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Introduction

Inverse trigonometric functions are the inverse functions of the basic trigonometric functions (sine, cosine, tangent, cotangent, secant, and cosecant). They play a crucial role in various fields such as engineering, physics, and mathematics. In the context of JEE Main Mathematics, understanding these functions is essential for solving problems related to trigonometry, calculus, and geometry.

Basic Concepts

Definition

Inverse trigonometric functions allow us to find the angle whose trigonometric function value is given. They are denoted as follows:

Inverse Sine: $ \sin^{-1}(x) $ or $ \arcsin(x) $
Inverse Cosine: $ \cos^{-1}(x) $ or $ \arccos(x) $
Inverse Tangent: $ \tan^{-1}(x) $ or $ \arctan(x) $
Inverse Cotangent: $ \cot^{-1}(x) $ or $ \arccot(x) $
Inverse Secant: $ \sec^{-1}(x) $ or $ \arcsec(x) $
Inverse Cosecant: $ \csc^{-1}(x) $ or $ \arccsc(x) $

Principal Value Branch

The principal value branch is the range of values for which the inverse trigonometric functions are defined. It ensures that each inverse function gives a unique output.

$ \sin^{-1}(x) $: $ -\frac{\pi}{2} \leq \sin^{-1}(x) \leq \frac{\pi}{2} $
$ \cos^{-1}(x) $: $ 0 \leq \cos^{-1}(x) \leq \pi $
$ \tan^{-1}(x) $: $ -\frac{\pi}{2}

< \tan^{-1}(x) < \frac{\pi}{2} $

$ \cot^{-1}(x) $: $ 0 < \cot^{-1}(x) < \pi $
$ \sec^{-1}(x) $: $ 0 \leq \sec^{-1}(x) \leq \pi $, $ \sec^{-1}(x) \neq \frac{\pi}{2} $
$ \csc^{-1}(x) $: $ -\frac{\pi}{2} \leq \csc^{-1}(x) \leq \frac{\pi}{2} $, $ \csc^{-1}(x) \neq 0 $

Note

The principal value branch is crucial for ensuring the one-to-one nature of inverse trigonometric functions.

Properties

Domain and Range

$ \sin^{-1}(x) $: Domain $[-1, 1]$, Range $[-\frac{\pi}{2}, \frac{\pi}{2}]$
$ \cos^{-1}(x) $: Domain $[-1, 1]$, Range $[0, \pi]$
$ \tan^{-1}(x) $: Domain $(-\infty, \infty)$, Range $(-\frac{\pi}{2}, \frac{\pi}{2})$
$ \cot^{-1}(x) $: Domain $(-\infty, \infty)$, Range $(0, \pi)$
$ \sec^{-1}(x) $: Domain $(-\infty, -1] \cup [1, \infty)$, Range $[0, \pi]$, $ \sec^{-1}(x) \neq \frac{\pi}{2}$
$ \csc^{-1}(x) $: Domain $(-\infty, -1] \cup [1, \infty)$, Range $[-\frac{\pi}{2}, \frac{\pi}{2}]$, $ \csc^{-1}(x) \neq 0$

Symmetry and Periodicity

Inverse trigonometric functions exhibit certain symmetries:

$ \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) $
$ \csc^{-1}(-x) = -\csc^{-1}(x) $

Important Formulas

Basic Identities

$ \sin(\sin^{-1}(x)) = x $ for $ x \in [-1, 1] $
$ \cos(\cos^{-1}(x)) = x $ for $ x \in [-1, 1] $
$ \tan(\tan^{-1}(x)) = x $ for $ x \in (-\infty, \infty) $
$ \cot(\cot^{-1}(x)) = x $ for $ x \in (-\infty, \infty) $
$ \sec(\sec^{-1}(x)) = x $ for $ x \in (-\infty, -1] \cup [1, \infty) $
$ \csc(\csc^{-1}(x)) = x $ for $ x \in (-\infty, -1] \cup [1, \infty) $

Composite Functions

$ \sin(\cos^{-1}(x)) = \sqrt{1 - x^2} $
$ \cos(\sin^{-1}(x)) = \sqrt{1 - x^2} $
$ \tan(\cos^{-1}(x)) = \frac{\sqrt{1 - x^2}}{x} $
$ \cot(\sin^{-1}(x)) = \frac{\sqrt{1 - x^2}}{x} $

Example

Find the value of $ \sin(\cos^{-1}(\frac{3}{5})) $.

Using the identity $ \sin(\cos^{-1}(x)) = \sqrt{1 - x^2} $: $$ \sin(\cos^{-1}(\frac{3}{5})) = \sqrt{1 - (\frac{3}{5})^2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5} $$

Derivatives and Integrals

Derivatives

The derivatives of inverse trigonometric functions are important in calculus. Here are the derivatives:

$ \frac{d}{dx} (\sin^{-1}(x)) = \frac{1}{\sqrt{1 - x^2}} $
$ \frac{d}{dx} (\cos^{-1}(x)) = -\frac{1}{\sqrt{1 - x^2}} $
$ \frac{d}{dx} (\tan^{-1}(x)) = \frac{1}{1 + x^2} $
$ \frac{d}{dx} (\cot^{-1}(x)) = -\frac{1}{1 + x^2} $
$ \frac{d}{dx} (\sec^{-1}(x)) = \frac{1}{|x|\sqrt{x^2 - 1}} $
$ \frac{d}{dx} (\csc^{-1}(x)) = -\frac{1}{|x|\sqrt{x^2 - 1}} $

Integrals

The integrals involving inverse trigonometric functions are also frequently encountered:

$ \int \frac{1}{\sqrt{1 - x^2}} dx = \sin^{-1}(x) + C $
$ \int \frac{1}{\sqrt{1 - x^2}} dx = -\cos^{-1}(x) + C $
$ \int \frac{1}{1 + x^2} dx = \tan^{-1}(x) + C $
$ \int \frac{1}{1 + x^2} dx = -\cot^{-1}(x) + C $
$ \int \frac{1}{x\sqrt{x^2 - 1}} dx = \sec^{-1}(x) + C $
$ \int \frac{1}{x\sqrt{x^2 - 1}} dx = -\csc^{-1}(x) + C $

Note

The constant of integration $C$ is added because the integral represents an indefinite integral.

Applications

Solving Equations

Inverse trigonometric functions are often used to solve trigonometric equations. For example:

Example

Solve $ \sin(x) = \frac{1}{2} $.

Using the inverse sine function: $$ x = \sin^{-1}(\frac{1}{2}) $$ From the principal value branch: $$ x = \frac{\pi}{6} \text{ or } x = \pi - \frac{\pi}{6} = \frac{5\pi}{6} $$

Calculus

Inverse trigonometric functions are used in calculus for finding derivatives and integrals, as shown in the previous sections.

Geometry

In geometry, these functions help in finding angles in right-angled triangles when the lengths of the sides are known.

Tip

Always remember the principal value branches to avoid confusion while solving problems involving inverse trigonometric functions.

Conclusion

Inverse trigonometric functions are a fundamental part of the JEE Main Mathematics syllabus. A thorough understanding of their properties, identities, derivatives, and integrals is crucial for solving a wide range of mathematical problems. Practice is key to mastering these concepts, and always pay attention to the principal value branches to avoid common mistakes.

Common Mistake

A common mistake is to forget the restrictions on the domain and range of inverse trigonometric functions, leading to incorrect solutions.

By breaking down complex ideas into smaller, manageable parts, and practicing regularly, you can gain a solid grasp of inverse trigonometric functions and excel in your JEE Main Mathematics exam.

Introduction

Basic Concepts

Definition

Inverse trigonometric functions allow us to find the angle whose trigonometric function value is given. They are denoted as follows:

Inverse Sine: $ \sin^{-1}(x) $ or $ \arcsin(x) $
Inverse Cosine: $ \cos^{-1}(x) $ or $ \arccos(x) $
Inverse Tangent: $ \tan^{-1}(x) $ or $ \arctan(x) $
Inverse Cotangent: $ \cot^{-1}(x) $ or $ \arccot(x) $
Inverse Secant: $ \sec^{-1}(x) $ or $ \arcsec(x) $
Inverse Cosecant: $ \csc^{-1}(x) $ or $ \arccsc(x) $

Principal Value Branch

The principal value branch is the range of values for which the inverse trigonometric functions are defined. It ensures that each inverse function gives a unique output.

$ \sin^{-1}(x) $: $ -\frac{\pi}{2} \leq \sin^{-1}(x) \leq \frac{\pi}{2} $
$ \cos^{-1}(x) $: $ 0 \leq \cos^{-1}(x) \leq \pi $
$ \tan^{-1}(x) $: $ -\frac{\pi}{2}

< \tan^{-1}(x) < \frac{\pi}{2} $

$ \cot^{-1}(x) $: $ 0 < \cot^{-1}(x) < \pi $
$ \sec^{-1}(x) $: $ 0 \leq \sec^{-1}(x) \leq \pi $, $ \sec^{-1}(x) \neq \frac{\pi}{2} $
$ \csc^{-1}(x) $: $ -\frac{\pi}{2} \leq \csc^{-1}(x) \leq \frac{\pi}{2} $, $ \csc^{-1}(x) \neq 0 $

Note

The principal value branch is crucial for ensuring the one-to-one nature of inverse trigonometric functions.

Properties

Domain and Range

$ \sin^{-1}(x) $: Domain $[-1, 1]$, Range $[-\frac{\pi}{2}, \frac{\pi}{2}]$
$ \cos^{-1}(x) $: Domain $[-1, 1]$, Range $[0, \pi]$
$ \tan^{-1}(x) $: Domain $(-\infty, \infty)$, Range $(-\frac{\pi}{2}, \frac{\pi}{2})$
$ \cot^{-1}(x) $: Domain $(-\infty, \infty)$, Range $(0, \pi)$
$ \sec^{-1}(x) $: Domain $(-\infty, -1] \cup [1, \infty)$, Range $[0, \pi]$, $ \sec^{-1}(x) \neq \frac{\pi}{2}$
$ \csc^{-1}(x) $: Domain $(-\infty, -1] \cup [1, \infty)$, Range $[-\frac{\pi}{2}, \frac{\pi}{2}]$, $ \csc^{-1}(x) \neq 0$

Symmetry and Periodicity

Inverse trigonometric functions exhibit certain symmetries:

$ \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) $
$ \csc^{-1}(-x) = -\csc^{-1}(x) $

Important Formulas

Basic Identities

$ \sin(\sin^{-1}(x)) = x $ for $ x \in [-1, 1] $
$ \cos(\cos^{-1}(x)) = x $ for $ x \in [-1, 1] $
$ \tan(\tan^{-1}(x)) = x $ for $ x \in (-\infty, \infty) $
$ \cot(\cot^{-1}(x)) = x $ for $ x \in (-\infty, \infty) $
$ \sec(\sec^{-1}(x)) = x $ for $ x \in (-\infty, -1] \cup [1, \infty) $
$ \csc(\csc^{-1}(x)) = x $ for $ x \in (-\infty, -1] \cup [1, \infty) $

Composite Functions

$ \sin(\cos^{-1}(x)) = \sqrt{1 - x^2} $
$ \cos(\sin^{-1}(x)) = \sqrt{1 - x^2} $
$ \tan(\cos^{-1}(x)) = \frac{\sqrt{1 - x^2}}{x} $
$ \cot(\sin^{-1}(x)) = \frac{\sqrt{1 - x^2}}{x} $

Example

Find the value of $ \sin(\cos^{-1}(\frac{3}{5})) $.

Using the identity $ \sin(\cos^{-1}(x)) = \sqrt{1 - x^2} $: $$ \sin(\cos^{-1}(\frac{3}{5})) = \sqrt{1 - (\frac{3}{5})^2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5} $$

Derivatives and Integrals

Derivatives

The derivatives of inverse trigonometric functions are important in calculus. Here are the derivatives:

$ \frac{d}{dx} (\sin^{-1}(x)) = \frac{1}{\sqrt{1 - x^2}} $
$ \frac{d}{dx} (\cos^{-1}(x)) = -\frac{1}{\sqrt{1 - x^2}} $
$ \frac{d}{dx} (\tan^{-1}(x)) = \frac{1}{1 + x^2} $
$ \frac{d}{dx} (\cot^{-1}(x)) = -\frac{1}{1 + x^2} $
$ \frac{d}{dx} (\sec^{-1}(x)) = \frac{1}{|x|\sqrt{x^2 - 1}} $
$ \frac{d}{dx} (\csc^{-1}(x)) = -\frac{1}{|x|\sqrt{x^2 - 1}} $

Integrals

The integrals involving inverse trigonometric functions are also frequently encountered:

$ \int \frac{1}{\sqrt{1 - x^2}} dx = \sin^{-1}(x) + C $
$ \int \frac{1}{\sqrt{1 - x^2}} dx = -\cos^{-1}(x) + C $
$ \int \frac{1}{1 + x^2} dx = \tan^{-1}(x) + C $
$ \int \frac{1}{1 + x^2} dx = -\cot^{-1}(x) + C $
$ \int \frac{1}{x\sqrt{x^2 - 1}} dx = \sec^{-1}(x) + C $
$ \int \frac{1}{x\sqrt{x^2 - 1}} dx = -\csc^{-1}(x) + C $

Note

The constant of integration $C$ is added because the integral represents an indefinite integral.

Applications

Solving Equations

Inverse trigonometric functions are often used to solve trigonometric equations. For example:

Example

Solve $ \sin(x) = \frac{1}{2} $.

Using the inverse sine function: $$ x = \sin^{-1}(\frac{1}{2}) $$ From the principal value branch: $$ x = \frac{\pi}{6} \text{ or } x = \pi - \frac{\pi}{6} = \frac{5\pi}{6} $$

Calculus

Inverse trigonometric functions are used in calculus for finding derivatives and integrals, as shown in the previous sections.

Geometry

In geometry, these functions help in finding angles in right-angled triangles when the lengths of the sides are known.

Tip

Always remember the principal value branches to avoid confusion while solving problems involving inverse trigonometric functions.

Conclusion

Common Mistake

A common mistake is to forget the restrictions on the domain and range of inverse trigonometric functions, leading to incorrect solutions.

By breaking down complex ideas into smaller, manageable parts, and practicing regularly, you can gain a solid grasp of inverse trigonometric functions and excel in your JEE Main Mathematics exam.

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Algebra11 subtopics

Calculus8 subtopics

Coordinate Geometry5 subtopics

Trigonometry4 subtopics

Differential Calculus1 subtopic

Inverse Trigonometric Functions Notes

Introduction

Basic Concepts

Definition

Principal Value Branch

Properties

Domain and Range

Symmetry and Periodicity

Important Formulas

Basic Identities

Composite Functions

Derivatives and Integrals

Derivatives

Integrals

Applications

Solving Equations

Calculus

Geometry

Conclusion

Algebra11 subtopics

Calculus8 subtopics

Coordinate Geometry5 subtopics

Trigonometry4 subtopics

Differential Calculus1 subtopic

Introduction

Basic Concepts

Definition

Principal Value Branch

Properties

Domain and Range

Symmetry and Periodicity

Important Formulas

Basic Identities

Composite Functions

Derivatives and Integrals

Derivatives

Integrals

Applications

Solving Equations

Calculus

Geometry

Conclusion

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