Question

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

Answer 1

Let ${\sin }^{-1}\frac{4}{5}=a,{\sin }^{-1}\frac{5}{13}=b$ & ${\sin }^{-1}\frac{16}{65}=c$

$\sin a=\frac{4}{5}$

$\sin b=\frac{5}{13}$

$\sin c=\frac{16}{65}$

$a+b+c=n$(Suppose)

$\sin m=\sin (a+b+c)$

$=\sin (a+b)\cos c+\cos (a+b)\sin c$

$=(\sin a\cos b+\cos a\sin b)\cos c+\cos a\cos b\sin c-\sin a\sin b\sin c$

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

$=\frac{4}{5}\times \frac{12}{13}\times \frac{63}{65}+\frac{3}{5}\times \frac{5}{13}\times \frac{63}{65}+\frac{3}{5}\times \frac{12}{13}\times \frac{16}{65}-\frac{4}{5}\times \frac{5}{13}\times \frac{16}{65}$

$=\frac{3024+945+576-320}{5\times 13\times 65}$

$=\frac{4225}{4225}=1$

$n={\sin }^{-1}1=\frac{\pi }{2}$

Hence, ${\sin }^{-1}\frac{4}{5}+{\sin }^{-1}\frac{5}{13}+{\sin }^{-1}\frac{16}{65}=1\cdot \frac{\pi }{2}$

Hence, $n=1$