Question

Prove that:
${\sin }^{-1}\frac{3}{5}+{\sin }^{-1}\frac{8}{17}={\sin }^{-1}\frac{77}{85}$

Answer 1

Let ${\sin }^{-1}\frac{3}{5}=A$
$\therefore \sin A=\frac{3}{5}$
$\cos A=\sqrt{1-{\sin }^{2}A}$
$=\sqrt{1-{\left(\frac{3}{5}\right)}^2}=\sqrt{\frac{16}{25}}=\frac{4}{5}$
Let ${\sin }^{-1}\frac{8}{17}=B$
$\therefore \sin B=\frac{8}{17}$
$\cos B=\sqrt{1-{\sin }^{2}B}=\sqrt{1-{\left(\frac{8}{17}\right)}^2}=\sqrt{\frac{225}{289}}=\frac{15}{17}$
we know that $\sin (A+B)=\sin A\cos B+\cos A\sin B$
$\sin (A+B)=\frac{3}{5}\times \frac{15}{17}+\frac{4}{5}\times \frac{8}{17}=\frac{77}{85}$
Putting the value of $A$ and $B$
${\sin }^{-1}\frac{3}{5}+{\sin }^{-1}\frac{8}{17}={\sin }^{-1}\frac{77}{85}$