Chapter 2 Inverse Trigonometric Functions (Concepts)

Inverse Trigonometric Functions

Trigonometric functions, also known as circular functions, are fundamental functions in mathematics that relate an angle of a right-angled triangle to the ratios of its side lengths. They are also widely used to model periodic phenomena like sound and light waves. There are six basic trigonometric functions.

1. Definitions based on a Right-Angled Triangle

For an acute angle $\theta$ in a right-angled triangle, the six trigonometric functions are defined as follows:

• Sine ($\sin \theta$): The ratio of the length of the side opposite the angle to the length of the hypotenuse.

  $\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}}$

• Cosine ($\cos \theta$): The ratio of the length of the adjacent side to the length of the hypotenuse.

  $\cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}}$

• Tangent ($\tan \theta$): The ratio of the length of the opposite side to the length of the adjacent side.

  $\tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{\sin \theta}{\cos \theta}$

• Cosecant ($\text{cosec } \theta$): The reciprocal of sine.

  $\text{cosec } \theta = \frac{\text{Hypotenuse}}{\text{Opposite}} = \frac{1}{\sin \theta}$

• Secant ($\sec \theta$): The reciprocal of cosine.

  $\sec \theta = \frac{\text{Hypotenuse}}{\text{Adjacent}} = \frac{1}{\cos \theta}$

• Cotangent ($\cot \theta$): The reciprocal of tangent.

  $\cot \theta = \frac{\text{Adjacent}}{\text{Opposite}} = \frac{1}{\tan \theta}$

2. Definitions based on the Unit Circle

The definitions can be extended to any angle using the unit circle (a circle with radius 1 centered at the origin). If we take a point P(x, y) on the unit circle that corresponds to an angle $\theta$, then:

• $\sin \theta = y$
• $\cos \theta = x$
• $\tan \theta = \frac{y}{x}, (x \neq 0)$
• $\text{cosec } \theta = \frac{1}{y}, (y \neq 0)$
• $\sec \theta = \frac{1}{x}, (x \neq 0)$
• $\cot \theta = \frac{x}{y}, (y \neq 0)$

3. Domain, Range, and Periodicity

The trigonometric functions are defined for most real numbers. Their key characteristic is their periodicity—their graphs repeat over regular intervals. This property is crucial for understanding why their inverses require special definitions.

In mathematics, the concept of an inverse function is crucial. A function $f: A \to B$ has an inverse function, denoted by $f^{-1}: B \to A$, if and only if $f$ is a bijective function. A bijective function must be both:

• One-to-one (Injective): Each element in the domain maps to a unique element in the range. No two inputs give the same output.
• Onto (Surjective): Every element in the range is mapped to by at least one element from the domain.

Let's consider the standard trigonometric functions (sine, cosine, tangent, etc.). These functions are inherently periodic. For instance, the sine function repeats its values every $2\pi$ radians. This means that many different input angles produce the same output value. For example, $\sin(0) = 0$, $\sin(\pi) = 0$, and $\sin(2\pi) = 0$. Since multiple inputs lead to the same output, the function is many-to-one over its natural domain ($R$), and therefore not one-to-one.

Because trigonometric functions are not one-to-one on their natural domains, they do not have inverse functions in the standard sense.

To overcome this, we restrict the domain of each trigonometric function to a specific interval where it becomes one-to-one, while still covering its entire original range (making it onto). This restricted, bijective function can now be inverted. The range of the resulting inverse function is this carefully chosen, restricted domain of the original function. This specific range is known as the principal value branch, and any value within this branch is called the principal value.

Definitions, Domain, and Principal Value Branch

Here we define each of the six inverse trigonometric functions by restricting the domain of their original counterparts.

1. Inverse Sine Function ($\sin^{-1} x$ or $\text{arcsin } x$)

• Original Function: $y = \sin x$ with Domain: $R$ and Range: $[-1, 1]$.
• Restricted Domain for Invertibility: $[-\frac{\pi}{2}, \frac{\pi}{2}]$. On this interval, $\sin x$ is one-to-one, and its range remains $[-1, 1]$.
• Definition: The inverse sine function, $y = \sin^{-1} x$, gives the angle $y$ in the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$ whose sine is $x$.

• Domain of $\sin^{-1} x$: $[-1, 1]$
• Principal Value Branch (Range) of $\sin^{-1} x$: $[-\frac{\pi}{2}, \frac{\pi}{2}]$

2. Inverse Cosine Function ($\cos^{-1} x$ or $\text{arccos } x$)

• Original Function: $y = \cos x$ with Domain: $R$ and Range: $[-1, 1]$.
• Restricted Domain for Invertibility: $[0, \pi]$. On this interval, $\cos x$ is one-to-one, and its range remains $[-1, 1]$.
• Definition: The inverse cosine function, $y = \cos^{-1} x$, gives the angle $y$ in the interval $[0, \pi]$ whose cosine is $x$.

• Domain of $\cos^{-1} x$: $[-1, 1]$
• Principal Value Branch (Range) of $\cos^{-1} x$: $[0, \pi]$

3. Inverse Tangent Function ($\tan^{-1} x$ or $\text{arctan } x$)

• Original Function: $y = \tan x$ with Domain: $R - \{ x : x = (n + \frac{1}{2})\pi, n \in Z \}$, Range: $R$.
• Restricted Domain for Invertibility: $(-\frac{\pi}{2}, \frac{\pi}{2})$. On this open interval, $\tan x$ is one-to-one, and its range is the entire set of real numbers, $R$.
• Definition: The inverse tangent function, $y = \tan^{-1} x$, gives the angle $y$ in the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$ whose tangent is $x$.

• Domain of $\tan^{-1} x$: $R$
• Principal Value Branch (Range) of $\tan^{-1} x$: $(-\frac{\pi}{2}, \frac{\pi}{2})$

4. Inverse Cosecant Function ($\text{cosec}^{-1} x$ or $\text{arccosec } x$)

• Original Function: $y = \text{cosec } x$, Range: $(-\infty, -1] \cup [1, \infty)$.
• Restricted Domain for Invertibility: $[-\frac{\pi}{2}, \frac{\pi}{2}] - \{0\}$. We exclude 0 because $\text{cosec } 0$ is undefined.
• Definition: The inverse cosecant function, $y = \text{cosec}^{-1} x$, gives the angle $y$ in $[-\frac{\pi}{2}, \frac{\pi}{2}] - \{0\}$ whose cosecant is $x$.

• Domain of $\text{cosec}^{-1} x$: $(-\infty, -1] \cup [1, \infty)$ or $|x| \ge 1$.
• Principal Value Branch (Range) of $\text{cosec}^{-1} x$: $[-\frac{\pi}{2}, \frac{\pi}{2}] - \{0\}$

5. Inverse Secant Function ($\sec^{-1} x$ or $\text{arcsec } x$)

• Original Function: $y = \sec x$, Range: $(-\infty, -1] \cup [1, \infty)$.
• Restricted Domain for Invertibility: $[0, \pi] - \{\frac{\pi}{2}\}$. We exclude $\frac{\pi}{2}$ because $\sec(\frac{\pi}{2})$ is undefined.
• Definition: The inverse secant function, $y = \sec^{-1} x$, gives the angle $y$ in $[0, \pi] - \{\frac{\pi}{2}\}$ whose secant is $x$.

• Domain of $\sec^{-1} x$: $(-\infty, -1] \cup [1, \infty)$ or $|x| \ge 1$.
• Principal Value Branch (Range) of $\sec^{-1} x$: $[0, \pi] - \{\frac{\pi}{2}\}$

6. Inverse Cotangent Function ($\cot^{-1} x$ or $\text{arccot } x$)

• Original Function: $y = \cot x$, Range: $R$.
• Restricted Domain for Invertibility: $(0, \pi)$.
• Definition: The inverse cotangent function, $y = \cot^{-1} x$, gives the angle $y$ in the interval $(0, \pi)$ whose cotangent is $x$.

• Domain of $\cot^{-1} x$: $R$
• Principal Value Branch (Range) of $\cot^{-1} x$: $(0, \pi)$

Summary Table of Inverse Trigonometric Functions

This table summarizes the domain and principal value branch (range) for all six inverse trigonometric functions. It is essential to memorize this for solving problems.

Graphs of Inverse Trigonometric Functions

The graph of an inverse function $f^{-1}$ is the reflection of the graph of the restricted function $f$ across the line $y=x$. Visualizing these graphs helps to solidify the understanding of their domains and ranges. To create an inverse function, the original trigonometric function must be restricted to a specific domain where it is one-to-one (i.e., passes the horizontal line test).

Graph of y = sin⁻¹(x)

To define the inverse sine function, we restrict the domain of the sine function, $y = \sin(x)$, to the interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$. In this interval, the function is one-to-one, and its range is $[-1, 1]$. The table of values for the restricted sine function is swapped to get the table for the inverse sine function.

Table of Values for y = sin⁻¹(x):

• The domain of $\sin^{-1}(x)$ is the range of the restricted $\sin(x)$, which is $[-1, 1]$.
• The range of $\sin^{-1}(x)$ (the principal value branch) is the restricted domain of $\sin(x)$, which is $[-\frac{\pi}{2}, \frac{\pi}{2}]$.

Graph of y = cos⁻¹(x)

To define the inverse cosine function, we restrict the domain of the cosine function, $y = \cos(x)$, to the interval $[0, \pi]$. In this interval, the function is one-to-one, and its range is $[-1, 1]$.

Table of Values for y = cos⁻¹(x):

• The domain of $\cos^{-1}(x)$ is $[-1, 1]$.
• The range of $\cos^{-1}(x)$ (the principal value branch) is $[0, \pi]$.

Graph of y = tan⁻¹(x)

To define the inverse tangent function, we restrict the domain of the tangent function, $y = \tan(x)$, to the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$. In this interval, the function is one-to-one, and its range is all real numbers, $R$. The vertical asymptotes of $\tan(x)$ at $x = \pm \frac{\pi}{2}$ become horizontal asymptotes for $\tan^{-1}(x)$ at $y = \pm \frac{\pi}{2}$.

Table of Values for y = tan⁻¹(x):

• The domain of $\tan^{-1}(x)$ is $R$ (all real numbers).
• The range of $\tan^{-1}(x)$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$.

Graph of y = cosec⁻¹(x)

To define the inverse cosecant function, we restrict the domain of $y = \text{cosec}(x)$ to $[-\frac{\pi}{2}, \frac{\pi}{2}] - \{0\}$. In this interval, the range is $(-\infty, -1] \cup [1, \infty)$.

Table of Values for y = cosec⁻¹(x):

• The domain of $\text{cosec}^{-1}(x)$ is $(-\infty, -1] \cup [1, \infty)$.
• The range of $\text{cosec}^{-1}(x)$ is $[-\frac{\pi}{2}, \frac{\pi}{2}] - \{0\}$.

Graph of y = sec⁻¹(x)

To define the inverse secant function, we restrict the domain of $y = \sec(x)$ to $[0, \pi] - \{\frac{\pi}{2}\}$. In this interval, the range is $(-\infty, -1] \cup [1, \infty)$.

Table of Values for y = sec⁻¹(x):

• The domain of $\sec^{-1}(x)$ is $(-\infty, -1] \cup [1, \infty)$.
• The range of $\sec^{-1}(x)$ is $[0, \pi] - \{\frac{\pi}{2}\}$.

Graph of y = cot⁻¹(x)

To define the inverse cotangent function, we restrict the domain of $y = \cot(x)$ to the interval $(0, \pi)$. In this interval, the range is all real numbers, $R$. The vertical asymptotes of $\cot(x)$ at $x=0$ and $x=\pi$ become horizontal asymptotes for $\cot^{-1}(x)$ at $y=0$ and $y=\pi$.

Table of Values for y = cot⁻¹(x):

• The domain of $\cot^{-1}(x)$ is $R$ (all real numbers).
• The range of $\cot^{-1}(x)$ is $(0, \pi)$.

Answer:

Answer:

Properties of Inverse Trigonometric Functions

Inverse trigonometric functions possess numerous properties and identities that are invaluable for simplifying expressions, solving equations, and evaluating definite integrals in higher mathematics. These properties are derived directly from the definitions of the inverse trigonometric functions and the properties of the original trigonometric functions, always keeping in mind the specific restricted domains and principal value branches.

Basic Properties (Inverse Identities)

These are fundamental properties that arise directly from the definition of a function and its inverse: $f^{-1}(f(x)) = x$ and $f(f^{-1}(x)) = x$, provided $x$ is within the appropriate domain.

However, for inverse trigonometric functions, the domain of the argument is crucial. The property $f^{-1}(f(x)) = x$ holds only if $x$ is in the principal value branch of the inverse function $f^{-1}$. The property $f(f^{-1}(x)) = x$ holds only if $x$ is in the domain of the inverse function $f^{-1}$.

Properties of the form $\text{inv}(\text{func}(x)) = x$:

For these properties to hold, the angle $x$ must lie within the principal value branch of the corresponding inverse trigonometric function.

1. $\sin^{-1}(\sin x) = x$, if $x \in [-\frac{\pi}{2}, \frac{\pi}{2}]$.

2. $\cos^{-1}(\cos x) = x$, if $x \in [0, \pi]$.

3. $\tan^{-1}(\tan x) = x$, if $x \in (-\frac{\pi}{2}, \frac{\pi}{2})$.

4. $\text{cosec}^{-1}(\text{cosec } x) = x$, if $x \in [-\frac{\pi}{2}, \frac{\pi}{2}] - \{0\}$.

5. $\sec^{-1}(\sec x) = x$, if $x \in [0, \pi] - \{\frac{\pi}{2}\}$.

6. $\cot^{-1}(\cot x) = x$, if $x \in (0, \pi)$.

Properties of the form $\text{func}(\text{inv}(x)) = x$:

For these properties to hold, the argument $x$ must lie within the domain of the corresponding inverse trigonometric function.

7. $\sin(\sin^{-1} x) = x$, if $x \in [-1, 1]$.

8. $\cos(\cos^{-1} x) = x$, if $x \in [-1, 1]$.

9. $\tan(\tan^{-1} x) = x$, if $x \in R$.

10. $\text{cosec}(\text{cosec}^{-1} x) = x$, if $x \in R - (-1, 1)$, i.e., $x \geq 1$ or $x \leq -1$.

11. $\sec(\sec^{-1} x) = x$, if $x \in R - (-1, 1)$, i.e., $x \geq 1$ or $x \leq -1$.

12. $\cot(\cot^{-1} x) = x$, if $x \in R$.

Answer:

Reciprocal Properties

These properties connect an inverse trigonometric function with the inverse of its reciprocal function. They are useful for converting between different inverse trigonometric forms.

1. $\sin^{-1} x = \text{cosec}^{-1} \frac{1}{x}$, for $x \in [-1, 1]$ and $x \neq 0$.
Equivalently, $\text{cosec}^{-1} x = \sin^{-1} \frac{1}{x}$, for $|x| \geq 1$.

2. $\cos^{-1} x = \sec^{-1} \frac{1}{x}$, for $x \in [-1, 1]$ and $x \neq 0$.
Equivalently, $\sec^{-1} x = \cos^{-1} \frac{1}{x}$, for $|x| \geq 1$.

3. $\tan^{-1} x = \cot^{-1} \frac{1}{x}$, for $x > 0$.

4. $\tan^{-1} x = \cot^{-1} \frac{1}{x} - \pi$, for $x < 0$.

5. $\cot^{-1} x = \tan^{-1} \frac{1}{x}$, for $x > 0$.

6. $\cot^{-1} x = \pi + \tan^{-1} \frac{1}{x}$, for $x < 0$.

Derivation of $\sin^{-1} x = \text{cosec}^{-1} \frac{1}{x}$:

To Prove: $\sin^{-1} x = \text{cosec}^{-1} \frac{1}{x}$, for $x \in [-1, 1], x \neq 0$.

Proof:

Let $y = \sin^{-1} x$.

By the definition of inverse sine, this implies:

Since $x \in [-1, 1]$ and $x \neq 0$, it implies $\sin y \neq 0$. The range of $\sin^{-1} x$ is $y \in [-\frac{\pi}{2}, \frac{\pi}{2}]$. Since $x \neq 0$, $y \neq \sin^{-1} 0 = 0$. So, $y \in [-\frac{\pi}{2}, \frac{\pi}{2}] - \{0\}$. This is the principal value branch for $\text{cosec}^{-1}$.

Taking the reciprocal of both sides of equation (i):

Using the reciprocal identity $\text{cosec } y = \frac{1}{\sin y}$:

Now, by the definition of the inverse cosecant function, $\text{cosec } y = \frac{1}{x}$ implies $y = \text{cosec}^{-1} \frac{1}{x}$, provided $y$ is in the principal value branch of $\text{cosec}^{-1} x$, which is $[-\frac{\pi}{2}, \frac{\pi}{2}] - \{0\}$.

As established earlier, $y \in [-\frac{\pi}{2}, \frac{\pi}{2}] - \{0\}$, so this condition is satisfied.

Since $y = \sin^{-1} x$ and $y = \text{cosec}^{-1} \frac{1}{x}$, we have $\sin^{-1} x = \text{cosec}^{-1} \frac{1}{x}$.

The derivation for $\text{cosec}^{-1} x = \sin^{-1} \frac{1}{x}$ for $|x| \geq 1$ follows similarly by letting $y = \text{cosec}^{-1} x$, so $\text{cosec } y = x$. This means $\sin y = 1/x$. Since $|x| \geq 1$, $1/x \in [-1, 1]$ and $1/x \neq 0$. The range of $\text{cosec}^{-1} x$ is $y \in [-\pi/2, \pi/2] - \{0\}$, which is the principal value branch of $\sin^{-1}$. So $y = \sin^{-1} \frac{1}{x}$.

Derivation of $\tan^{-1} x = \cot^{-1} \frac{1}{x}$ for $x > 0$:

To Prove: $\tan^{-1} x = \cot^{-1} \frac{1}{x}$, for $x > 0$.

Proof:

Let $y = \tan^{-1} x$.

By the definition of inverse tangent, this implies:

Since $x > 0$, it implies $\tan y > 0$. The range of $\tan^{-1} x$ is $y \in (-\frac{\pi}{2}, \frac{\pi}{2})$. For $\tan y$ to be positive, $y$ must be in the first quadrant, i.e., $y \in (0, \frac{\pi}{2})$. This interval $(0, \frac{\pi}{2})$ is a subset of the principal value branch for $\cot^{-1} x$, which is $(0, \pi)$.

Taking the reciprocal of both sides of equation (ii):

Using the reciprocal identity $\cot y = \frac{1}{\tan y}$:

Now, by the definition of the inverse cotangent function, $\cot y = \frac{1}{x}$ implies $y = \cot^{-1} \frac{1}{x}$, provided $y$ is in the principal value branch of $\cot^{-1} x$, which is $(0, \pi)$.

As established earlier, $y \in (0, \frac{\pi}{2})$, which is a subset of $(0, \pi)$. So, this condition is satisfied.

Since $y = \tan^{-1} x$ and $y = \cot^{-1} \frac{1}{x}$, we have $\tan^{-1} x = \cot^{-1} \frac{1}{x}$ for $x > 0$.

The property for $x < 0$ requires more careful consideration of the principal value branches, as $\tan^{-1} x$ will be in $(-\pi/2, 0)$ while $\cot^{-1} \frac{1}{x}$ (with $1/x < 0$) will be in $(\pi/2, \pi)$. They differ by $\pi$.

Complementary Angle Properties

These properties relate inverse trigonometric functions of complementary angles. They stem from the complementary angle identities of trigonometric functions (like $\sin(\pi/2 - \theta) = \cos \theta$).

1. $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$, for $x \in [-1, 1]$.

2. $\tan^{-1} x + \cot^{-1} x = \frac{\pi}{2}$, for $x \in R$.

3. $\text{cosec}^{-1} x + \sec^{-1} x = \frac{\pi}{2}$, for $|x| \geq 1$.

Derivation of $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$:

To Prove: $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$, for $x \in [-1, 1]$.

Proof:

Let $y = \sin^{-1} x$.

Since $x \in [-1, 1]$, by the definition of inverse sine, the value of $y$ lies in its principal value branch:

Also, from the definition, $\sin y = x$.

We know the complementary angle identity $\cos(\frac{\pi}{2} - \theta) = \sin \theta$. Using $\theta = y$:

Substitute $\sin y = x$ from equation (iii):

Now, for this equation to imply $\frac{\pi}{2} - y = \cos^{-1} x$, the angle $\frac{\pi}{2} - y$ must lie in the principal value branch of $\cos^{-1} x$, which is $[0, \pi]$. Let's check the range of $\frac{\pi}{2} - y$ based on the range of $y$:

Multiply the inequality by -1 and reverse the signs:

Add $\frac{\pi}{2}$ to all parts of the inequality:

The angle $\frac{\pi}{2} - y$ is indeed in the interval $[0, \pi]$. Therefore, from $\cos(\frac{\pi}{2} - y) = x$, we can take the inverse cosine on both sides:

Substitute $y = \sin^{-1} x$ back into the equation:

Rearrange the terms:

This proves the property for $x \in [-1, 1]$. Similar methods can be used to derive the other complementary angle properties, always ensuring the resulting angle lies in the appropriate principal value branch.

Negative Argument Properties

These properties deal with the inverse trigonometric function values for negative arguments. The behaviour depends on whether the function is 'odd-like' (symmetric about the origin in terms of the angle) or 'even-like' (symmetric about the y-axis in terms of the input value magnitude but related to $\pi$ in the angle). Note that only $\cos x, \sec x$ are even functions and $\sin x, \text{cosec } x, \tan x, \cot x$ are odd functions, but the inverse functions behave differently due to restricted domains.

1. $\sin^{-1}(-x) = -\sin^{-1} x$, for $x \in [-1, 1]$.

2. $\tan^{-1}(-x) = -\tan^{-1} x$, for $x \in R$.

3. $\text{cosec}^{-1}(-x) = -\text{cosec}^{-1} x$, for $|x| \geq 1$.

These three functions have principal value branches that are symmetric about 0 (or include 0), allowing the negative sign to simply come out.

4. $\cos^{-1}(-x) = \pi - \cos^{-1} x$, for $x \in [-1, 1]$.

5. $\cot^{-1}(-x) = \pi - \cot^{-1} x$, for $x \in R$.

6. $\sec^{-1}(-x) = \pi - \sec^{-1} x$, for $|x| \geq 1$.

These three functions have principal value branches that are in $[0, \pi]$ or $(0, \pi)$ or subsets thereof. When the input is negative, the corresponding angle is found in the second quadrant, which is why the form $\pi - \theta$ appears.

Derivation of $\cos^{-1}(-x) = \pi - \cos^{-1} x$:

To Prove: $\cos^{-1}(-x) = \pi - \cos^{-1} x$, for $x \in [-1, 1]$.

Proof:

Let $y = \cos^{-1}(-x)$.

Since $x \in [-1, 1]$, $-x \in [-1, 1]$. By the definition of inverse cosine, the value of $y$ lies in its principal value branch:

Also, from the definition, $\cos y = -x$.

From equation (iv), we have $x = -\cos y$.

We know the trigonometric identity $\cos(\pi - \theta) = -\cos \theta$. Using $\theta = y$:

Substitute $-\cos y = x$:

Now, for this equation to imply $\pi - y = \cos^{-1} x$, the angle $\pi - y$ must lie in the principal value branch of $\cos^{-1} x$, which is $[0, \pi]$. Let's check the range of $\pi - y$ based on the range of $y$:

Multiply the inequality by -1 and reverse the signs:

Add $\pi$ to all parts of the inequality:

The angle $\pi - y$ is indeed in the interval $[0, \pi]$. Therefore, from $\cos(\pi - y) = x$, we can take the inverse cosine on both sides:

Substitute $y = \cos^{-1}(-x)$ back into the equation:

Rearrange the terms:

This proves the property for $x \in [-1, 1]$. Similar methods apply to $\cot^{-1}(-x)$ and $\sec^{-1}(-x)$.

Conversion Properties

These properties allow us to express one inverse trigonometric function in terms of another. They are particularly useful for simplifying expressions or evaluating them. For positive values of $x$, these conversions can be easily visualized and derived using a right-angled triangle.

Derivation using a Right-Angled Triangle (for x > 0)

The right-angled triangle method is a powerful visual tool for finding conversion formulas when the argument $x$ is positive. Let's see how it works.

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

Let $\theta = \sin^{-1}(x)$. By definition, this means $\sin(\theta) = x$. We can write $x$ as a fraction, $x = \frac{x}{1}$.

Since $\sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}}$, we can construct a right-angled triangle where:

• Opposite side = $x$
• Hypotenuse = $1$

Using the Pythagorean theorem (Adjacent$^2$ + Opposite$^2$ = Hypotenuse$^2$), we can find the adjacent side:

Now, from this triangle, we can find the other trigonometric ratios for the angle $\theta$:

Since $\theta = \sin^{-1}(x)$, we can substitute it back. For $x \in (0, 1)$, $\theta$ is in $(0, \frac{\pi}{2})$, which is within the principal value range for $\cos^{-1}$ and $\tan^{-1}$. Thus, we can write:

2. Converting from $\tan^{-1}(x)$

Let $\theta = \tan^{-1}(x)$. This means $\tan(\theta) = x = \frac{x}{1}$.

Since $\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}}$, we can construct a right-angled triangle where:

• Opposite side = $x$
• Adjacent side = $1$

Using the Pythagorean theorem, we find the hypotenuse:

Now, we find the other trigonometric ratios for $\theta$:

Since $\theta = \tan^{-1}(x)$, we can conclude (for $x > 0$):

Note on Negative Values: The triangle method works best for $x>0$. For negative values of $x$, one must be careful with the signs and the principal value ranges. It is often easier to use identities like $\sin^{-1}(-x) = -\sin^{-1}(x)$ and $\tan^{-1}(-x) = -\tan^{-1}(x)$ before applying the conversion.

Summary of Conversion Formulas

The following table lists the most common conversion formulas. It is important to be mindful of the valid range of $x$ for each formula, as the expression might change outside that range to keep the result within the principal value branch.

Sum and Difference Formulas

These properties are used to combine or simplify expressions involving the sum or difference of inverse trigonometric functions. They are analogous to the sum and difference formulas in standard trigonometry (e.g., $\sin(A+B)$). The conditions placed on the variables $x$ and $y$ are extremely important, as they determine which version of the formula to use. These conditions ensure that the final result lies within the correct principal value branch of the resulting inverse function.

Formulas involving $\tan^{-1}(x)$ and $\tan^{-1}(y)$

These are the most commonly used and important sum and difference formulas in this topic.

1. $\tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}\left(\frac{x+y}{1-xy}\right)$, if $xy < 1$.

2. $\tan^{-1}(x) + \tan^{-1}(y) = \pi + \tan^{-1}\left(\frac{x+y}{1-xy}\right)$, if $x > 0, y > 0, xy > 1$.

3. $\tan^{-1}(x) + \tan^{-1}(y) = -\pi + \tan^{-1}\left(\frac{x+y}{1-xy}\right)$, if $x < 0, y < 0, xy > 1$.

4. $\tan^{-1}(x) - \tan^{-1}(y) = \tan^{-1}\left(\frac{x-y}{1+xy}\right)$, if $xy > -1$.

Derivation of the Primary Formula (Case 1)

To Prove: $\tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}\left(\frac{x+y}{1-xy}\right)$, for the condition $xy < 1$.

Proof:

Let $A = \tan^{-1}(x)$ and $B = \tan^{-1}(y)$.

From trigonometry, we know the tangent addition formula:

Substituting $\tan(A) = x$ and $\tan(B) = y$ into this identity:

Now, we can take the inverse tangent of both sides. However, the identity $\tan^{-1}(\tan(\theta)) = \theta$ is only true if $\theta$ is within the principal value branch of $\tan^{-1}$, which is $(-\frac{\pi}{2}, \frac{\pi}{2})$.

The condition $xy < 1$ is precisely what ensures that the sum of the angles, $A+B$, falls within this required range. When this condition is met, we can directly apply $\tan^{-1}$ to both sides of equation (i):

Now, substitute the original definitions of A and B back into the equation:

Why are the other cases needed? (e.g., Formula 2)

Let's see what happens when the condition $xy < 1$ is not met. Consider $x > 0, y > 0$ and $xy > 1$. For example, let $x=2$ and $y=3$.

$A = \tan^{-1}(2)$ is in $(0, \frac{\pi}{2})$. Since $\tan(\frac{\pi}{2})$ is undefined, $A < \frac{\pi}{2}$.

$B = \tan^{-1}(3)$ is in $(0, \frac{\pi}{2})$. Since $\tan(\frac{\pi}{2})$ is undefined, $B < \frac{\pi}{2}$.

Since $\tan(\frac{\pi}{4}) = 1$, we know that $\tan^{-1}(2) > \frac{\pi}{4}$ and $\tan^{-1}(3) > \frac{\pi}{4}$.

Therefore, their sum $A+B = \tan^{-1}(2) + \tan^{-1}(3) > \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}$.

The sum $A+B$ is now in the interval $(\frac{\pi}{2}, \pi)$, which is outside the principal branch $(-\frac{\pi}{2}, \frac{\pi}{2})$. So we cannot simply take the inverse tangent. We use the fact that $\tan(\theta) = \tan(\theta - \pi)$.

From equation (i), we have $\tan(A+B) = \frac{x+y}{1-xy}$.

So, $\tan(A+B-\pi) = \tan(A+B) = \frac{x+y}{1-xy}$.

Since $A+B$ is in $(\frac{\pi}{2}, \pi)$, the new angle $(A+B-\pi)$ is in $(-\frac{\pi}{2}, 0)$, which is inside the principal branch. Now we can take the inverse tangent:

$A+B-\pi = \tan^{-1}\left(\frac{x+y}{1-xy}\right)$, which rearranges to $A+B = \pi + \tan^{-1}\left(\frac{x+y}{1-xy}\right)$. This explains Formula 2.

Formulas involving $\sin^{-1}(x)$ and $\cos^{-1}(x)$

These formulas are derived using a similar process based on the identities for $\sin(A \pm B)$ and $\cos(A \pm B)$. The conditions are often more complex due to the bounded ranges $[-\frac{\pi}{2}, \frac{\pi}{2}]$ and $[0, \pi]$. Below are the primary versions of the formulas, which are valid under the most common conditions.

Sum and Difference for $\sin^{-1}$ (for $x, y \in [-1, 1]$):

Sum and Difference for $\cos^{-1}$ (for $x, y \in [-1, 1]$):

Note: For most problems, you will encounter values of $x$ and $y$ for which these primary formulas are sufficient. Always double-check if the conditions are met before applying them.

Double and Triple Angle Formulas

These properties are derived directly from the double and triple angle identities in trigonometry (e.g., $\sin(2\theta)$, $\cos(3\theta)$). They are used to simplify expressions like $2\tan^{-1}(x)$ or $3\sin^{-1}(x)$. As with the sum and difference formulas, verifying that the resulting angle lies in the appropriate principal value branch is crucial for the validity of the formula, leading to different formulas for different ranges of $x$.

Double Angle Formulas

The most versatile double angle formulas are those for $2\tan^{-1}(x)$, as they can be converted into $\sin^{-1}$, $\cos^{-1}$, and $\tan^{-1}$ forms.

1. Formulas for $2\tan^{-1}(x)$

• $2\tan^{-1}(x) = \sin^{-1}\left(\frac{2x}{1+x^2}\right)$, if $|x| \le 1$.
• $2\tan^{-1}(x) = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)$, if $x \ge 0$.
• $2\tan^{-1}(x) = \tan^{-1}\left(\frac{2x}{1-x^2}\right)$, if $|x| < 1$.

(Note: There are other forms of these formulas for cases like $x > 1$ or $x < -1$, which involve adding or subtracting $\pi$.)

Derivation of $2\tan^{-1}(x) = \sin^{-1}\left(\frac{2x}{1+x^2}\right)$ for $|x| \le 1$:

To Prove: $2\tan^{-1}(x) = \sin^{-1}\left(\frac{2x}{1+x^2}\right)$, if $|x| \le 1$.

Proof:

Let $\theta = \tan^{-1}(x)$, which implies $\tan(\theta) = x$. The condition $|x| \le 1$ means $-1 \le x \le 1$.

Since $\tan(\frac{\pi}{4}) = 1$ and $\tan(-\frac{\pi}{4}) = -1$, this means $-\frac{\pi}{4} \le \theta \le \frac{\pi}{4}$.

Multiplying by 2, we get the range for $2\theta$:

This range for $2\theta$ is exactly the principal value branch of the $\sin^{-1}$ function.

Now, consider the trigonometric identity for $\sin(2\theta)$ in terms of $\tan(\theta)$:

Substitute $\tan(\theta) = x$:

Since $2\theta$ lies in the principal branch of $\sin^{-1}$, we can take the inverse sine of both sides:

Finally, substitute back $\theta = \tan^{-1}(x)$:

This proves the formula for the given condition.

2. Other Double Angle Formulas

• $2\sin^{-1}(x) = \sin^{-1}(2x\sqrt{1-x^2})$, if $|x| \le \frac{1}{\sqrt{2}}$.
• $2\cos^{-1}(x) = \cos^{-1}(2x^2 - 1)$, if $0 \le x \le 1$.

Triple Angle Formulas

These are derived from the triple angle identities ($\sin(3\theta)$, $\cos(3\theta)$, $\tan(3\theta)$) using a similar method of substitution and checking if the resulting angle ($3\theta$) lies within the principal value branch.

1. Formulas for $3\sin^{-1}(x)$, $3\cos^{-1}(x)$, $3\tan^{-1}(x)$

• $3\sin^{-1}(x) = \sin^{-1}(3x - 4x^3)$, if $|x| \le \frac{1}{2}$.
• $3\cos^{-1}(x) = \cos^{-1}(4x^3 - 3x)$, if $x \in [\frac{1}{2}, 1]$.
• $3\tan^{-1}(x) = \tan^{-1}\left(\frac{3x-x^3}{1-3x^2}\right)$, if $|x| < \frac{1}{\sqrt{3}}$.

Derivation of $3\sin^{-1}(x) = \sin^{-1}(3x - 4x^3)$ for $|x| \le \frac{1}{2}$:

To Prove: $3\sin^{-1}(x) = \sin^{-1}(3x - 4x^3)$, if $|x| \le \frac{1}{2}$.

Proof:

Let $\theta = \sin^{-1}(x)$, which implies $\sin(\theta) = x$.

The condition $|x| \le \frac{1}{2}$ means $-\frac{1}{2} \le x \le \frac{1}{2}$.

Applying $\sin^{-1}$ to this inequality:

Multiplying by 3, we get the range for $3\theta$:

This range for $3\theta$ is the principal value branch of the $\sin^{-1}$ function.

Now, consider the trigonometric identity for $\sin(3\theta)$:

Substitute $\sin(\theta) = x$:

Since $3\theta$ lies in the principal branch of $\sin^{-1}$, we can take the inverse sine of both sides:

Finally, substitute back $\theta = \sin^{-1}(x)$:

This proves the property for the given condition.

Answer: