Question

Evaluate the given expression:

$\cos ({\sin }^{-1}\frac{3}{5}+{\sin }^{-1}\frac{5}{13})$

Answer 1

$\cos ({\sin }^{-1}\frac{3}{5}+{\sin }^{-1}\frac{5}{13})$

Let ${\sin }^{-1}\frac{3}{5}=a,{\sin }^{-1}\frac{5}{13}=b$

$\Rightarrow \sin a=\frac{3}{5},\sin b=\frac{5}{13}$

$\Rightarrow {\sin }^{2}a=\frac{9}{25},{\sin }^{2}b=\frac{25}{169}$

$\Rightarrow {\cos }^{2}a=1-{\sin }^{2}a=1-\frac{9}{25}=\frac{25-9}{25}=\frac{16}{25},{\cos }^{2}b=1-{\sin }^{2}b=1-\frac{25}{169}=\frac{169-25}{169}=\frac{144}{169}$

$\Rightarrow \cos a=\frac{4}{5},\cos b=\frac{12}{13}$

Now,$\cos ({\sin }^{-1}\frac{3}{5}+{\sin }^{-1}\frac{5}{13})$

$=\cos (a+b)$

$=\cos a\cos b-\sin a\sin b$

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

$=\frac{48}{65}-\frac{15}{65}$

$=\frac{48-15}{65}=\frac{33}{65}$