Question

Prove that
${\cos }^{-1}\frac{12}{13}+{\sin }^{-1}\frac{3}{5}={\sin }^{-1}\frac{56}{65}$

Answer 1

Let $a={\cos }^{-1}\frac{12}{13}$ or $\cos a=\frac{12}{13}$
$\Rightarrow \sin a=\sqrt{1-{\cos }^{2}a}=\sqrt{1-\frac{144}{169}}=\sqrt{\frac{25}{169}}=\frac{5}{13}$

Let $b={\sin }^{-1}\frac{3}{5}$ or $\sin b=\frac{3}{5}$
$\Rightarrow \cos b=\sqrt{1-{\sin }^{2}b}=\sqrt{1-\frac{9}{25}}=\sqrt{\frac{16}{25}}=\frac{4}{5}$

Now, we know that
$\sin (a+b)=\sin a\cos b+\cos a\sin b$
$\sin (a+b)=\frac{5}{13}\times \frac{4}{5}+\frac{12}{13}\times \frac{3}{5}$
$=\frac{20}{65}+\frac{36}{65}=\frac{56}{65}$
$\therefore \sin (a+b)=\frac{56}{65}$
$\Rightarrow (a+b)={\sin }^{-1}\left(\frac{56}{65}\right)$
$\Rightarrow {\cos }^{-1}\frac{12}{13}+{\sin }^{-1}\frac{3}{5}={\sin }^{-1}\left(\frac{56}{65}\right)$
Hence proved.