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Properties of Inverse Trigonometric Functions

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Properties of Inverse Trigonometric Functions

In last page,we cover about what are Inverse Trigonometric Functions and how they are used to find the angles corresponding to given trigonometric ratios, which is an essential concept in mathematics, physics, and engineering
In this article, we will discuss the properties of these inverse trigonometric functions.
It may be mentioned here that these results are valid within the principal value branches of the corresponding inverse trigonometric functions and wherever they are defined.
Some results may not be valid for all values of the domains of inverse trigonometric functions

Inverse of Negative x

$sin^{-1} (-x) = -sin^{-1} (x)$
$cos^{-1} (-x) = \pi - cos^{-1} (x)$
$tan^{-1} (-x) = -tan^{-1} (x)$
$sec^{-1} (-x) = \pi - sec^{-1} (x)$
$cosec^{-1} (-x) = -cosec^{-1} (x)$
$cot^{-1} (-x) = \pi - cot^{-1} (x)$
Lets see the proof of one of them
Proof on Sin
Let $sin^{-1} (-x) = y$
Then $-x = sin y$
or $x = - sin y$
or $x = sin (-y)$
Hence $sin^{-1}x = -y = - sin^{-1} (-x)$
Similary we can prove for others

Reciprocal of x

$sin^{-1} (\frac {1}{x}) = cosec^{-1} (x)$
$cos^{-1} (\frac {1}{x}) = sec^{-1} (x)$
$tan^{-1} (\frac {1}{x}) = cot^{-1} (x)$

Complementary Inverse function

$sin^{-1} (x) + cos^ {-1} (x) = \frac {\pi}{2}$
$sec^{-1} (x) + cosec^ {-1} (x) = \frac {\pi}{2}$
$tan^{-1} (x) + cot^ {-1} (x) = \frac {\pi}{2}$

Addition of Same Inverse function

$sin^{-1} (x) + sin^ {-1} (y) = sin ^{-1} (x \sqrt {1-y^2} + y \sqrt {1-x^2})$ if $x,y \geq 0 $, $x^2 + y^2 \leq 1$
$sin^{-1} (x) + sin^ {-1} (y) = \pi - sin ^{-1} (x \sqrt {1-y^2} + y \sqrt {1-x^2})$ if $x,y \geq 0 $, $x^2 + y^2 > 1$
$sin^{-1} (x) - sin^ {-1} (y) = sin ^{-1} (x \sqrt {1-y^2} - y \sqrt {1-x^2)}$ if $x,y \geq 0 $, $x^2 + y^2 \leq 1$
$sin^{-1} (x) - sin^ {-1} (y) = \pi - sin ^{-1} (x \sqrt {1-y^2} - y \sqrt {1-x^2})$ if $x,y \geq 0 $, $x^2 + y^2 > 1$
$cos^{-1} (x) + cos^ {-1} (y) = cos ^{-1} (x y - \sqrt {1-y^2} \sqrt {1-x^2})$ if $x,y \geq 0 $, $x^2 + y^2 \leq 1$
$cos^{-1} (x) + cos^ {-1} (y) = \pi - cos ^{-1} ((x y - \sqrt {1-y^2} \sqrt {1-x^2})$ if $x,y \geq 0 $, $x^2 + y^2 > 1$
$cos^{-1} (x) - cos^ {-1} (y) = cos ^{-1} (x y + \sqrt {1-y^2} \sqrt {1-x^2})$ if $x,y \geq 0 $, $x^2 + y^2 \leq 1$
$cos^{-1} (x) - cos^ {-1} (y) = \pi - cos ^{-1} (x y + \sqrt {1-y^2} \sqrt {1-x^2})$ if $x,y \geq 0 $, $x^2 + y^2 > 1$
$tan^{-1} (x) + tan^ {-1} (y)= tan^{-1} (\frac {x+y}{1-xy})$ , if $x,y > 0 $, $xy < 1$
$tan^{-1} (x) + tan^ {-1} (y)= \pi + tan^{-1} (\frac {x+y}{1-xy})$ , if $x,y > 0 $, $xy > 1$
$tan^{-1} (x) + tan^ {-1} (y)= tan^{-1} (\frac {x+y}{1-xy}) - \pi$ , if $x < 0, y > 0 $, $xy > 1$
$tan^{-1} (x) - tan^ {-1} (y)= tan^{-1} (\frac {x-y}{1+xy}) - \pi$ , if $xy > -1$
$tan^{-1} (x) + tan^ {-1} (y) + tan^ {-1} (z) = tan^{-1} (\frac {x+y+z - xyz}{1-xy-yz-xz})$

Double angle identity for Inverse functions

$ 2 sin^{-1} (x) = sin^{-1} (2x \sqrt {1-x^2})$ if $ -\frac {1}{\sqrt {2}} \leq x \frac {1}{\sqrt {2}} $
$ 2 cos^{-1} (x) = cos^{-1} (2x^2 -1)$
$2 tan^{-1} (x) = tan^{-1} (\frac {2x}{1-x^2})$ if $ -1 $2 tan^{-1} (x) = sin^{-1} (\frac {2x}{1+x^2})$ if $ |x| \leq 1$
$2 tan^{-1} (x) = cos^{-1} (\frac {1 -x^2}{1+x^2})$ if $ x \geq 0$
$3 sin^{-1} (x) = sin^{-1} (3x -4x^3)$
$3 cos^{-1} (x) = cos^{-1} (4x^3 - 3x)$
$3 tan^{-1} (x) = tan^{-1} (\frac {3x -x^3}{1-3x^2})$