Question

Sin inverse(4/5)+sin inverse( 5/13)+ sin inverse( 16/65) =π/2

Answer 1

To prove:

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

Solution:

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

$\Rightarrow \sin ^{-1}\left(\frac{4}{5}\right)+\sin ^{-1}\left(\frac{5}{13}\right)=\frac{\pi}{2}-\sin ^{-1}\left(\frac{16}{65}\right)$

$\Rightarrow \sin ^{-1}\left(\frac{4}{5}\right)+\sin ^{-1}\left(\frac{5}{13}\right)=\cos ^{-1}\left(\frac{16}{65}\right)$

We just prove $\bold{\sin ^{-1}\left(\frac{4}{5}\right)+\sin ^{-1}\left(\frac{5}{13}\right)=\cos ^{-1}\left(\frac{16}{65}\right)}$

We take left hand side $=\sin ^{-1}\left(\frac{4}{5}\right)+\sin ^{-1}\left(\frac{5}{13}\right)$

Let $\sin ^{-1}\left(\frac{4}{5}\right)=x \text { and } \sin ^{-1}\left(\frac{5}{13}\right)=y$

$\Rightarrow \sin x=\frac{4}{5} \text { and } \sin y=\frac{5}{13}$

Now, $\cos x=\sqrt{1-\sin ^{2} x}=\sqrt{1-\frac{16}{25}}=\frac{3}{5}$

Now, $\cos y=\sqrt{1-\sin ^{2} y}=\sqrt{1-\frac{25}{169}}=\frac{12}{13}$

Now, $\bold{\cos (x+y)=\cos x \cdot \cos y-\sin x \cdot \sin y}$

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

$\Rightarrow \cos (x+y)=\frac{36}{65}-\frac{20}{65}$

$\Rightarrow \cos (x+y)=\frac{16}{65}$

$\Rightarrow x+y=\cos ^{-1}\left(\frac{16}{65}\right)$

$\bold{\Rightarrow \sin ^{-1}\left(\frac{4}{5}\right)+\sin ^{-1}\left(\frac{5}{13}\right)=\cos ^{-1}\left(\frac{16}{65}\right)}$ = Right hand side.

Hence proved.