# Trigonometric and Inverse Trigonometric Functions

Let us now establish some elementary but useful results involving trigonometric and inverse trigonometric functions.

a. For a given angle $\theta$,

1. $\theta=\sin^{-1}\sin\theta$

Let $\sin\theta=x$, then, $\theta=\sin^{-1}x$ $\therefore\theta=\sin^{-1}\sin\theta$

2. $\theta=\sin\sin^{-1}\theta$

Let $y=\sin^{-1}\theta$, then, $\theta=\sin y$ $\theta=\sin\sin^{-1}y$ The rest of the trigonometric functions; cosine, tangent, etc. follows similarly.

b. For a given numerical value $x$, $\operatorname{cosec}^{-1}x=\sin^{-1}\frac{1}{x}$

Let $\theta=\operatorname{cosec}^{-1}x$, then, $\operatorname{cosec}\theta=x$ $\sin\theta=\frac{1}{x}$ $\theta=\sin^{-1}\frac{1}{x}$ $\therefore\operatorname{cosec}^{-1}x=\sin^{-1}\frac{1}{x}$

Similarly, we can show, $\sin^{-1}x=\operatorname{cosec}^{-1}\frac{1}{x}$ $\sec^{-1}x=\cos^{-1}\frac{1}{x}$ $\cot^{-1}x=\tan^{-1}\frac{1}{x},\;\;\text{etc.}$

c. Expressions of a given inverse trigonometric function in terms of the remaining inverse trigonometric functions

1. $\sin^{-1}x=\cos^{-1}\sqrt{1-x^2}$

Let $\theta=\sin^{-1}x$, then, $\sin\theta=x$. Now, $\cos\theta=\sqrt{1-\sin^2\theta}=\sqrt{1-x^2}$ $\theta=\cos^{-1}\sqrt{1-x^2}$ $\therefore\sin^{-1}x=\cos^{-1}\sqrt{1-x^2}$

2. $\sin^{-1}x=\tan^{-1}\frac{x}{\sqrt{1-x^2}}$

Let $\theta=\sin^{-1}x$, then, $\sin\theta=x$. Now, $\tan\theta=\frac{\sin\theta}{\cos\theta}=\frac{x}{\sqrt{1-x^2}}$ $\theta=\tan^{-1}\frac{x}{\sqrt{1-x^2}}$ $\therefore\sin^{-1}x=\tan^{-1}\frac{x}{\sqrt{1-x^2}}$

Similarly, we can prove, $\sin^{-1}x=\cot^{-1}\frac{\sqrt{1-x^2}}{x}$ $\sin^{-1}x=\sec^{-1}\frac{1}{\sqrt{1-x^2}}$

Expressions of other inverse trigonometric functions can also be deduced in the same way.

d. For a given numerical value $x$, $\sin^{-1}x+\cos^{-1}x=\frac{π}{2}$

Let $\theta=\sin^{-1}x$, then, $x=\sin\theta$. Hence, $x=\cos\left(\frac{π}{2}-\theta\right)$ $\cos^{-1}x=\frac{π}{2}-\theta$ $\therefore\sin^{-1}x+\cos^{-1}x=\frac{π}{2}$

Similarly, we can prove,

$\tan^{-1}x+\cot^{-1}x=\frac{π}{2}$ $\operatorname{cosec}^{-1}x+\sec^{-1}x=\frac{π}{2}$

e. For a given numerical value $x$,

1. $\sin^{-1}(-x)=-\sin^{-1}x$

Let $\sin^{-1}x=\theta$, then, $x=\sin\theta$. Now, $-x=-\sin\theta$ $-x=\sin(-\theta)$ $\sin^{-1}(-x)=-\theta$ $\therefore\sin^{-1}(-x)=-\sin^{-1}x$

2. $\cos^{-1}(-x)=π-\cos^{-1}x$

Let $\cos^{-1}x=\theta$, then, $x=\cos\theta$. Now, $-x=-\cos\theta$ $-x=\cos(π-\theta)$ $\cos^{-1}(-x)=π-\theta$ $\therefore\cos^{-1}(-x)=π-\cos^{-1}x$

Similarly, we can prove, $\tan^{-1}(-x)=-\tan^{-1}x$ $\cot^{-1}(-x)=π-\cot^{-1}x$

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