Sum of two Inverse sine functions

Proof

$\alpha$ and $\beta$ are two angles of the two right angled triangles and the compound of angle of them is $\alpha + \beta$. The sine of the sum of two angles can be expanded in terms of sine and cosine of both angles.

$\sin(\alpha+\beta)$ $\,=\,$ $\sin \alpha \cos \beta + \cos \alpha \sin \beta$

One the basis of this trigonometric formula, the sum of two arcsin functions can be expressed in mathematical form.

Take $\sin \alpha = x$ and $\sin \beta = y$.

$(1) \,\,\,\,\,\,$ $\sin \alpha = x \implies \alpha = \sin^{-1} x$

$(2) \,\,\,\,\,\,$ $\sin \beta = y \implies \beta = \sin^{-1} y$

According to the Pythagorean identity of sine and cosine functions, express $\cos \alpha$ and $\cos \beta$ in square root form of $\sin \alpha$ and $\sin \beta$ respectively.

$\cos \alpha = \sqrt{1-\sin^2 \alpha} \,\,\,$ and $\,\,\, \cos \beta = \sqrt{1-\sin^2 \beta}$

As per our consideration, $\sin \alpha = x$ and $\sin \beta = y$. Now, transform the above two equations in terms of $x$ and $y$ purely.

$(3) \,\,\,\,\,\,$ $\cos \alpha = \sqrt{1-x^2}$

$(4) \,\,\,\,\,\,$ $\cos \beta = \sqrt{1-y^2}$

Transform the sine of sum of angles rule purely in terms of x and y to obtain the formula of sum of two inverse sine functions.

$\sin(\alpha+\beta) \,=\, \sin \alpha \cos \beta + \cos \alpha \sin \beta$

$\implies \sin(\alpha+\beta)$ $\,=\,$ $(x)(\sqrt{1-y^2})$ $\,+\,$ $(\sqrt{1-x^2})(y)$

$\implies \sin(\alpha+\beta)$ $\,=\,$ $(x)(\sqrt{1-y^2})$ $\,+\,$ $(y)(\sqrt{1-x^2})$

$\implies \sin(\alpha+\beta)$ $\,=\,$ $x\sqrt{1-y^2}$ $\,+\,$ $y\sqrt{1-x^2}$

Express sine function in its inverse form.

$\implies \alpha+\beta$ $\,=\,$ $\sin^{-1} [x\sqrt{1-y^2}$ $\,+\,$ $y\sqrt{1-x^2}]$

Replace, $\alpha$ and $\beta$ in terms of $x$ and $y$.

$\therefore \,\,\,\,\,\, \sin^{-1} x + \sin^{-1} y$ $\,=\,$ $\sin^{-1} [x\sqrt{1-y^2}$ $\,+\,$ $y\sqrt{1-x^2}]$

It is a property of the inverse trigonometry to write summation of two inverse sine functions in the form of same inverse sine function.

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