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

$\alpha$ 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.

01

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

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

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

02

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.

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

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

03

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.

Latest Math Topics

Jul 24, 2024

Dec 13, 2023

Jul 20, 2023

Latest Math Problems

Jan 30, 2024

Oct 15, 2023

Copyright © 2012 - 2023 Math Doubts, All Rights Reserved