Inverse Trigonometric Functions: Introduction and Properties
Introduction to Inverse Trigonometric Functions ($\sin^{-1}x$, $\cos^{-1}x$, etc.)
We have studied trigonometric functions, such as $ y = \sin x $, $ y = \cos x $, $ y = \tan x $, etc. These functions take an angle $x$ as input and give a unique real value $y$ as output. For example, if $ x = 30^\circ $ (or $ \pi/6 $ radians), then $ y = \sin(\pi/6) = 1/2 $. In this case, the angle $ \pi/6 $ is mapped to the value $ 1/2 $ by the sine function.
Inverse trigonometric functions are the inverse operations of trigonometric functions. They take a real value (which is the output of a trigonometric function) as input and return the corresponding angle whose trigonometric function equals that value. For example, the inverse sine function takes the value $1/2$ as input and returns an angle whose sine is $1/2$. If $ y = \sin x $, then the inverse relation is $ x = \text{angle whose sine is } y $. This inverse relation is denoted as $ x = \sin^{-1} y $.
The notation $ \sin^{-1} y $ reads as "sine inverse of y" or "arcsine of y".
Need for Domain Restriction (Principal Value Branches)
For a function to have an inverse that is also a function, the original function must be one-to-one (injective). A function is one-to-one if every distinct input maps to a distinct output. Graphically, a one-to-one function passes the horizontal line test (any horizontal line intersects the graph at most once).
Consider the graphs of the basic trigonometric functions:
As you can see from the graph of $y = \sin x$, any horizontal line (like $y = 1/2$) intersects the graph at infinitely many points (e.g., $x = \pi/6, 5\pi/6, 13\pi/6, -7\pi/6$, etc.). This means that different angles ($ \pi/6, 5\pi/6, \dots $) have the same sine value ($1/2$). Therefore, the standard trigonometric functions are not one-to-one over their entire domains ($\mathbb{R}$). This prevents us from defining a unique inverse function over their full domains.
To make the inverse trigonometric functions well-defined as functions, we must restrict the domain of the original trigonometric function to an interval where it is one-to-one. Within such a restricted domain, the function becomes bijective (one-to-one and onto its restricted range).
For each trigonometric function, there are infinitely many intervals where it is one-to-one. By convention, specific intervals are chosen as the principal value branches. These chosen restricted domains of the original functions become the ranges of the inverse trigonometric functions.
Notation for Inverse Trigonometric Functions
There are two common notations for inverse trigonometric functions:
Using the superscript $-1$: $ \sin^{-1} x $, $ \cos^{-1} x $, $ \tan^{-1} x $, $ \text{cosec}^{-1} x $, $ \sec^{-1} x $, $ \cot^{-1} x $.
Using the prefix "arc": $ \arcsin x $, $ \arccos x $, $ \arctan x $, $ \text{arccsc} x $, $ \text{arcsec} x $, $ \text{arccot} x $.
Both notations are used interchangeably and represent the same concept: the angle whose trigonometric function is $x$. For example, $ \sin^{-1} (1/2) $ and $ \arcsin (1/2) $ both mean "the angle whose sine is $1/2$".
Important Note: The notation $ \sin^{-1} x $ should not be confused with $ (\sin x)^{-1} $, which means the reciprocal of $ \sin x $. $ (\sin x)^{-1} = \frac{1}{\sin x} = \text{cosec} \, x $. The superscript $-1$ in $ \sin^{-1} x $ denotes the inverse function, not exponentiation.
Formal Definition
If a trigonometric function $f$ has a domain $D$ and range $R$, and we restrict $f$ to a subdomain $D'$ where it is bijective onto $R$, then its inverse function, denoted by $f^{-1}$, is defined as:
$f^{-1}(y) = x \iff f(x) = y$, for all $x \in D'$ and for all $y \in R$.
For example, let's consider the sine function $ y = \sin x $. The standard domain is $ \mathbb{R} $, and the range is $ [-1, 1] $. To define the inverse sine function, we restrict the domain of $ \sin x $ to the interval $ [-\frac{\pi}{2}, \frac{\pi}{2}] $. In this interval, $ \sin x $ is one-to-one and its range is still $ [-1, 1] $.
The inverse sine function, $ y = \sin^{-1} x $ (or $ y = \arcsin x $), has the domain $ [-1, 1] $ and its range is the restricted domain of sine, $ [-\frac{\pi}{2}, \frac{\pi}{2}] $. The definition is:
$ \sin^{-1} x = y \iff \sin y = x $, provided $ x \in [-1, 1] $ and $ y \in [-\frac{\pi}{2}, \frac{\pi}{2}] $.
This means that the output of $ \sin^{-1} x $ is an angle (in radians or degrees) within the specific interval $ [-\frac{\pi}{2}, \frac{\pi}{2}] $.
For example, $ \sin^{-1} (1/2) = \pi/6 $ because $ \sin(\pi/6) = 1/2 $ and $ \pi/6 $ is in the interval $ [-\frac{\pi}{2}, \frac{\pi}{2}] $. We do not say $ \sin^{-1} (1/2) = 5\pi/6 $ because $ 5\pi/6 $ is not in the restricted range $ [-\frac{\pi}{2}, \frac{\pi}{2}] $, even though $ \sin(5\pi/6) = 1/2 $.
Similarly, $ \sin^{-1} (-1/2) = -\pi/6 $ because $ \sin(-\pi/6) = -1/2 $ and $ -\pi/6 $ is in $ [-\frac{\pi}{2}, \frac{\pi}{2}] $.
Example 1: Evaluating Inverse Sine
Example 1. Evaluate $ \sin^{-1} (\sqrt{3}/2) $.
Answer:
Solution:
Let $ y = \sin^{-1} (\sqrt{3}/2) $. By the definition, this means $ \sin y = \sqrt{3}/2 $. The output $y$ must be in the range of $\sin^{-1}$, which is $ [-\frac{\pi}{2}, \frac{\pi}{2}] $. We need to find an angle $y$ in this interval such that its sine is $ \sqrt{3}/2 $.
We know that $ \sin (\pi/3) = \sqrt{3}/2 $. Since $ \pi/3 $ is in the interval $ [-\frac{\pi}{2}, \frac{\pi}{2}] $, $ y = \pi/3 $ is the principal value.
Therefore, $ \sin^{-1} (\sqrt{3}/2) = \pi/3 $.
Note for Competitive Exams
Inverse trigonometric functions are essential for solving trigonometric equations and evaluating certain integrals in calculus. The key concept is the domain restriction of the original trigonometric functions to make them invertible. This leads to the definition of the principal value branch for each inverse function's range. Understand that $ \sin^{-1} x $ (or $ \arcsin x $) gives the unique angle in $ [-\pi/2, \pi/2] $ whose sine is $x$. Similarly for other inverse functions with their specific ranges. Never confuse $ \sin^{-1} x $ with $ 1/\sin x $. Practice evaluating inverse trig functions for standard values and understanding their definitions.
Domain and Range of Inverse Trigonometric Functions
The domain of an inverse function is the range of the original function, and the range of an inverse function is the restricted domain of the original function. For inverse trigonometric functions, this restricted range is specifically called the principal value branch.
The standard definitions for the domains and principal value branches of the six inverse trigonometric functions are crucial to remember.
Summary Table: Domain and Range (Principal Value Branch)
The table below lists the domain (possible input values of $x$) and the range (possible output angle values of $y$) for each inverse trigonometric function based on the standard conventions:
| Function ($y = f^{-1}(x)$) | Domain of $f^{-1}$ (Input $x$) | Range of $f^{-1}$ (Output $y$, Principal Value Branch) | Original Function ($x = f(y)$) Definition |
|---|---|---|---|
| $y = \sin^{-1} x$ | $[-1, 1]$ | $[-\frac{\pi}{2}, \frac{\pi}{2}]$ ($[-90^\circ, 90^\circ]$) | $x = \sin y$, where $y \in [-\frac{\pi}{2}, \frac{\pi}{2}]$ |
| $y = \cos^{-1} x$ | $[-1, 1]$ | $[0, \pi]$ ($[0^\circ, 180^\circ]$) | $x = \cos y$, where $y \in [0, \pi]$ |
| $y = \tan^{-1} x$ | $\mathbb{R}$ (all real numbers) | $(-\frac{\pi}{2}, \frac{\pi}{2})$ ($-90^\circ < y < 90^\circ$) | $x = \tan y$, where $y \in (-\frac{\pi}{2}, \frac{\pi}{2})$ |
| $y = \text{cosec}^{-1} x$ | $(-\infty, -1] \cup [1, \infty)$ or $|x| \ge 1$ | $[-\frac{\pi}{2}, 0) \cup (0, \frac{\pi}{2}]$ ($[-90^\circ, 0^\circ) \cup (0^\circ, 90^\circ]$) | $x = \text{cosec} \, y$, where $y \in [-\frac{\pi}{2}, 0) \cup (0, \frac{\pi}{2}]$ |
| $y = \sec^{-1} x$ | $(-\infty, -1] \cup [1, \infty)$ or $|x| \ge 1$ | $[0, \frac{\pi}{2}) \cup (\frac{\pi}{2}, \pi]$ ($[0^\circ, 90^\circ) \cup (90^\circ, 180^\circ]$) | $x = \sec y$, where $y \in [0, \frac{\pi}{2}) \cup (\frac{\pi}{2}, \pi]$ |
| $y = \cot^{-1} x$ | $\mathbb{R}$ (all real numbers) | $(0, \pi)$ ($0^\circ < y < 180^\circ$) | $x = \cot y$, where $y \in (0, \pi)$ |
Explanation of Domains and Ranges
The domain of an inverse trigonometric function $f^{-1}(x)$ is the set of all possible values that can be input for $x$. The range (principal value branch) is the set of all possible output angle values $y = f^{-1}(x)$.
$y = \sin^{-1} x$: The range of $ \sin y $ is $ [-1, 1] $, so the domain of $ \sin^{-1} x $ is $ [-1, 1] $. The standard restricted domain for $ \sin y $ to be invertible is $ [-\frac{\pi}{2}, \frac{\pi}{2}] $, which is the range of $ \sin^{-1} x $. This range covers Quadrant IV and Quadrant I.
$y = \cos^{-1} x$: The range of $ \cos y $ is $ [-1, 1] $, so the domain of $ \cos^{-1} x $ is $ [-1, 1] $. The standard restricted domain for $ \cos y $ is $ [0, \pi] $, which is the range of $ \cos^{-1} x $. This range covers Quadrant I and Quadrant II.
$y = \tan^{-1} x$: The range of $ \tan y $ is $ \mathbb{R} $ (all real numbers), so the domain of $ \tan^{-1} x $ is $ \mathbb{R} $. The standard restricted domain for $ \tan y $ is $ (-\frac{\pi}{2}, \frac{\pi}{2}) $, which is the range of $ \tan^{-1} x $. This range covers Quadrant IV and Quadrant I, excluding the vertical axis values where $\tan$ is undefined.
$y = \text{cosec}^{-1} x$: The range of $ \text{cosec} \, y $ is $ (-\infty, -1] \cup [1, \infty) $, so the domain of $ \text{cosec}^{-1} x $ is $ (-\infty, -1] \cup [1, \infty) $. The standard restricted domain for $ \text{cosec} \, y $ is $ [-\frac{\pi}{2}, \frac{\pi}{2}] $, excluding $ y = 0 $ where $ \text{cosec} \, y $ is undefined. This is the range of $ \text{cosec}^{-1} x $: $ [-\frac{\pi}{2}, 0) \cup (0, \frac{\pi}{2}] $. It covers parts of Quadrant IV and Quadrant I.
$y = \sec^{-1} x$: The range of $ \sec y $ is $ (-\infty, -1] \cup [1, \infty) $, so the domain of $ \sec^{-1} x $ is $ (-\infty, -1] \cup [1, \infty) $. The standard restricted domain for $ \sec y $ is $ [0, \pi] $, excluding $ y = \frac{\pi}{2} $ where $ \sec y $ is undefined. This is the range of $ \sec^{-1} x $: $ [0, \frac{\pi}{2}) \cup (\frac{\pi}{2}, \pi] $. It covers parts of Quadrant I and Quadrant II.
$y = \cot^{-1} x$: The range of $ \cot y $ is $ \mathbb{R} $, so the domain of $ \cot^{-1} x $ is $ \mathbb{R} $. The standard restricted domain for $ \cot y $ is $ (0, \pi) $, which is the range of $ \cot^{-1} x $. This range covers Quadrant I and Quadrant II, excluding the horizontal axis values where $\cot$ is undefined.
It is essential to memorise the domains and ranges (principal value branches) of all six inverse trigonometric functions, as they are fundamental for solving problems and evaluating expressions correctly. The output of an inverse trigonometric function must always lie within its defined principal value branch.
Example 1: Domain Check
Example 1. For what values of $x$ is $ \sin^{-1} (2x - 3) $ defined?
Answer:
Solution:
The domain of $ \sin^{-1} u $ is $ [-1, 1] $. This means that the input to the $ \sin^{-1} $ function must be between -1 and 1, inclusive.
In this case, the input is $ 2x - 3 $. So, for $ \sin^{-1} (2x - 3) $ to be defined, we must have:
$-1 \le 2x - 3 \le 1$
This is a compound inequality. We can solve it by applying the same operation to all three parts.
Add 3 to all parts:
$ -1 + 3 \le 2x - 3 + 3 \le 1 + 3 $
$2 \le 2x \le 4$
Divide all parts by 2:
$ \frac{2}{2} \le \frac{2x}{2} \le \frac{4}{2} $
$1 \le x \le 2$
So, $ \sin^{-1} (2x - 3) $ is defined for values of $x$ in the interval $[1, 2]$.
Example 2: Range Check
Example 2. What are the possible values of $ \cos^{-1} (0) $ and $ \cos^{-1} (-1) $?
Answer:
Solution:
The possible values of $ \cos^{-1} x $ are restricted to its range (principal value branch), which is $ [0, \pi] $. We need to find the angle $y$ in this interval such that $ \cos y = x $.
a) For $ \cos^{-1} (0) $:
Let $ y = \cos^{-1} (0) $. This means $ \cos y = 0 $, and $y$ must be in $ [0, \pi] $. We know that $ \cos (\pi/2) = 0 $. Since $ \pi/2 $ is in the interval $ [0, \pi] $, the unique principal value is $ \pi/2 $.
Therefore, $ \cos^{-1} (0) = \pi/2 $.
b) For $ \cos^{-1} (-1) $:
Let $ y = \cos^{-1} (-1) $. This means $ \cos y = -1 $, and $y$ must be in $ [0, \pi] $. We know that $ \cos (\pi) = -1 $. Since $ \pi $ is in the interval $ [0, \pi] $, the unique principal value is $ \pi $.
Therefore, $ \cos^{-1} (-1) = \pi $.
(Note: $ \cos (\pi + 2n\pi) $ or $ \cos (\pi) $ also equals -1 for other integer $n$, but these angles are not in the principal value range $ [0, \pi] $, so they are not the output of $ \cos^{-1} (-1) $).
Note for Competitive Exams
Questions often test your knowledge of the domains and ranges of inverse trigonometric functions. You might be asked to find the domain of a composite function involving inverse trig functions (as in Example 1) or evaluate expressions by keeping the range restrictions in mind. It's crucial to distinguish between the infinite solutions of a trigonometric equation ($ \sin x = k $) and the unique principal value output of an inverse trigonometric function ($ \sin^{-1} k $). Memorising the table of domains and principal value branches is essential.
Principal Value Branch of Inverse Trigonometric Functions
As we've discussed, to define the inverse trigonometric functions as true functions, we must restrict the domain of the original trigonometric functions to an interval where they are one-to-one. The specific interval chosen for this restriction defines the Principal Value Branch for the range of the inverse function. The output of an inverse trigonometric function, for a given input value, is always the unique angle that lies within this defined principal value branch and satisfies the definition.
When we talk about "the value" of $ \sin^{-1} x $, $ \cos^{-1} x $, etc., we are referring to this unique principal value.
Summary of Principal Value Branches (Ranges)
The table from the previous section is crucial here. The third column lists the standard Principal Value Branches (Ranges) for each inverse trigonometric function:
| Function ($y = f^{-1}(x)$) | Domain of $f^{-1}$ (Input $x$) | Range of $f^{-1}$ (Output $y$, Principal Value Branch) |
|---|---|---|
| $y = \sin^{-1} x$ | $[-1, 1]$ | $[-\frac{\pi}{2}, \frac{\pi}{2}]$ |
| $y = \cos^{-1} x$ | $[-1, 1]$ | $[0, \pi]$ |
| $y = \tan^{-1} x$ | $\mathbb{R}$ | $(-\frac{\pi}{2}, \frac{\pi}{2})$ |
| $y = \text{cosec}^{-1} x$ | $(-\infty, -1] \cup [1, \infty)$ | $[-\frac{\pi}{2}, 0) \cup (0, \frac{\pi}{2}]$ |
| $y = \sec^{-1} x$ | $(-\infty, -1] \cup [1, \infty)$ | $[0, \frac{\pi}{2}) \cup (\frac{\pi}{2}, \pi]$ |
| $y = \cot^{-1} x$ | $\mathbb{R}$ | $(0, \pi)$ |
Finding Principal Values
To find the principal value of an inverse trigonometric function $f^{-1}(x)$ for a given value of $x$ (within the domain of $f^{-1}$), we follow these steps:
Let $y = f^{-1}(x)$.
Rewrite this using the definition of the inverse function: $ f(y) = x $.
Identify the principal value branch (range) for $ f^{-1} $. The angle $y$ must belong to this interval.
Find the angle $y$ that satisfies $ f(y) = x $ and lies within the principal value branch identified in Step 3.
For values of $x$ that are positive, the principal value is usually the acute angle in Quadrant I. For negative values of $x$, the principal value lies in the specific negative range for $ \sin^{-1}, \tan^{-1}, \text{cosec}^{-1} $ or in Quadrant II for $ \cos^{-1}, \sec^{-1}, \cot^{-1} $.
Example 1: Principal Value of $\sin^{-1}$
Example 1. Find the principal value of $\sin^{-1} \left(\frac{1}{2}\right)$.
Answer:
Solution:
Let $y = \sin^{-1} \left(\frac{1}{2}\right)$.
By the definition of the inverse sine function, this means $ \sin y = \frac{1}{2} $.
The principal value branch for $ \sin^{-1} x $ is $ [-\frac{\pi}{2}, \frac{\pi}{2}] $. We are looking for an angle $y$ in this interval such that its sine is $ 1/2 $.
We know that $ \sin \frac{\pi}{6} = \frac{1}{2} $.
Check if $ \frac{\pi}{6} $ is in the interval $ [-\frac{\pi}{2}, \frac{\pi}{2}] $. Yes, $ -\frac{\pi}{2} \le \frac{\pi}{6} \le \frac{\pi}{2} $.
Since there is only one such angle in the principal value branch, $ \frac{\pi}{6} $ is the principal value.
Therefore, $ \sin^{-1} \left(\frac{1}{2}\right) = \frac{\pi}{6} $.
(Note: Although $ \sin \frac{5\pi}{6} = \frac{1}{2} $ is also true, $ \frac{5\pi}{6} $ is not in the principal value branch $ [-\frac{\pi}{2}, \frac{\pi}{2}] $, so it is not the principal value of $ \sin^{-1} (1/2) $).
Example 2: Principal Value of $\cos^{-1}$ (Negative Input)
Example 2. Find the principal value of $\cos^{-1} \left(-\frac{\sqrt{3}}{2}\right)$.
Answer:
Solution:
Let $y = \cos^{-1} \left(-\frac{\sqrt{3}}{2}\right)$.
By definition, this means $ \cos y = -\frac{\sqrt{3}}{2} $.
The principal value branch for $ \cos^{-1} x $ is $ [0, \pi] $. We are looking for an angle $y$ in this interval such that its cosine is $ -\frac{\sqrt{3}}{2} $.
Since $ \cos y $ is negative, the angle $y$ must lie in Quadrant II (within the range $ [0, \pi] $).
First, find the reference angle $\alpha$, which is the acute angle such that $ \cos \alpha = |-\frac{\sqrt{3}}{2}| = \frac{\sqrt{3}}{2} $. We know that $ \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2} $, so $ \alpha = \frac{\pi}{6} $.
For an angle in Quadrant II with reference angle $\alpha$, the angle is $ \pi - \alpha $.
$ y = \pi - \frac{\pi}{6} = \frac{6\pi - \pi}{6} = \frac{5\pi}{6} $.
Check if $ \frac{5\pi}{6} $ is in the interval $ [0, \pi] $. Yes, $ 0 \le \frac{5\pi}{6} \le \pi $.
Since this is the unique angle in the principal value branch satisfying the condition, $ \frac{5\pi}{6} $ is the principal value.
Therefore, $ \cos^{-1} \left(-\frac{\sqrt{3}}{2}\right) = \frac{5\pi}{6} $.
Example 3: Principal Value of $\tan^{-1}$ (Negative Input)
Example 3. Find the principal value of $\tan^{-1} (-\sqrt{3})$.
Answer:
Solution:
Let $y = \tan^{-1} (-\sqrt{3})$.
By definition, this means $ \tan y = -\sqrt{3} $.
The principal value branch for $ \tan^{-1} x $ is $ (-\frac{\pi}{2}, \frac{\pi}{2}) $. We are looking for an angle $y$ in this interval such that its tangent is $ -\sqrt{3} $.
Since $ \tan y $ is negative, the angle $y$ must lie in Quadrant IV (within the range $ (-\frac{\pi}{2}, \frac{\pi}{2}) $).
First, find the reference angle $\alpha$, the acute angle such that $ \tan \alpha = |-\sqrt{3}| = \sqrt{3} $. We know that $ \tan \frac{\pi}{3} = \sqrt{3} $, so $ \alpha = \frac{\pi}{3} $.
For an angle in Quadrant IV, the angle in the range $ (-\frac{\pi}{2}, \frac{\pi}{2}) $ is $ -\alpha $.
$ y = -\alpha = -\frac{\pi}{3} $.
Check if $ -\frac{\pi}{3} $ is in the interval $ (-\frac{\pi}{2}, \frac{\pi}{2}) $. Yes, $ -\frac{\pi}{2} < -\frac{\pi}{3} < \frac{\pi}{2} $.
Since this is the unique angle in the principal value branch, $ -\frac{\pi}{3} $ is the principal value.
Therefore, $ \tan^{-1} (-\sqrt{3}) = -\frac{\pi}{3} $.
Note for Competitive Exams
The principal value branch is the cornerstone of inverse trigonometric functions. Always state the principal value branch when working with inverse trig functions, especially when evaluating them. Understand how to find the principal value for both positive and negative inputs by first finding the reference angle and then adjusting it based on the required quadrant within the principal range. Pay close attention to the open vs. closed intervals for the ranges (e.g., $ [-\pi/2, \pi/2] $ vs. $ (-\pi/2, \pi/2) $ vs. $ [-\pi/2, 0) \cup (0, \pi/2] $).
Basic Properties of Inverse Trigonometric Functions
The basic properties of inverse trigonometric functions are derived from the fundamental definitions of inverse functions and the specific principal value branches chosen. These properties are essential for simplifying expressions and solving problems involving inverse trigonometric functions.
Property Set 1: Composition of Original Function with its Inverse
These properties show what happens when a trigonometric function is applied to the output of its inverse function.
$ \sin(\sin^{-1} x) = x $, for all $ x \in [-1, 1] $
$ \cos(\cos^{-1} x) = x $, for all $ x \in [-1, 1] $
$ \tan(\tan^{-1} x) = x $, for all $ x \in \mathbb{R} $
$ \text{cosec}(\text{cosec}^{-1} x) = x $, for all $ x \in (-\infty, -1] \cup [1, \infty) $ (i.e., $ |x| \ge 1 $)
$ \sec(\sec^{-1} x) = x $, for all $ x \in (-\infty, -1] \cup [1, \infty) $ (i.e., $ |x| \ge 1 $)
$ \cot(\cot^{-1} x) = x $, for all $ x \in \mathbb{R} $
Explanation: These identities are true because the input $x$ is within the domain of the inverse function, and the inverse function outputs an angle $y$ such that $ f(y) = x $. Applying the original function $f$ to this angle $y$ simply returns the original input $x$. The domain restriction of the inverse function ensures that $f^{-1}(x)$ is a valid angle for $f$ to operate on.
Property Set 2: Composition of Inverse Function with Original Function
These properties show what happens when an inverse trigonometric function is applied to the output of the original function. Here, the restriction on the angle is crucial.
$ \sin^{-1}(\sin y) = y $, if and only if $ y \in [-\frac{\pi}{2}, \frac{\pi}{2}] $
$ \cos^{-1}(\cos y) = y $, if and only if $ y \in [0, \pi] $
$ \tan^{-1}(\tan y) = y $, if and only if $ y \in (-\frac{\pi}{2}, \frac{\pi}{2}) $
$ \text{cosec}^{-1}(\text{cosec} \, y) = y $, if and only if $ y \in [-\frac{\pi}{2}, 0) \cup (0, \frac{\pi}{2}] $
$ \sec^{-1}(\sec y) = y $, if and only if $ y \in [0, \frac{\pi}{2}) \cup (\frac{\pi}{2}, \pi] $
$ \cot^{-1}(\cot y) = y $, if and only if $ y \in (0, \pi) $
Explanation: The output of an inverse trigonometric function must lie within its principal value branch. If the input angle $y$ is within this branch, then $f^{-1}(f(y))$ simply returns $y$. However, if $y$ is outside the principal value branch, $f(y)$ will have a specific value $k$. Then $f^{-1}(k)$ will return the unique angle *within the principal value branch* that has the same trigonometric value $k$. This angle will not be equal to the original angle $y$.
Example 1: Composition with Angle Outside Principal Branch
Example 1. Find the value of $\cos^{-1}\left(\cos \frac{7\pi}{6}\right)$.
Answer:
Solution:
We are asked to find $ \cos^{-1}\left(\cos \frac{7\pi}{6}\right) $. The angle $y = \frac{7\pi}{6}$ is $210^\circ$. The principal value branch for $ \cos^{-1} x $ is $ [0, \pi] $. Since $ \frac{7\pi}{6} $ is greater than $ \pi $, it is not in the principal value branch.
Therefore, $ \cos^{-1}\left(\cos \frac{7\pi}{6}\right) \neq \frac{7\pi}{6} $.
To find the correct value, first evaluate the inner expression $ \cos \frac{7\pi}{6} $.
The angle $ \frac{7\pi}{6} $ is in Quadrant III. The reference angle is $ \frac{7\pi}{6} - \pi = \frac{\pi}{6} $.
$ \cos \frac{7\pi}{6} = \cos\left(\pi + \frac{\pi}{6}\right) = -\cos \frac{\pi}{6} $
From the table of specific angles, $ \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2} $. So, $ \cos \frac{7\pi}{6} = -\frac{\sqrt{3}}{2} $.
Now the problem is equivalent to finding $ \cos^{-1}\left(-\frac{\sqrt{3}}{2}\right) $.
Let $z = \cos^{-1}\left(-\frac{\sqrt{3}}{2}\right)$. By definition, this means $ \cos z = -\frac{\sqrt{3}}{2} $, and $z$ must be in the principal value branch of $ \cos^{-1} $, which is $ [0, \pi] $. Since $ \cos z $ is negative and $ z \in [0, \pi] $, $z$ must be in Quadrant II.
The reference angle $\alpha$ such that $ \cos \alpha = |-\frac{\sqrt{3}}{2}| = \frac{\sqrt{3}}{2} $ is $ \alpha = \frac{\pi}{6} $.
The angle in Quadrant II is $ z = \pi - \alpha = \pi - \frac{\pi}{6} = \frac{5\pi}{6} $.
Check if $ \frac{5\pi}{6} $ is in the interval $ [0, \pi] $. Yes, $ 0 \le \frac{5\pi}{6} \le \pi $.
Therefore, $ \cos^{-1}\left(\cos \frac{7\pi}{6}\right) = \frac{5\pi}{6} $.
This example highlights the importance of the principal value branch for Property Set 2. $ \cos^{-1}(\cos y) = y $ is only true if $ y $ is already in the range $ [0, \pi] $.
Property Set 3: Inverse Functions of Negative Arguments
These properties relate the value of an inverse trigonometric function for a negative input $x$ to its value for the positive input $x$. They depend on the symmetry of the principal value branches.
For functions with principal branches symmetric about the origin ($ [-\pi/2, \pi/2] $ or $ (-\pi/2, \pi/2) $, excluding endpoints):
$ \sin^{-1}(-x) = -\sin^{-1} x $, for $ x \in [-1, 1] $
$ \tan^{-1}(-x) = -\tan^{-1} x $, for $ x \in \mathbb{R} $
$ \text{cosec}^{-1}(-x) = -\text{cosec}^{-1} x $, for $ |x| \ge 1 $
For functions with principal branches in Quadrant I and II ($ [0, \pi] $ or $ (0, \pi) $, excluding points):
$ \cos^{-1}(-x) = \pi - \cos^{-1} x $, for $ x \in [-1, 1] $
$ \sec^{-1}(-x) = \pi - \sec^{-1} x $, for $ |x| \ge 1 $
$ \cot^{-1}(-x) = \pi - \cot^{-1} x $, for $ x \in \mathbb{R} $
Derivation for $\cos^{-1}(-x) = \pi - \cos^{-1} x$:
To Prove: $ \cos^{-1}(-x) = \pi - \cos^{-1} x $, for $ x \in [-1, 1] $.
Proof:
Let $ y = \cos^{-1}(-x) $. By the definition of inverse cosine:
$\cos y = -x$
and $y \in [0, \pi]$
Since $ x \in [-1, 1] $, the input $ -x $ is also in $ [-1, 1] $, so $ \cos^{-1}(-x) $ is defined.
From $ \cos y = -x $, we can write $ x = -\cos y $.
We know from the property of supplementary angles that $ \cos(\pi - \theta) = -\cos \theta $. So, $ -\cos y = \cos(\pi - y) $.
Thus, $ x = \cos(\pi - y) $.
Now consider the range of the angle $ (\pi - y) $. Since $ y \in [0, \pi] $:
$ 0 \le y \le \pi $
Multiplying by -1 and reversing the inequalities:
$ -\pi \le -y \le 0 $
Adding $\pi$ to all parts:
$ \pi - \pi \le \pi - y \le \pi + 0 $
$ 0 \le \pi - y \le \pi $.
The angle $ (\pi - y) $ lies in the interval $ [0, \pi] $. This interval is the principal value branch of $ \cos^{-1} $.
Since $ x = \cos(\pi - y) $ and $ (\pi - y) $ is in the principal range of $ \cos^{-1} $, we can apply the inverse cosine function to both sides:
$ \cos^{-1}(x) = \cos^{-1}(\cos(\pi - y)) $
Using Property Set 2 (since $ (\pi - y) $ is in the principal range):
$ \cos^{-1}(x) = \pi - y $.
Finally, substitute back our initial definition $ y = \cos^{-1}(-x) $:
$ \cos^{-1}(x) = \pi - \cos^{-1}(-x) $
Rearranging to solve for $ \cos^{-1}(-x) $:
$ \cos^{-1}(-x) = \pi - \cos^{-1} x $. (Proved)
Example 2: Using Negative Argument Property
Example 2. Evaluate $ \sin^{-1} (-1/2) $ using properties.
Answer:
Solution:
We use the property $ \sin^{-1}(-x) = -\sin^{-1} x $. Here $x = 1/2$.
$ \sin^{-1} (-1/2) = -\sin^{-1} (1/2) $
We know that $ \sin^{-1} (1/2) = \pi/6 $ (since $ \sin(\pi/6) = 1/2 $ and $ \pi/6 \in [-\pi/2, \pi/2] $).
Therefore, $ \sin^{-1} (-1/2) = -(\pi/6) = -\pi/6 $.
Let's verify: $ \sin (-\pi/6) = - \sin (\pi/6) = -1/2 $. And $ -\pi/6 $ is in the principal value branch $ [-\pi/2, \pi/2] $.
Property Set 4: Reciprocal Arguments
These properties relate the inverse trigonometric functions of reciprocal values. They arise from the definitions of the reciprocal trigonometric functions.
$ \sin^{-1}\left(\frac{1}{x}\right) = \text{cosec}^{-1} x $, for $ |x| \ge 1 $
$ \cos^{-1}\left(\frac{1}{x}\right) = \sec^{-1} x $, for $ |x| \ge 1 $
$ \tan^{-1}\left(\frac{1}{x}\right) = \cot^{-1} x $, for $ x > 0 $
$ \tan^{-1}\left(\frac{1}{x}\right) = \cot^{-1} x - \pi $, for $ x < 0 $ (or $ \pi + \cot^{-1} x $ depending on definition/range of $\cot^{-1}$ used by some texts; use $(0, \pi)$ range for $\cot^{-1}$ consistently).
$ \cot^{-1}\left(\frac{1}{x}\right) = \tan^{-1} x $, for $ x > 0 $
$ \cot^{-1}\left(\frac{1}{x}\right) = \tan^{-1} x + \pi $, for $ x < 0 $.
Explanation: The conditions on $x$ (like $x>0$ or $|x| \ge 1$) are necessary to ensure that both sides of the identity are defined and that the output angle falls within the correct principal value branch. For example, $ \sin^{-1}(1/x) $ requires $ |1/x| \le 1 $, which implies $ |x| \ge 1 $. The range of $ \sin^{-1}(1/x) $ is $ [-\pi/2, 0) \cup (0, \pi/2] $, which matches the range of $ \text{cosec}^{-1} x $.
For $ \tan^{-1}(1/x) = \cot^{-1} x $, when $ x > 0 $, both $ 1/x $ and $ x $ are positive. $ \tan^{-1}(1/x) $ gives an angle in $ (0, \pi/2) $. $ \cot^{-1} x $ also gives an angle in $ (0, \pi/2) $. They match. When $ x < 0 $, $ 1/x $ is negative. $ \tan^{-1}(1/x) $ gives an angle in $ (-\pi/2, 0) $. $ \cot^{-1} x $ gives an angle in $ (\pi/2, \pi) $. They differ by $\pi$.
Derivation for $\sin^{-1}(1/x) = \text{cosec}^{-1} x$:
To Prove: $ \sin^{-1}(1/x) = \text{cosec}^{-1} x $, for $ |x| \ge 1 $.
Proof:
Let $ y = \text{cosec}^{-1} x $. By the definition of inverse cosecant:
$\text{cosec} y = x$
and $y \in [-\frac{\pi}{2}, 0) \cup (0, \frac{\pi}{2}]$
Since $ \text{cosec} y = \frac{1}{\sin y} $, we have $ \frac{1}{\sin y} = x $.
Taking the reciprocal of both sides (since $ x \ne 0 $ for $ |x| \ge 1 $ and $ \sin y \ne 0 $ since $ y \ne 0 $):
$\sin y = \frac{1}{x}$
Now, consider the range of $y$, which is $ [-\frac{\pi}{2}, 0) \cup (0, \frac{\pi}{2}] $. This interval is exactly the principal value branch for $ \sin^{-1} $, excluding $0$.
Since $ \sin y = \frac{1}{x} $ and $y$ is in the principal range for $ \sin^{-1} $, we can apply the $ \sin^{-1} $ function to both sides:
$ \sin^{-1}(\sin y) = \sin^{-1}\left(\frac{1}{x}\right) $
Using Property Set 2 (since $y$ is in the principal range):
$ y = \sin^{-1}\left(\frac{1}{x}\right) $
Substitute back our initial definition $ y = \text{cosec}^{-1} x $:
$ \text{cosec}^{-1} x = \sin^{-1}\left(\frac{1}{x}\right) $. (Proved)
Note for Competitive Exams
These basic properties are essential for simplifying complex expressions involving inverse trigonometric functions and proving identities. Property Set 1 is straightforward, but Property Set 2 requires careful attention to whether the input angle is within the principal value branch. Property Set 3 (negative arguments) helps simplify expressions like $ \sin^{-1}(-1/2) $. Property Set 4 (reciprocal arguments) helps convert between functions, e.g., changing a $ \text{cosec}^{-1} $ problem into a $ \sin^{-1} $ problem. Memorise these key properties and understand the conditions (domains/ranges) under which they apply directly.
Properties of Inverse Trigonometric Functions (Sum, Difference, Conversion)
Inverse trigonometric functions have several important properties derived from their definitions, their restricted domains/ranges (principal value branches), and the relationships between the original trigonometric functions. These properties are invaluable for simplifying expressions, evaluating values, and solving equations involving inverse trigonometric functions.
Property Set 5: Complementary Function Identities
These properties relate pairs of inverse trigonometric functions whose corresponding original functions are complementary (sine/cosine, tangent/cotangent, secant/cosecant). They sum up to $ \pi/2 $.
$ \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} $, for $ x \in [-1, 1] $
$ \tan^{-1} x + \cot^{-1} x = \frac{\pi}{2} $, for $ x \in \mathbb{R} $
$ \sec^{-1} x + \text{cosec}^{-1} x = \frac{\pi}{2} $, for $ |x| \ge 1 $ (i.e., $ x \in (-\infty, -1] \cup [1, \infty) $)
The domain conditions are the intersection of the domains of the two inverse functions involved.
Derivation for $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$:
To Prove: $ \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} $, for $ x \in [-1, 1] $.
Proof:
Let $ y = \sin^{-1} x $. By the definition of inverse sine:
$\sin y = x$
and $y \in [-\frac{\pi}{2}, \frac{\pi}{2}]$
We know from the complementary angle identities for the original trigonometric functions that $ \sin y = \cos\left(\frac{\pi}{2} - y\right) $.
So, substitute this into the equation $ \sin y = x $:
$x = \cos\left(\frac{\pi}{2} - y\right)$
Now, consider the range of the angle $ \left(\frac{\pi}{2} - y\right) $. Since $ y \in [-\frac{\pi}{2}, \frac{\pi}{2}] $:
$ -\frac{\pi}{2} \le y \le \frac{\pi}{2} $
Multiplying the inequality by -1 (and reversing the inequality signs):
$ \frac{\pi}{2} \ge -y \ge -\frac{\pi}{2} $ or $ -\frac{\pi}{2} \le -y \le \frac{\pi}{2} $.
Adding $ \frac{\pi}{2} $ to all parts of the inequality:
$ \frac{\pi}{2} - \frac{\pi}{2} \le \frac{\pi}{2} - y \le \frac{\pi}{2} + \frac{\pi}{2} $
$ 0 \le \frac{\pi}{2} - y \le \pi $.
The angle $ \left(\frac{\pi}{2} - y\right) $ lies in the interval $ [0, \pi] $. This interval is the principal value branch (range) of the $ \cos^{-1} $ function.
Since $ x = \cos\left(\frac{\pi}{2} - y\right) $ and $ \left(\frac{\pi}{2} - y\right) $ is within the principal range of $ \cos^{-1} $, we can apply the $ \cos^{-1} $ function to both sides of the equation $ x = \cos\left(\frac{\pi}{2} - y\right) $:
$ \cos^{-1}(x) = \cos^{-1}\left(\cos\left(\frac{\pi}{2} - y\right)\right) $
Using the property $ \cos^{-1}(\cos \theta) = \theta $ when $ \theta $ is in the principal range $ [0, \pi] $ (Property Set 2):
$\cos^{-1}(x) = \frac{\pi}{2} - y$
[Since $\frac{\pi}{2} - y \in [0, \pi]$]
Finally, substitute back our initial definition $ y = \sin^{-1} x $:
$ \cos^{-1}(x) = \frac{\pi}{2} - \sin^{-1} x $.
Rearranging the terms to get the desired identity:
$ \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} $. (Proved)
Example 1: Using Complementary Function Property
Example 1. Evaluate $ \sin^{-1} (1/2) + \cos^{-1} (1/2) $.
Answer:
Solution:
We can evaluate each term separately:
$ \sin^{-1}(1/2) = \pi/6 $ (since $ \sin(\pi/6) = 1/2 $ and $ \pi/6 \in [-\pi/2, \pi/2] $)
$ \cos^{-1}(1/2) = \pi/3 $ (since $ \cos(\pi/3) = 1/2 $ and $ \pi/3 \in [0, \pi] $)
Summing these values:
$ \sin^{-1} (1/2) + \cos^{-1} (1/2) = \frac{\pi}{6} + \frac{\pi}{3} $
Find a common denominator (6):
$ = \frac{\pi}{6} + \frac{2\pi}{6} = \frac{\pi + 2\pi}{6} = \frac{3\pi}{6} = \frac{\pi}{2} $.
Alternatively, directly using the property $ \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} $, with $ x = 1/2 $. Since $ 1/2 \in [-1, 1] $, the property applies.
$ \sin^{-1} (1/2) + \cos^{-1} (1/2) = \frac{\pi}{2} $.
Both methods give the value $ \frac{\pi}{2} $. The property provides a shortcut when the input values are the same.
Property Set 6: Sum and Difference Formulas for Inverse Tangent
These formulas allow us to combine sums or differences of inverse tangent functions. They are analogous to the tangent sum and difference formulas for the original trigonometric functions ($ \tan(A \pm B) $).
$ \tan^{-1} x + \tan^{-1} y = \tan^{-1}\left(\frac{x+y}{1-xy}\right) $, if $ xy < 1 $
$ \tan^{-1} x + \tan^{-1} y = \pi + \tan^{-1}\left(\frac{x+y}{1-xy}\right) $, if $ xy > 1 $ and $ x > 0, y > 0 $
$ \tan^{-1} x + \tan^{-1} y = -\pi + \tan^{-1}\left(\frac{x+y}{1-xy}\right) $, if $ xy > 1 $ and $ x < 0, y < 0 $
$ \tan^{-1} x - \tan^{-1} y = \tan^{-1}\left(\frac{x-y}{1+xy}\right) $, if $ xy > -1 $
(Note: There are additional cases for $ \tan^{-1} x - \tan^{-1} y $ when $ xy < -1 $, similar to the sum formula, but the $ xy > -1 $ case is most common.)
Derivation for $\tan^{-1} x + \tan^{-1} y = \tan^{-1}\left(\frac{x+y}{1-xy}\right)$ (Case $xy < 1$):
To Prove: $ \tan^{-1} x + \tan^{-1} y = \tan^{-1}\left(\frac{x+y}{1-xy}\right) $, if $ xy < 1 $.
Proof:
Let $ A = \tan^{-1} x $ and $ B = \tan^{-1} y $.
By the definition of inverse tangent: $ \tan A = x $ and $ \tan B = y $.
The range of $ \tan^{-1} $ is $ (-\frac{\pi}{2}, \frac{\pi}{2}) $. So, $ A \in (-\frac{\pi}{2}, \frac{\pi}{2}) $ and $ B \in (-\frac{\pi}{2}, \frac{\pi}{2}) $.
Consider the tangent sum formula for the original trigonometric function:
$\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$
[Tangent Sum Formula (7.5)]
Substitute $ \tan A = x $ and $ \tan B = y $ into the right side:
$\tan(A+B) = \frac{x+y}{1-xy}$
Now, apply the $ \tan^{-1} $ function to both sides:
$ \tan^{-1}(\tan(A+B)) = \tan^{-1}\left(\frac{x+y}{1-xy}\right) $
For the left side $ \tan^{-1}(\tan \theta) $ to simplify directly to $ \theta $, the angle $ \theta = A+B $ must lie within the principal value branch of $ \tan^{-1} $, which is $ (-\frac{\pi}{2}, \frac{\pi}{2}) $. Since $ A \in (-\frac{\pi}{2}, \frac{\pi}{2}) $ and $ B \in (-\frac{\pi}{2}, \frac{\pi}{2}) $, the sum $ A+B $ is in the interval $ (-\pi, \pi) $.
$ -\frac{\pi}{2} < A < \frac{\pi}{2} $
$ -\frac{\pi}{2} < B < \frac{\pi}{2} $
Adding these inequalities: $ -\frac{\pi}{2} - \frac{\pi}{2} < A + B < \frac{\pi}{2} + \frac{\pi}{2} $
$ -\pi < A + B < \pi $.
We need to show that when $ xy < 1 $, the angle $ A+B $ is actually restricted to the interval $ (-\frac{\pi}{2}, \frac{\pi}{2}) $.
Consider the value of $ \tan(A+B) = \frac{x+y}{1-xy} $.
If $ xy < 1 $, then $ 1 - xy > 0 $.
If $ A+B $ were outside $ (-\frac{\pi}{2}, \frac{\pi}{2}) $ but within $ (-\pi, \pi) $, it would be in $ (-\pi, -\pi/2] \cup [\pi/2, \pi) $. In these intervals, $ \tan(A+B) $ is $\ge 0$ or $\le 0$ (or undefined at $\pm \pi/2$). Specifically, $ \tan(A+B) $ is negative in $ (-\pi, -\pi/2) $ and positive in $ (\pi/2, \pi) $. The condition $ xy < 1 $ relates to the sign of the denominator. If $ xy < 1 \iff 1 - xy > 0 $, then the sign of $ \tan(A+B) $ is determined by the sign of $ x+y $.
A more rigorous proof involves considering the cases based on the signs of $x$ and $y$ when $xy < 1$ or using the properties of the $\tan^{-1}$ function. For $xy < 1$, it can be shown that $ -\frac{\pi}{2} < A+B < \frac{\pi}{2} $. Thus, $ \tan^{-1}(\tan(A+B)) = A+B $.
So, when $ xy < 1 $:
$ A+B = \tan^{-1}\left(\frac{x+y}{1-xy}\right) $.
Substitute back $ A = \tan^{-1} x $ and $ B = \tan^{-1} y $:
$ \tan^{-1} x + \tan^{-1} y = \tan^{-1}\left(\frac{x+y}{1-xy}\right) $. (Proved for $xy < 1$)
The other cases for $ \tan^{-1} x + \tan^{-1} y $ (when $ xy > 1 $) arise because $ A+B $ falls outside the interval $ (-\pi/2, \pi/2) $, and we need to add or subtract $ \pi $ to bring the angle $ A+B $ to the equivalent angle in the principal value branch that has the same tangent value.
Example 2: Using Inverse Tangent Sum Formula
Example 2. Evaluate $ \tan^{-1} (2) + \tan^{-1} (3) $.
Answer:
Solution:
We want to evaluate $ \tan^{-1} (2) + \tan^{-1} (3) $. Let $ x = 2 $ and $ y = 3 $. The product $ xy = 2 \times 3 = 6 $. Since $ xy = 6 > 1 $ and $ x > 0, y > 0 $, we use the formula:
$ \tan^{-1} x + \tan^{-1} y = \pi + \tan^{-1}\left(\frac{x+y}{1-xy}\right) $
Substitute $ x = 2 $ and $ y = 3 $:
$ \tan^{-1} (2) + \tan^{-1} (3) = \pi + \tan^{-1}\left(\frac{2+3}{1-(2)(3)}\right) $
$ = \pi + \tan^{-1}\left(\frac{5}{1-6}\right) = \pi + \tan^{-1}\left(\frac{5}{-5}\right) $
$ = \pi + \tan^{-1}(-1) $
Now evaluate $ \tan^{-1}(-1) $. The principal value branch for $ \tan^{-1} $ is $ (-\pi/2, \pi/2) $. The angle in this range whose tangent is -1 is $ -\pi/4 $. ($ \tan(-\pi/4) = -1 $ and $ -\pi/4 \in (-\pi/2, \pi/2) $).
$ = \pi + (-\frac{\pi}{4}) $
$ = \pi - \frac{\pi}{4} = \frac{4\pi - \pi}{4} = \frac{3\pi}{4} $.
Therefore, $ \tan^{-1} (2) + \tan^{-1} (3) = \frac{3\pi}{4} $. Note that $ \frac{3\pi}{4} $ is in the interval $ (\pi/2, \pi) $. This is consistent with the fact that the sum of two angles in $ (0, \pi/2) $ (since $ \tan^{-1} 2 $ and $ \tan^{-1} 3 $ are both positive angles) can result in an angle greater than $ \pi/2 $.
Property Set 7: Double Angle Formulas for Inverse Tangent
These formulas express $ 2 \tan^{-1} x $ in terms of $ \tan^{-1} $, $ \sin^{-1} $, and $ \cos^{-1} $. They are derived from the tangent sum formula ($ \tan^{-1} x + \tan^{-1} x $) or the double angle formulas for sine and cosine in terms of tangent.
$ 2\tan^{-1} x = \tan^{-1}\left(\frac{2x}{1-x^2}\right) $, if $ |x| < 1 $ (i.e., $ -1 < x < 1 $).
$ 2\tan^{-1} x = \sin^{-1}\left(\frac{2x}{1+x^2}\right) $, if $ |x| \le 1 $ (i.e., $ -1 \le x \le 1 $).
$ 2\tan^{-1} x = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) $, if $ x \ge 0 $.
(Note: There are other forms for these identities with different conditions on $x$ that result in adding/subtracting $\pi$ or $2\pi$ to the right side to match the range of the left side).
Derivation for $2\tan^{-1} x = \sin^{-1}\left(\frac{2x}{1+x^2}\right)$:
To Prove: $ 2\tan^{-1} x = \sin^{-1}\left(\frac{2x}{1+x^2}\right) $, if $ |x| \le 1 $.
Proof:
Let $ \theta = \tan^{-1} x $. By definition, $ \tan \theta = x $.
The principal value branch for $ \tan^{-1} x $ is $ (-\frac{\pi}{2}, \frac{\pi}{2}) $, so $ \theta \in (-\frac{\pi}{2}, \frac{\pi}{2}) $.
Consider the double angle formula for sine in terms of tangent: $ \sin 2\theta = \frac{2 \tan \theta}{1 + \tan^2 \theta} $ (Equation 7.10).
Substitute $ \tan \theta = x $ into the right side:
$ \sin 2\theta = \frac{2x}{1+x^2} $.
Now, apply the $ \sin^{-1} $ function to both sides:
$ \sin^{-1}(\sin 2\theta) = \sin^{-1}\left(\frac{2x}{1+x^2}\right) $.
For the left side $ \sin^{-1}(\sin \phi) $ to simplify directly to $ \phi $, the angle $ \phi = 2\theta $ must lie within the principal value branch of $ \sin^{-1} $, which is $ [-\frac{\pi}{2}, \frac{\pi}{2}] $.
Since $ \theta \in (-\frac{\pi}{2}, \frac{\pi}{2}) $, multiplying by 2 gives the range of $ 2\theta $:
$ 2 \times (-\frac{\pi}{2}) < 2\theta < 2 \times (\frac{\pi}{2}) $
$ -\pi < 2\theta < \pi $.
For $ \sin^{-1}(\sin 2\theta) = 2\theta $, we need $ 2\theta \in [-\frac{\pi}{2}, \frac{\pi}{2}] $. This means:
$ -\frac{\pi}{2} \le 2\theta \le \frac{\pi}{2} $.
Divide the inequality by 2:
$ -\frac{\pi}{4} \le \theta \le \frac{\pi}{4} $.
Since $ \theta = \tan^{-1} x $, this condition on $ \theta $ means that $ \tan(-\frac{\pi}{4}) \le \tan(\theta) \le \tan(\frac{\pi}{4}) $, which simplifies to $ -1 \le x \le 1 $.
So, if $ |x| \le 1 $ (which means $ -1 \le x \le 1 $), then the angle $ 2\theta = 2\tan^{-1} x $ lies in the interval $ [-\pi/2, \pi/2] $, and therefore $ \sin^{-1}(\sin 2\theta) = 2\theta $.
Thus, when $ |x| \le 1 $:
$ 2\theta = \sin^{-1}\left(\frac{2x}{1+x^2}\right) $.
Substitute back $ \theta = \tan^{-1} x $:
$ 2\tan^{-1} x = \sin^{-1}\left(\frac{2x}{1+x^2}\right) $. (Proved for $ |x| \le 1 $)
Example 3: Using Double Angle Formula for Inverse Tangent
Example 3. Evaluate $ \sin^{-1} \left(\frac{2 \times 0.5}{1 + (0.5)^2}\right) $.
Answer:
Solution:
The expression is in the form $ \sin^{-1}\left(\frac{2x}{1+x^2}\right) $, where $ x = 0.5 $.
We check the condition for the identity $ 2\tan^{-1} x = \sin^{-1}\left(\frac{2x}{1+x^2}\right) $. The condition is $ |x| \le 1 $. Here $ x = 0.5 $, and $ |0.5| = 0.5 \le 1 $. So, the identity applies.
$ \sin^{-1}\left(\frac{2 \times 0.5}{1 + (0.5)^2}\right) = 2 \tan^{-1}(0.5) $.
We can evaluate this further if needed, but expressing it in terms of $ \tan^{-1}(0.5) $ is a simplification.
$ 2 \tan^{-1}(0.5) = 2 \tan^{-1}\left(\frac{1}{2}\right) $.
Let's verify the value by calculating the inside of the $\sin^{-1}$ directly:
$ \frac{2 \times 0.5}{1 + (0.5)^2} = \frac{1}{1 + 0.25} = \frac{1}{1.25} = \frac{1}{5/4} = \frac{4}{5} $.
So, we are evaluating $ \sin^{-1}(4/5) $. Let $ \phi = \sin^{-1}(4/5) $. Then $ \sin \phi = 4/5 $ and $ \phi \in [-\pi/2, \pi/2] $. Since $ 4/5 > 0 $, $\phi$ is in $(0, \pi/2)$.
If $ \tan \theta = 1/2 $, we can find $ \sin \theta $ and $ \cos \theta $ using a right triangle. Opposite = 1, Adjacent = 2. Hypotenuse = $ \sqrt{1^2 + 2^2} = \sqrt{5} $. So $ \sin \theta = 1/\sqrt{5} $ and $ \cos \theta = 2/\sqrt{5} $.
Then $ \sin 2\theta = 2 \sin \theta \cos \theta = 2 \times \frac{1}{\sqrt{5}} \times \frac{2}{\sqrt{5}} = \frac{4}{5} $. So $ 2\theta = \sin^{-1}(4/5) $ if $ 2\theta \in [-\pi/2, \pi/2] $. Since $ \tan^{-1}(1/2) $ is in $(0, \pi/2)$, $ 2\tan^{-1}(1/2) $ is in $(0, \pi) $. Wait, the condition $|x| \le 1$ for $2\tan^{-1} x = \sin^{-1}(\frac{2x}{1+x^2})$ ensures $2\tan^{-1} x$ is in $[-\pi/2, \pi/2]$. Since $x=0.5 \in [-1,1]$, $2\tan^{-1}(0.5)$ is indeed in $[-\pi/2, \pi/2]$.
So, $ \sin^{-1}(4/5) = 2 \tan^{-1}(1/2) $.
Property Set 8: Conversion between Inverse Functions
It is often useful to express one inverse trigonometric function in terms of another. This can be done using the definitions of the inverse functions and by referencing a right-angled triangle. This is similar to how we derived relationships between the original trigonometric ratios.
Consider the identity $ \sin^{-1} x = \cos^{-1} (\sqrt{1 - x^2}) $ for $ x \in [0, 1] $. Let $ y = \sin^{-1} x $. By definition, $ \sin y = x $ and $ y \in [-\pi/2, \pi/2] $. Since we assume $ x \in [0, 1] $, the angle $ y $ must be in $ [0, \pi/2] $.
Draw a right-angled triangle where $ y $ is one of the acute angles. Since $ \sin y = x = \frac{x}{1} $, we can label the side opposite to $ y $ as $x$ and the hypotenuse as 1.
Using the Pythagorean theorem, the adjacent side is $ \sqrt{1^2 - x^2} = \sqrt{1 - x^2} $.
Now, from this triangle, we can find the cosine and tangent of the angle $y$:
$ \cos y = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{\sqrt{1 - x^2}}{1} = \sqrt{1 - x^2} $
$ \tan y = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{x}{\sqrt{1 - x^2}} $
Since $ y = \sin^{-1} x $ and $ y \in [0, \pi/2] $ (which is in the principal range of $ \cos^{-1} $ and $ \tan^{-1} $ for positive arguments):
From $ \cos y = \sqrt{1 - x^2} $, applying $ \cos^{-1} $ gives $ y = \cos^{-1} (\sqrt{1 - x^2}) $. So, $ \sin^{-1} x = \cos^{-1} (\sqrt{1 - x^2}) $ for $ x \in [0, 1] $ (since $ y \in [0, \pi/2] \subset [0, \pi] $).
From $ \tan y = \frac{x}{\sqrt{1 - x^2}} $, applying $ \tan^{-1} $ gives $ y = \tan^{-1} \left(\frac{x}{\sqrt{1 - x^2}}\right) $. So, $ \sin^{-1} x = \tan^{-1} \left(\frac{x}{\sqrt{1 - x^2}}\right) $ for $ x \in [0, 1) $ (since $ y \in [0, \pi/2) $, $\tan y$ is defined, and range of $\tan^{-1}$ is $(-\pi/2, \pi/2)$).
Similar derivations can be performed starting with other inverse functions. It is important to consider the range of the initial inverse function's output and the target inverse function's range to determine the correct domain conditions for $x$ and whether any $\pi$ adjustments are needed, especially when $x$ is negative.
Some common conversion formulas (mostly for $ x \ge 0 $ or $ x > 0 $):
$\sin^{-1} x = \cos^{-1} (\sqrt{1 - x^2}) = \tan^{-1} \left(\frac{x}{\sqrt{1 - x^2}}\right) = \text{cosec}^{-1} \left(\frac{1}{x}\right) = \sec^{-1} \left(\frac{1}{\sqrt{1 - x^2}}\right) = \cot^{-1} \left(\frac{\sqrt{1 - x^2}}{x}\right)$ (with appropriate conditions on $x$ for each equality and the range of the angles).
$\cos^{-1} x = \sin^{-1} (\sqrt{1 - x^2}) = \tan^{-1} \left(\frac{\sqrt{1 - x^2}}{x}\right) = \dots$
$\tan^{-1} x = \sin^{-1} \left(\frac{x}{\sqrt{1 + x^2}}\right) = \cos^{-1} \left(\frac{1}{\sqrt{1 + x^2}}\right) = \dots$
Example 4: Conversion between Inverse Functions
Example 4. Express $\cos^{-1}(1/2)$ in terms of $\sin^{-1}$.
Answer:
Solution:
We want to express $ \cos^{-1}(1/2) $ in terms of $ \sin^{-1} $. Let $ y = \cos^{-1}(1/2) $. By definition, $ \cos y = 1/2 $ and $ y \in [0, \pi] $. Since $ 1/2 > 0 $, $ y $ is in $ [0, \pi/2] $.
We know that $ \cos(\pi/3) = 1/2 $, and $ \pi/3 \in [0, \pi] $, so $ y = \pi/3 $.
We need to find a value $z$ such that $ \sin^{-1} z = \pi/3 $. By definition, $ \sin(\pi/3) = z $.
$ z = \sin(\pi/3) = \frac{\sqrt{3}}{2} $.
So, $ \cos^{-1}(1/2) = \pi/3 $ and $ \sin^{-1}(\sqrt{3}/2) = \pi/3 $.
Therefore, $ \cos^{-1}(1/2) = \sin^{-1}(\sqrt{3}/2) $.
This is an instance of the general conversion $ \cos^{-1} x = \sin^{-1}(\sqrt{1-x^2}) $ for $ x \in [0, 1] $. Here $ x=1/2 $, $ \sqrt{1-x^2} = \sqrt{1-(1/2)^2} = \sqrt{1-1/4} = \sqrt{3/4} = \sqrt{3}/2 $.
Note for Competitive Exams
The properties in this section, especially the complementary identities (summing to $\pi/2$) and the sum/difference formulas for $\tan^{-1}$, are very important. Memorise them, paying close attention to the conditions on the variables (like $xy<1$ or $|x|\le 1$). The conversion formulas using right triangles are versatile; if you forget a specific identity, you can often derive it quickly by setting up a triangle based on the definition of one inverse function and then finding the other trigonometric ratios of that angle. Remember to always verify the result against the principal value branches. Practice using these properties to simplify expressions, evaluate values, and solve equations. The $\tan^{-1}$ sum/difference formulas are particularly common in simplification and equation problems.