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

Then $\sin x=\frac{3}{5},$ where $0<x<\frac{\pi }{2}$

$\cos y=\frac{12}{13}$, where $0<y<\frac{\pi }{2}$

and $\sin z=\frac{56}{65},$ where $0<z<\frac{\pi }{2}$

$\therefore \cos x>0,\sin y>0$

Now, $\cos x=\sqrt{1-{\sin }^{2}x}$

$=\sqrt{1-\frac{9}{25}}$

$=\sqrt{\frac{16}{25}}=\frac{4}{5}$

and $\sin y=\sqrt{1-co{s}^{2}y}$

$=\sqrt{1-\frac{144}{169}}$

$\sqrt{\frac{25}{169}}=\frac{5}{13}$

We have to prove, that, $x+y=z$

Now, $\sin (x+y)=\sin x\cos y+\cos x\sin y$

$=\left(\frac{3}{5}\right)\left(\frac{12}{13}\right)+\left(\frac{4}{5}\right)\left(\frac{5}{13}\right)$

$=\frac{36}{65}+\frac{20}{65}=\frac{56}{65}$

$\therefore \sin (x+y)=\sin z$

$\therefore x+y=z$

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