Basic Concepts of Inverse Trigonometric Functions: Domain, Range, Principal Value, and other Properties

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Inverse trigonometric functions (ITF) are special functions that provide the measurement of angle for any given trigonometric ratios. Often, we convert provided numerical values, algebraic, exponential, or any other type of values, into well-known trigonometric ratios. Using inverse trig functions, we get a measure of an angle for the given value.

For Example, $$\sin^{-1}\left(\frac{1}{2}\right) = 30^\circ$$We first need to convert 1/2 in the sine ratio. Since,

$$\sin(30^\circ) = \frac{1}{2}$$

So by substituting the value we get,

expression:

$${\sin }^{-1}\left(\frac{1}{2}\right)={\sin }^{-1}(\sin ({30}^{\circ }))\; =\; 30$$

However, an important point here is that these trigonometric ratios are not one-one and onto functions. This means we can't have an inverse trigonometric function for any of the trig ratios until we fix the range. This fixed range is called the principal value. This principal value ensures a unique output for each input, making the inverse function invertible. 

$$\sin^{-1}\left(\frac{1}{2}\right) = \sin^{-1}\left(\sin(30^\circ)\right)$$

Trigonometric basics are really important to develop a good understanding in class 11 Maths, Inverse Trigonometric Functions. Students should have a comprehensive and deep understanding of all the inverse trig formulas and fundamental concepts. We will discuss these given concepts below in detail.

• Principle Value
• Domain and Range for All Inverse Trig Functions
• Key Properties of Inverse Trigonometric Functions and Formulas

All trigonometric functions are periodic, which means all trig ratios repeat their value after a given period. This means trigonometric ratios have an infinite domain. To understand in simple terms, sin (2nπ + θ) will provide the same value for any value of n (n being an integer). 

This domain of trigonometric functions becomes the range for inverse trig functions. The restricted range of inverse trig functions is called the principal value branch of the functions. Students can check the principal value for all functions below.

Inverse Trigonometric Function	Principal Value  in Radians
$\sin^{-1}(x)$	$[- \frac{\pi}{2}, \frac{\pi}{2}]$
$\cos^{-1}(x)$	$[0, \pi]$
$\tan^{-1}(x)$	$(- \frac{\pi}{2}, \frac{\pi}{2})$
$\cot^{-1}(x)$	$(0, \pi)$
$\sec^{-1}(x)$	$[0, \pi] \setminus \left( \frac{\pi}{2} \right)$
$\csc^{-1}(x)$	$[- \frac{\pi}{2}, \frac{\pi}{2}] \setminus \{0\}$

Like any other function, these inverse trigonometric functions also have specific domains and ranges. Students must know, and if possible memorise, the domain and range of these functions to quickly understand and solve questions. You can check the table below for the domain and range of all the ITFs in class 11 Maths.

Inverse Trigonometric Function	Domain	Range in Degrees(Principal value)
$\sin^{-1}x$	$[-1, 1]$	$[-90^\circ, 90^\circ]$
$\cos^{-1}x$	$[-1, 1]$	$[0^\circ, 180^\circ]$
$\tan^{-1}x$	$(-\infty, \infty)$	$(-90^\circ, 90^\circ)$
$\cot^{-1}x$	$(-\infty, \infty)$	$(0^\circ, 180^\circ)$
$\sec^{-1}x$	$(-\infty, -1] \cup [1, \infty)$	$[0^\circ, 180^\circ] \setminus \{90^\circ\}$
$\csc^{-1}x$	$(-\infty, -1] \cup [1, \infty)$	$[-90^\circ, 90^\circ] \setminus \{0^\circ\}$

  

Inverse trigonometry includes many important properties and formulas. These inverse trig formulas are very important not only for this chapter, and other topics, including integration, differentiation. We have provided a detailed Key Properties of Inverse Trigonometric Function page, which discusses all these key properties in depth. Read the properties below.

• Negative Values 

sin⁻¹(-x) = -sin⁻¹(x) for x ∈ [-1, 

cos⁻¹(-x) = π - cos⁻¹(x) for x ∈ [-1, 

tan⁻¹(-x) = -tan⁻¹(x) for x ∈ R 

• Reciprocal Values

sin⁻¹(x) = cosec⁻¹(1/x)

cos⁻¹(x) = sec⁻¹(1/x) 

tan⁻¹(x) = cot⁻¹(1/x)

• sum and difference inverse trigonometric formulas 

sin⁻¹(x) + cos⁻¹(x) = π/2 

tan⁻¹(x) + cot⁻¹(x) = π/2 

• Conversion Identities of Inverse Trigonometric Formulas 

$\sin^{-1}(x) = \cos^{-1}\left(\sqrt{1 - x^2}\right)$

$\cos^{-1}(x) = \sin^{-1}\left(\sqrt{1 - x^2}\right)$

$\tan^{-1}(x) = \sin^{-1}\left(\dfrac{x}{\sqrt{1 + x^2}}\right)$

• Various important Inverse trigonometric formulas ( ${\mathrm{2sin}}^{-1}(x)$, ${\mathrm{3sin}}^{-1}(x)$)

$2{\sin }^{-1}(x)={\sin }^{-1}\left(2x\sqrt{1-x^2}\right)$

$3\sin^{-1}(x) = \sin^{-1}\left(3x - 4x^3\right)$

• Derivatives of Inverse trig formulas

 

• $\dfrac{d}{dx}[\sin^{-1}x] = \dfrac{1}{\sqrt{1 - x^2}}, \quad |x| < 1$
  

• $\dfrac{d}{dx}[\cos^{-1}x] = -\dfrac{1}{\sqrt{1 - x^2}}, \quad |x| < 1$
  

• $\dfrac{d}{dx}[\tan^{-1}x] = \dfrac{1}{1 + x^2}, \quad x \in \mathbb{R}$
  

 

Let's take a few examples to understand how to solve basic inverse trig questions.

Q.1: Find the principal value of $\sin^{-1}\left(\frac{1}{\sqrt{2}}\right)$

Let's assume, ${\sin }^{-1}(\frac{1}{})\; =\; y$

Then it is clear,  $<br>\frac{\mathrm{sin\; y}}{}=\; 1$

As you have read above the principal value (range) of $\sin^{-1}$is: $$[-\frac{\pi }{2},\frac{\pi }{2}]$$

Also, $$\sin \left(\frac{\pi }{2}\right)=\frac{1}{}$$

by comparison,

$$\sin \left(\frac{\pi }{2}\right)= \; sin\; y\frac{\mathrm{<br>}}{}$$

So, y=1, which ultimately gives answer; ${\sin }^{-1}(\frac{1}{})\; = \; \pi /2$

Q.2: Find the principal value of $\tan^{-1}\left(\frac{-1}{\sqrt{3}}\right)$

Let's assume

$${\tan }^{-1}\left(\frac{-1}{\sqrt{3}}\right)=y$$

Then,

$$\tan y=\frac{-1}{\sqrt{3}}$$

Now we know,

$$\mathrm{tan\; y\; =\; tan}(-\frac{\pi }{6})=-\tan \left(\frac{\pi }{6}\right)=-\frac{1}{\sqrt{3}}$$

So the final answer,

$$\tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) = -\frac{\pi}{6}$$​