alpha =$

• A. $\frac{33}{56}$
• B. $\frac{56}{33}$
• C. $\frac{16}{33}$
• D. none

Answer 1

The correct option is B $\frac{56}{33}$

$\cos (\alpha +\beta )=\frac{4}{5},\sin (\alpha -\beta )=\frac{5}{13}$

$\sin (\alpha +\beta )=\sqrt{1-{\cos }^2(\alpha +\beta )}=\sqrt{1-{\left(\frac{4}{5}\right)}^2}=\frac{3}{5}$

$\cos (\alpha -\beta )=\sqrt{1-{\sin }^2(\alpha -\beta )}=\sqrt{1-{\left(\frac{5}{13}\right)}^2}=\frac{12}{13}$

$\sin \left(2\alpha \right)=\sin (\alpha +\beta +\alpha -\beta )=\sin (\alpha +\beta )\cos (\alpha -\beta )+\sin (\alpha -\beta )\cos (\alpha +\beta )$

$=\frac{3}{5}\times \frac{12}{13}+\frac{5\times 4}{5\times 13}=\frac{56}{65}$

$\cos \left(2\alpha \right)=\cos (\alpha +\beta +\alpha -\beta )=\cos (\alpha +\beta )\cos (\alpha -\beta )-\sin (\alpha -\beta )\sin (\alpha -\beta )$

$=\frac{4\times 12}{5\times 13}-\frac{5\times 13}{13\times 5}=\frac{33}{65}$

$\tan 2\alpha =\frac{\sin 2\alpha }{\cos 2\alpha }=\frac{56}{65}\times \frac{65}{33}=\frac{56}{33}$