Question

Show that sin^-14/5 +sin^-1 5/13 +sin^-116/65=π/2

Answer 1

$$\begin{gathered} \mathrm{LHS}={\sin }^{-1}\frac{4}{5}+{\sin }^{-1}\frac{5}{13}+{\sin }^{-1}\frac{16}{65} \\ ={\sin }^{-1}\left[\frac{4}{5}\sqrt{1-{\left(\frac{5}{13}\right)}^2}+\frac{5}{13}\sqrt{1-{\left(\frac{4}{5}\right)}^2}\right]+{\sin }^{-1}\frac{16}{65}\left[{\sin }^{-1}x+{\sin }^{-1}y={\sin }^{-1}\left(x\sqrt{1-y^2}+y\sqrt{1-x^2}\right)\right] \\ ={\sin }^{-1}\left(\frac{4}{5}\times \frac{12}{13}+\frac{5}{13}\times \frac{3}{5}\right)+{\sin }^{-1}\frac{16}{65} \\ ={\sin }^{-1}\left(\frac{48}{65}+\frac{15}{65}\right)+{\sin }^{-1}\frac{16}{65} \\ ={\sin }^{-1}\left(\frac{63}{65}\right)+{\sin }^{-1}\frac{16}{65} \\ ={\sin }^{-1}\left(\frac{63}{65}\sqrt{1-{\left(\frac{16}{65}\right)}^2}+\frac{16}{65}\sqrt{1-{\left(\frac{63}{65}\right)}^2}\right) \\ ={\sin }^{-1}\left(\frac{63}{65}\times \frac{63}{65}+\frac{16}{65}\times \frac{16}{65}\right) \\ ={\sin }^{-1}\left(\frac{3969}{4225}+\frac{256}{4225}\right) \\ ={\sin }^{-1}\left(1\right) \\ =\frac{\mathrm{\pi }}{2}=\mathrm{RHS} \\ \mathrm{hence},\mathrm{proved}. \end{gathered}$$