Question

Prove that ${\sin }^{-1}\left(\frac{8}{17}\right)+{\sin }^{-1}\left(\frac{3}{5}\right)={\cos }^{-1}\left(\frac{36}{85}\right)$

Answer 1

Let $si{n}^{-1}\left(\frac{8}{17}\right)=\alpha$ and $si{n}^{-1}\left(\frac{3}{5}\right)=\beta$
$sin\alpha =\frac{8}{17}$ and $sin\beta =\frac{3}{5}$
$\Rightarrow cos\alpha =\sqrt{1-si{n}^2\alpha }$ and $cos\beta =\sqrt{1-si{n}^2\beta }$
$\Rightarrow cos\alpha =\sqrt{1-\frac{64}{289}}$ and $cos\beta =\sqrt{1-\frac{9}{25}}$
$\Rightarrow cos\alpha =\sqrt{\frac{289-64}{289}}$ and $cos\beta =\sqrt{\frac{25-9}{25}}$
$\Rightarrow cos\alpha =\sqrt{\frac{225}{289}}$ and $cos\beta =\sqrt{\frac{16}{25}}$
$\Rightarrow cos\alpha =\frac{15}{17}$ and $cos\beta =\frac{4}{5}$
Now, $cos(\alpha +\beta )=cos\alpha .cos\beta -sin\alpha .sin\beta$
$\Rightarrow cos(\alpha +\beta )=\frac{15}{17}\times \frac{4}{5}-\frac{8}{17}\times \frac{3}{5}$
$\Rightarrow cos(\alpha +\beta )=\frac{60}{85}-\frac{24}{85}$
$\Rightarrow cos(\alpha +\beta )=\frac{36}{85}$
$\Rightarrow \alpha +\beta =co{s}^{-1}\frac{36}{85}$
Therefore, $\Rightarrow si{n}^{-1}\frac{8}{17}+si{n}^{-1}\frac{3}{5}=co{s}^{-1}\frac{36}{85}$
Hence proved.