Chapter 2 Inverse Trigonometric Functions

Notes

Inverse Trigonometric Functions

Inverse trigonometric functions are the inverse functions of trigonometric functions when their domains are suitably restricted. The principal value branches are used to make the inverse functions single-valued.

1. Definition and Principal Value Ranges

1.1 Inverse Sine Function

If

\[ \sin\theta=x, \]

then

\[ \sin^{-1}x=\theta. \]

Here,

\[ x\in[-1,1] \]

and the principal value of \(\theta\) lies in

\[ \theta\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]. \]

1.2 Inverse Cosine Function

If

\[ \cos\theta=x, \]

then

\[ \cos^{-1}x=\theta. \]

Here,

\[ x\in[-1,1] \]

and

\[ \theta\in[0,\pi]. \]

1.3 Inverse Tangent Function

If

\[ \tan\theta=x, \]

then

\[ \tan^{-1}x=\theta. \]

Here,

\[ x\in\mathbb{R} \]

and

\[ \theta\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right). \]

1.4 Inverse Cotangent Function

If

\[ \cot\theta=x, \]

then

\[ \cot^{-1}x=\theta. \]

Here,

\[ x\in\mathbb{R} \]

and

\[ \theta\in(0,\pi). \]

1.5 Inverse Secant Function

If

\[ \sec\theta=x, \]

then

\[ \sec^{-1}x=\theta. \]

Here,

\[ x\in\mathbb{R}-(-1,1) \]

and

\[ \theta\in[0,\pi]-\left\{\frac{\pi}{2}\right\}. \]

1.6 Inverse Cosecant Function

If

\[ cosec\theta=x, \]

then

\[ cosec^{-1}x=\theta. \]

Here,

\[ x\in\mathbb{R}-(-1,1) \]

and

\[ \theta\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]-\{0\}. \]

2. Important Formulae for Negative Arguments

The following formulae are useful while simplifying inverse trigonometric expressions.

\[ \sin^{-1}(-x)=-\sin^{-1}x \]

\[ \cos^{-1}(-x)=\pi-\cos^{-1}x \]

\[ \tan^{-1}(-x)=-\tan^{-1}x \]

\[ \cot^{-1}(-x)=\pi-\cot^{-1}x \]

\[ \sec^{-1}(-x)=\pi-\sec^{-1}x \]

\[ cosec^{-1}(-x)=-cosec^{-1}x \]

3. Proof of Important Negative Argument Formulae

3.1 Prove that \(\sin^{-1}(-x)=-\sin^{-1}x\)

Let

\[ \sin^{-1}(-x)=\theta. \]

Then,

\[ \sin\theta=-x. \]

Therefore,

\[ -\sin\theta=x \]

or

\[ \sin(-\theta)=x. \]

Hence,

\[ \sin^{-1}x=-\theta. \]

Therefore,

\[ \boxed{\sin^{-1}(-x)=-\sin^{-1}x} \]

3.2 Prove that \(\cos^{-1}(-x)=\pi-\cos^{-1}x\)

Let

\[ \cos^{-1}(-x)=\theta. \]

Then,

\[ \cos\theta=-x. \]

Therefore,

\[ -\cos\theta=x. \]

Using

\[ \cos(\pi-\theta)=-\cos\theta, \]