TOPIC 4 OF 16

Definitions and Principal Value Branches of Inverse Trig Functions

Q: What are inverse trigonometric functions?

Inverse trigonometric functions reverse trigonometric operations. Since trig functions are not one-one over their full domain, we restrict domains to make them invertible with specific principal value branches.

Q: What is the principal value branch?

The principal value branch is the restricted range that makes an inverse trig function a proper function. For sin-inverse it is [-pi/2, pi/2]; for cos-inverse it is [0, pi]; for tan-inverse it is (-pi/2, pi/2).

Q: What are the domains and ranges of inverse trig functions?

sin-inverse: domain [-1,1], range [-pi/2, pi/2]. cos-inverse: domain [-1,1], range [0, pi]. tan-inverse: domain R, range (-pi/2, pi/2). Similar for cosec, sec, and cot inverse.

Q: How do you find principal values?

Find the angle in the principal branch whose trig value matches the given number. For example, sin-inverse(1/2) = pi/6 because sin(pi/6) = 1/2 and pi/6 is in [-pi/2, pi/2].

Q: Why do we need restricted domains for inverse trig functions?

Trigonometric functions are periodic and not one-one, so they cannot have inverses. Restricting to one period where the function is bijective creates an invertible function.

🎓 Class 12 Mathematics CBSE Theory Ch 2 — Inverse Trigonometric Functions ⏱ ~25 min

🌐 Language: [gtranslate]

🧠 AI-Powered MCQ Assessment ▲

This MCQ module is based on: Definitions and Principal Value Branches of Inverse Trig Functions

No. of Questions:

Difficulty Level:

📐 Maths Assessment ▲

This mathematics assessment will be based on: Definitions and Principal Value Branches of Inverse Trig Functions
Targeting Class 12 level in Trigonometry, with Advanced difficulty.

Question Type:

Number of Questions:

Upload Additional Content (Optional):

Upload images, PDFs, or Word documents to include their content in assessment generation.

2.1 Introduction

In Chapter 1, we studied that the inverse of a function \(f\), denoted by \(f^{-1}\), exists if \(f\) is one-one and onto. Trigonometric functions are not one-one and onto over their natural domains and ranges, so their inverses do not exist in the usual sense. However, by restricting the domains (and correspondingly the ranges) of trigonometric functions, we can make them one-one and onto, and thereby define their inverses.

The inverse trigonometric functions^? play a crucial role in calculus since they help define many important integrals. They are also widely used in science and engineering.

2.2 Basic Concepts

In Class XI, we studied the following trigonometric functions and their domains/ranges:

Function	Domain	Range
\(\sin\)	\(\mathbf{R}\)	\([-1, 1]\)
\(\cos\)	\(\mathbf{R}\)	\([-1, 1]\)
\(\tan\)	\(\mathbf{R} - \{(2n+1)\frac{\pi}{2} : n \in \mathbf{Z}\}\)	\(\mathbf{R}\)
\(\cot\)	\(\mathbf{R} - \{n\pi : n \in \mathbf{Z}\}\)	\(\mathbf{R}\)
\(\sec\)	\(\mathbf{R} - \{(2n+1)\frac{\pi}{2} : n \in \mathbf{Z}\}\)	\(\mathbf{R} - (-1, 1)\)
\(\csc\)	\(\mathbf{R} - \{n\pi : n \in \mathbf{Z}\}\)	\(\mathbf{R} - (-1, 1)\)

Inverse of Sine Function (sin¹)

The sine function is not one-one on \(\mathbf{R}\) (since, for example, \(\sin 0 = \sin \pi = 0\)). If we restrict its domain to \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\), then it becomes one-one and onto with range \([-1, 1]\). We can then define the inverse:

Definition — sin¹ (arc sine)

\[\sin^{-1}: [-1, 1] \to \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\] The branch with range \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) is called the principal value branch^? of \(\sin^{-1}\).

From the definition: \(\sin(\sin^{-1} x) = x\) for \(-1 \leq x \leq 1\), and \(\sin^{-1}(\sin x) = x\) for \(-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}\).

Fig 2.1 — Graphs of \(y = \sin x\) (principal branch in red) and \(y = \sin^{-1} x\)

Inverse of Cosine Function (cos¹)

Restricting cosine to \([0, \pi]\) makes it one-one and onto with range \([-1, 1]\).

Definition — cos¹ (arc cosine)

\[\cos^{-1}: [-1, 1] \to [0, \pi]\] The principal value branch of \(\cos^{-1}\) has range \([0, \pi]\).

Inverse of Cosecant Function (cosec¹)

Restricting cosecant to \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] - \{0\}\) makes it bijective onto \(\mathbf{R} - (-1, 1)\).

Definition — cosec¹

\[\csc^{-1}: \mathbf{R} - (-1, 1) \to \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] - \{0\}\]

Inverse of Secant Function (sec¹)

Definition — sec¹

\[\sec^{-1}: \mathbf{R} - (-1, 1) \to [0, \pi] - \left\{\frac{\pi}{2}\right\}\]

Inverse of Tangent Function (tan¹)

Restricting tangent to \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\) makes it bijective onto \(\mathbf{R}\).

Definition — tan¹ (arc tangent)

\[\tan^{-1}: \mathbf{R} \to \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\]

Inverse of Cotangent Function (cot¹)

Definition — cot¹

\[\cot^{-1}: \mathbf{R} \to (0, \pi)\]

Summary Table of All Six Inverse Trigonometric Functions

Function	Domain	Range (Principal Value Branch)
\(\sin^{-1}\)	\([-1, 1]\)	\(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\)
\(\cos^{-1}\)	\([-1, 1]\)	\([0, \pi]\)
\(\csc^{-1}\)	\(\mathbf{R} - (-1, 1)\)	\(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] - \{0\}\)
\(\sec^{-1}\)	\(\mathbf{R} - (-1, 1)\)	\([0, \pi] - \left\{\frac{\pi}{2}\right\}\)
\(\tan^{-1}\)	\(\mathbf{R}\)	\(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\)
\(\cot^{-1}\)	\(\mathbf{R}\)	\((0, \pi)\)

Important Notes

\(\sin^{-1} x\) should NOT be confused with \((\sin x)^{-1} = \frac{1}{\sin x} = \csc x\). In fact, \((\sin x)^{-1} = \frac{1}{\sin x}\) and similarly for other trigonometric functions.
Whenever no branch of an inverse trigonometric function is mentioned, we mean the principal value branch of that function.
The value of an inverse trigonometric function which lies in its principal value range is called the principal value of that inverse trigonometric function.

Worked Example 1

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

Solution

Let \(\sin^{-1}\left(\frac{1}{2}\right) = y\). Then \(\sin y = \frac{1}{2}\).

We know that \(\sin\frac{\pi}{6} = \frac{1}{2}\) and \(\frac{\pi}{6} \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\).

Therefore, the principal value of \(\sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6}\).

Worked Example 2

Find the principal value of \(\cot^{-1}\left(-\frac{1}{\sqrt{3}}\right)\).

Solution

Let \(\cot^{-1}\left(-\frac{1}{\sqrt{3}}\right) = y\). Then \(\cot y = -\frac{1}{\sqrt{3}}\).

We know that \(\cot\frac{\pi}{3} = \frac{1}{\sqrt{3}}\). Since cotangent is negative, and the range of \(\cot^{-1}\) is \((0, \pi)\), we need \(y = \pi - \frac{\pi}{3} = \frac{2\pi}{3}\).

Therefore, the principal value = \(\frac{2\pi}{3}\).

Interactive: Inverse Trig Principal Value Calculator

Select a function, enter a value, and compute the principal value

Function:

Value (x):

Exercise 2.1

Find the principal values of the following:

Q1. \(\sin^{-1}\left(-\frac{1}{2}\right)\)

\(\sin^{-1}\left(-\frac{1}{2}\right) = -\sin^{-1}\left(\frac{1}{2}\right) = -\frac{\pi}{6}\)
(Since \(\sin^{-1}(-x) = -\sin^{-1}(x)\) for \(x \in [-1, 1]\).)

Q2. \(\cos^{-1}\left(\frac{\sqrt{3}}{2}\right)\)

\(\cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}\) and \(\frac{\pi}{6} \in [0, \pi]\).
So \(\cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6}\).

Q3. \(\csc^{-1}(2)\)

\(\csc^{-1}(2) = \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6}\).

Q4. \(\tan^{-1}(-\sqrt{3})\)

\(\tan\frac{\pi}{3} = \sqrt{3}\). Since \(\tan^{-1}(-x) = -\tan^{-1}(x)\):
\(\tan^{-1}(-\sqrt{3}) = -\tan^{-1}(\sqrt{3}) = -\frac{\pi}{3}\).

Q5. \(\cos^{-1}\left(-\frac{1}{2}\right)\)

\(\cos^{-1}\left(-\frac{1}{2}\right) = \pi - \cos^{-1}\left(\frac{1}{2}\right) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}\).

Q6. \(\tan^{-1}(-1)\)

\(\tan^{-1}(-1) = -\tan^{-1}(1) = -\frac{\pi}{4}\).

Q7. \(\sec^{-1}\left(\frac{2}{\sqrt{3}}\right)\)

\(\sec^{-1}\left(\frac{2}{\sqrt{3}}\right) = \cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6}\).

Q8. \(\cot^{-1}(\sqrt{3})\)

\(\cot\frac{\pi}{6} = \sqrt{3}\) and \(\frac{\pi}{6} \in (0, \pi)\).
So \(\cot^{-1}(\sqrt{3}) = \frac{\pi}{6}\).

Q9. \(\cos^{-1}\left(-\frac{1}{\sqrt{2}}\right)\)

\(\cos^{-1}\left(-\frac{1}{\sqrt{2}}\right) = \pi - \cos^{-1}\left(\frac{1}{\sqrt{2}}\right) = \pi - \frac{\pi}{4} = \frac{3\pi}{4}\).

Q10. \(\csc^{-1}(-\sqrt{2})\)

\(\csc^{-1}(-\sqrt{2}) = \sin^{-1}\left(-\frac{1}{\sqrt{2}}\right) = -\sin^{-1}\left(\frac{1}{\sqrt{2}}\right) = -\frac{\pi}{4}\).

Find the values of the following:

Q11. \(\tan^{-1}(1) + \cos^{-1}\left(-\frac{1}{2}\right) + \sin^{-1}\left(-\frac{1}{2}\right)\)

\(\tan^{-1}(1) = \frac{\pi}{4}\), \(\cos^{-1}\left(-\frac{1}{2}\right) = \frac{2\pi}{3}\), \(\sin^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6}\).

Sum = \(\frac{\pi}{4} + \frac{2\pi}{3} - \frac{\pi}{6} = \frac{3\pi + 8\pi - 2\pi}{12} = \frac{9\pi}{12} = \frac{3\pi}{4}\).

Q12. \(\cos^{-1}\left(\frac{1}{2}\right) + 2\sin^{-1}\left(\frac{1}{2}\right)\)

\(\cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3}\), \(\sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6}\).

\(\frac{\pi}{3} + 2 \cdot \frac{\pi}{6} = \frac{\pi}{3} + \frac{\pi}{3} = \frac{2\pi}{3}\).

Q13. If \(\sin^{-1} x = y\), then
(A) \(0 \leq y \leq \pi\) (B) \(-\frac{\pi}{2} \leq y \leq \frac{\pi}{2}\)
(C) \(0 < y < \pi\) (D) \(-\frac{\pi}{2} < y < \frac{\pi}{2}\)

The range of \(\sin^{-1}\) (principal value branch) is \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\).
Answer: (B)

Q14. \(\tan^{-1}\sqrt{3} - \sec^{-1}(-2)\) is equal to
(A) \(\pi\) (B) \(-\frac{\pi}{3}\) (C) \(\frac{\pi}{3}\) (D) \(\frac{2\pi}{3}\)

\(\tan^{-1}\sqrt{3} = \frac{\pi}{3}\)
\(\sec^{-1}(-2) = \cos^{-1}\left(-\frac{1}{2}\right) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}\)

\(\frac{\pi}{3} - \frac{2\pi}{3} = -\frac{\pi}{3}\).
Answer: (B) \(-\frac{\pi}{3}\)

Activity: Why Do We Need Principal Value Branches?

Predict: If \(\sin \theta = \frac{1}{2}\), how many values of \(\theta\) exist? Can a function have multiple outputs for one input?

Consider the equation \(\sin \theta = \frac{1}{2}\). List a few solutions: \(\theta = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{13\pi}{6}, -\frac{11\pi}{6}, ...\)
There are infinitely many solutions! But a function must give exactly one output for each input.
To make sin invertible, we restrict the domain to \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\). In this interval, \(\sin \theta = \frac{1}{2}\) has exactly one solution: \(\theta = \frac{\pi}{6}\).
Similarly, different intervals give different branches of \(\sin^{-1}\).

Observation: The principal value branch is chosen so that (a) the inverse function is well-defined (single-valued), and (b) the range is a convenient, continuous interval.

Explanation: The principal value branch of \(\sin^{-1}\) uses \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) because sin is strictly increasing on this interval, making the inverse easy to work with. For \(\cos^{-1}\), we use \([0, \pi]\) because cos is strictly decreasing there.

Competency-Based Questions

A surveyor uses a clinometer to measure the angle of elevation of a tower. The tower is 50 m tall and is at a horizontal distance \(d\) from the surveyor. The angle of elevation \(\theta\) satisfies \(\tan \theta = \frac{50}{d}\), so \(\theta = \tan^{-1}\left(\frac{50}{d}\right)\).

Q1. If the surveyor stands 50 m from the base, what is the angle of elevation (in degrees)?

L3 Apply

\(\theta = \tan^{-1}\left(\frac{50}{50}\right) = \tan^{-1}(1) = \frac{\pi}{4} = 45^\circ\).

Q2. At what distance must the surveyor stand so that the angle of elevation is \(30^\circ\)?

L3 Apply

\(30^\circ = \tan^{-1}\left(\frac{50}{d}\right)\), so \(\tan 30^\circ = \frac{50}{d}\), giving \(\frac{1}{\sqrt{3}} = \frac{50}{d}\), hence \(d = 50\sqrt{3} \approx 86.6\) m.

Q3. As the surveyor walks towards the tower (d decreases from a very large value), the angle \(\theta\) increases. What is the maximum possible value of \(\theta\)?

L4 Analyse

As \(d \to 0^+\), \(\frac{50}{d} \to \infty\), so \(\theta = \tan^{-1}\left(\frac{50}{d}\right) \to \frac{\pi}{2} = 90^\circ\).
The maximum value approaches but never reaches \(90^\circ\). This makes sense because the range of \(\tan^{-1}\) is \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\) (open interval).

Q4. Why does the principal value branch of \(\tan^{-1}\) use the open interval \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\) rather than a closed interval?

L5 Evaluate

The tangent function is undefined at \(\pm\frac{\pi}{2}\) (it approaches \(\pm\infty\)). So the restricted domain of tan cannot include these endpoints, and consequently the range of \(\tan^{-1}\) is the open interval \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\). This ensures the inverse is defined for all real numbers without any discontinuity.

Assertion–Reason Questions

Assertion (A): \(\sin^{-1}\left(\frac{1}{2}\right) = \frac{5\pi}{6}\)
Reason (R): \(\sin\frac{5\pi}{6} = \frac{1}{2}\)

(a) Both true, R explains A.

(b) Both true, R doesn't explain A.

(c) A true, R false.

(d) A false, R true.

Answer: (d) — R is true: \(\sin\frac{5\pi}{6} = \sin(\pi - \frac{\pi}{6}) = \sin\frac{\pi}{6} = \frac{1}{2}\). But A is false: the principal value of \(\sin^{-1}\) lies in \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\), and \(\frac{5\pi}{6}\) is outside this range. The correct answer is \(\sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6}\).

Assertion (A): The domain of \(\cos^{-1}\) is \([-1, 1]\).
Reason (R): The range of the cosine function is \([-1, 1]\), and the domain of an inverse function equals the range of the original function.

(a) Both true, R explains A.

(b) Both true, R doesn't explain A.

(c) A true, R false.

(d) A false, R true.

Answer: (a) — Both are true. The domain of \(\cos^{-1}\) is exactly the range of \(\cos\), which is \([-1, 1]\). R correctly explains A.

Assertion (A): \(\tan^{-1}(1) + \tan^{-1}(1) = \tan^{-1}(2)\)
Reason (R): \(\tan^{-1} x + \tan^{-1} y = \tan^{-1}\left(\frac{x+y}{1-xy}\right)\) when \(xy < 1\).

(a) Both true, R explains A.

(b) Both true, R doesn't explain A.

(c) A true, R false.

(d) A false, R true.

Answer: (d) — A is false. Using R with \(x = y = 1\): \(xy = 1\), so the formula's condition \(xy < 1\) is not satisfied. In fact, \(\tan^{-1}(1) + \tan^{-1}(1) = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}\), which is NOT \(\tan^{-1}(2) \approx 1.107\). R is true as a formula (with the stated condition).

Frequently Asked Questions

What are inverse trigonometric functions?

Inverse trigonometric functions reverse trigonometric operations. Since trig functions are not one-one over their full domain, we restrict domains to make them invertible with specific principal value branches.

What is the principal value branch?

The principal value branch is the restricted range that makes an inverse trig function a proper function. For sin-inverse it is [-pi/2, pi/2]; for cos-inverse it is [0, pi]; for tan-inverse it is (-pi/2, pi/2).

What are the domains and ranges of inverse trig functions?

sin-inverse: domain [-1,1], range [-pi/2, pi/2]. cos-inverse: domain [-1,1], range [0, pi]. tan-inverse: domain R, range (-pi/2, pi/2). Similar for cosec, sec, and cot inverse.

How do you find principal values?

Find the angle in the principal branch whose trig value matches the given number. For example, sin-inverse(1/2) = pi/6 because sin(pi/6) = 1/2 and pi/6 is in [-pi/2, pi/2].

Why do we need restricted domains for inverse trig functions?

Trigonometric functions are periodic and not one-one, so they cannot have inverses. Restricting to one period where the function is bijective creates an invertible function.

Frequently Asked Questions — Inverse Trigonometric Functions

What is Definitions and Principal Value Branches of Inverse Trig Functions in NCERT Class 12 Mathematics?

Definitions and Principal Value Branches of Inverse Trig Functions is a key concept covered in NCERT Class 12 Mathematics, Chapter 2: Inverse Trigonometric Functions. This lesson builds the student's foundation in the chapter by explaining the core ideas with worked examples, definitions, and step-by-step methods aligned to the CBSE curriculum.

How do I solve problems on Definitions and Principal Value Branches of Inverse Trig Functions step by step?

To solve problems on Definitions and Principal Value Branches of Inverse Trig Functions, follow the NCERT method: identify the given quantities, choose the relevant formula or theorem, substitute values carefully, and simplify. Class 12 exercises gradually increase in difficulty — start with solved NCERT examples before attempting exercise questions, and always verify your answer by substitution or diagram.

What are the most important formulas for Chapter 2: Inverse Trigonometric Functions?

The essential formulas of Chapter 2 (Inverse Trigonometric Functions) are listed in the chapter summary and highlighted throughout the lesson in formula boxes. Memorise them and practise at least 2–3 problems per formula. CBSE board exams frequently test direct application as well as combined use of multiple formulas from this chapter.

Is Definitions and Principal Value Branches of Inverse Trig Functions important for the Class 12 board exam?

Definitions and Principal Value Branches of Inverse Trig Functions is part of the NCERT Class 12 Mathematics syllabus and appears in CBSE board exams. Questions typically include short-answer, long-answer, and competency-based items. Review the NCERT examples, exercise questions, and previous-year board problems on this topic to prepare confidently.

What mistakes should students avoid in Definitions and Principal Value Branches of Inverse Trig Functions?

Common mistakes in Definitions and Principal Value Branches of Inverse Trig Functions include skipping steps, misapplying formulas, sign errors, and losing track of units. Write each step clearly, double-check algebraic manipulations, and re-read the question after solving to verify that your answer matches what was asked.

Where can I find more NCERT practice questions on Definitions and Principal Value Branches of Inverse Trig Functions?

End-of-chapter NCERT exercises for Definitions and Principal Value Branches of Inverse Trig Functions cover all difficulty levels tested in CBSE exams. After completing them, try the examples again without looking at the solutions, attempt the NCERT Exemplar questions for Chapter 2, and solve at least one previous-year board paper to consolidate your understanding.

AI Tutor

Mathematics Class 12 — Part I

Ready

Hi! 👋 I'm Gaura, your AI Tutor for Definitions and Principal Value Branches of Inverse Trig Functions. Take your time studying the lesson — whenever you have a doubt, just ask me! I'm here to help.