Question

Prove that: ${\sin }^{-1}\left(\frac{3}{5}\right)+{\cos }^{-1}\left(\frac{12}{13}\right)={\sin }^{-1}\left(\frac{56}{65}\right)$.

Answer 1

Let ${\cos }^{-1}\left(\frac{12}{13}\right)=x$

$\Rightarrow \frac{12}{13}=\cos x$

$\therefore \sin x=\frac{5}{13}$

Let ${\sin }^{-1}\left(\frac{3}{5}\right)=y$

$\Rightarrow \frac{3}{5}=\sin y$

$\therefore \cos y=\frac{4}{5}$

We know that, $\sin (x+y)=\sin x\cos y+\cos x\sin y$

$\sin (x+y)=\frac{5}{13}\times \frac{4}{5}+\frac{12}{13}\times \frac{3}{5}$

$\sin (x+y)=\frac{56}{65}$

$\therefore x+y={\sin }^{-1}\left(\frac{56}{65}\right)$

${\cos }^{-1}\left(\frac{12}{13}\right)+{\sin }^{-1}\left(\frac{3}{5}\right)={\sin }^{-1}\left(\frac{56}{65}\right)$

Hence proved.