Question

What is ${\sin }^{-1}x+{\sin }^{-1}y$ formula?

Answer 1

Inverse trigonometric function formula:

The formula for ${\sin }^{-1}x+{\sin }^{-1}y$ can be derived as,

Let us consider ${\sin }^{-1}x=A$

So that, $\sin A=x$

$\Rightarrow \cos A=\sqrt{1-x^2}\left[{\sin }^{2}A+{\cos }^{2}A=1\right]$

Let us consider ${\sin }^{-1}y=B$

So that, $\sin B=y$

$\Rightarrow \cos B=\sqrt{1-y^2}\left[{\sin }^{2}B+{\cos }^{2}B=1\right]$

As we know,

$$\begin{gathered} \sin (A+B)=\sin A·\cos B+\sin B·\cos A \\ \Rightarrow \sin (A+B)=x·\sqrt{1-y^2}+y·\sqrt{1-x^2} \\ \Rightarrow A+B={\sin }^{-1}(x·\sqrt{1-y^2}+y·\sqrt{1-x^2}) \\ \Rightarrow {\sin }^{-1}x+{\sin }^{-1}y={\sin }^{-1}(x·\sqrt{1-y^2}+y·\sqrt{1-x^2})\left[\because A={\sin }^{-1}xandB={\sin }^{-1}y\right] \end{gathered}$$

Hence,

$$\begin{gathered} \mathbf{}{\mathbf{sin}}^{\mathbf{-}\mathbf{1}}\mathbf{}\mathit{x}\mathbf{+}{\mathbf{sin}}^{\mathbf{-}\mathbf{1}}\mathbf{}\mathit{y}\mathbf{=}{\mathbf{sin}}^{\mathbf{-}\mathbf{1}}\mathbf{}\mathbf{(}\mathit{x}\mathbf{·}\sqrt{\mathbf{1}\mathbf{-}{\mathbf{y}}^{\mathbf{2}}}\mathbf{+}\mathit{y}\mathbf{·}\sqrt{\mathbf{1}\mathbf{-}{\mathbf{x}}^{\mathbf{2}}}\mathbf{)}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{if}\mathbf{}\mathit{x}\mathbf{,}\mathbf{}\mathit{y}\mathbf{}\mathbf{\ge }\mathbf{}\mathbf{0}\mathbf{}\mathbf{and}\mathbf{}{\mathit{x}}^{\mathbf{2}}\mathbf{+}{\mathit{y}}^{\mathbf{2}}\mathbf{\le }\mathbf{1} \\ \mathbf{}{\mathbf{sin}}^{\mathbf{-}\mathbf{1}}\mathbf{}\mathit{x}\mathbf{+}{\mathbf{sin}}^{\mathbf{-}\mathbf{1}}\mathbf{}\mathit{y}\mathbf{=}\mathbf{\pi }\mathbf{-}{\mathbf{sin}}^{\mathbf{-}\mathbf{1}}\mathbf{}\mathbf{(}\mathit{x}\mathbf{·}\sqrt{\mathbf{1}\mathbf{-}{\mathbf{y}}^{\mathbf{2}}}\mathbf{+}\mathit{y}\mathbf{·}\sqrt{\mathbf{1}\mathbf{-}{\mathbf{x}}^{\mathbf{2}}}\mathbf{)}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{if}\mathbf{}\mathit{x}\mathbf{,}\mathbf{}\mathit{y}\mathbf{}\mathbf{\ge }\mathbf{}\mathbf{0}\mathbf{}\mathbf{and}\mathbf{}{\mathit{x}}^{\mathbf{2}}\mathbf{+}{\mathit{y}}^{\mathbf{2}}\mathbf{>}\mathbf{1} \end{gathered}$$