Question

Show that

${\sin }^{-1}\frac{4}{5}+{\sin }^{-1}\frac{5}{13}+{\sin }^{-1}\frac{16}{65}=\frac{\pi }{2}$

Answer 1

${\sin }^{-1}\frac{4}{5}+{\sin }^{-1}\frac{5}{13}+{\sin }^{-1}\frac{16}{65}=\frac{\pi }{2}$

${\sin }^{-1}\frac{4}{5}+{\sin }^{-1}\frac{5}{13}=\frac{\pi }{2}-{\sin }^{-1}\frac{16}{65}$

${\sin }^{-1}\frac{4}{5}+{\sin }^{-1}\frac{5}{13}={\cos }^{-1}\frac{16}{65}$

${\sin }^{-1}\frac{4}{5}=A\Rightarrow \sin A=\frac{4}{5}\Rightarrow \cos A=\frac{3}{5}$

${\sin }^{-1}\frac{5}{13}=B\Rightarrow \sin B=\frac{5}{13}\Rightarrow \cos B=\frac{12}{13}$

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

$=\frac{3}{5}\times \frac{12}{13}-\frac{4}{5}\times \frac{5}{13}$

$=\frac{36}{65}-\frac{20}{65}=\frac{16}{65}$

$\cos (A+B)=\frac{16}{65}$

$A+B={\cos }^{-1}\frac{16}{65}$

${\sin }^{-1}\frac{4}{5}+{\sin }^{-1}\frac{5}{13}={\cos }^{-1}\frac{16}{65}$

${\sin }^{-1}\frac{4}{5}+{\sin }^{-1}\frac{5}{13}=\frac{\pi }{2}-{\sin }^{-1}\frac{16}{65}$

$\therefore {\sin }^{-1}\frac{4}{5}+{\sin }^{-1}\frac{5}{13}+{\sin }^{-1}\frac{16}{65}=\frac{\pi }{2}$