- Radian-Degree Relation: \( \pi \text{ radians} = 180° \), so \( 1° = \frac{\pi}{180} \) rad and \( 1 \text{ rad} = \frac{180°}{\pi} \)
- Arc Length Formula: \( l = r\theta \) where \( \theta \) is in radians
- Quadrant Sign Rule (ASTC): All positive (Q1), Sin/Cosec positive (Q2), Tan/Cot positive (Q3), Cos/Sec positive (Q4)
- Pythagorean Identities: \( \sin^2 x + \cos^2 x = 1 \), \( 1 + \tan^2 x = \sec^2 x \), \( 1 + \cot^2 x = \csc^2 x \)
- Sum Formula: \( \sin(A+B) = \sin A \cos B + \cos A \sin B \)
- Difference Formula: \( \cos(A-B) = \cos A \cos B + \sin A \sin B \)
- Periodicity: sin and cos repeat every \( 2\pi \); tan and cot repeat every \( \pi \)
- Removed from CBSE 2026-27: Ex 3.4 (Trigonometric Equations) is NOT in the current rationalised syllabus
The NCERT solutions CBSE free for class 11th maths chapter 3 trigonometric functions on this page are fully updated for the 2026-27 academic year and cover every question in Exercises 3.1, 3.2, 3.3, and the Miscellaneous Exercise with clear step-by-step working. You can also explore all subjects on our NCERT Solutions hub or go directly to the NCERT Solutions for Class 11 page for other chapters. These solutions are based on the official NCERT official textbook and align with the latest CBSE rationalised syllabus.
Table of Contents
Chapter Overview — NCERT Solutions CBSE Free for Class 11th Maths Chapter 3 Trigonometric Functions
Chapter 3 of the NCERT Class 11 Mathematics textbook extends your understanding of trigonometry beyond acute angles to any real number. You will study radian measure, the unit circle definition of trigonometric functions, their signs in different quadrants, periodicity, and the powerful sum-and-difference identities. This chapter forms the backbone of calculus, coordinate geometry, and physics topics you will encounter in Class 12 and competitive exams like JEE.
For the CBSE 2026-27 board exam, Trigonometric Functions falls under the Algebra unit which carries approximately 13 marks. Questions appear as 2-mark short answers (prove identities, find values) and 5-mark long answers (multi-step proofs). MCQs based on quadrant signs and standard values are common in internal assessments.
Before starting this chapter, make sure you are comfortable with basic right-triangle trigonometry from Class 10, the Pythagorean theorem, and algebraic manipulation of fractions. The chapter is covered in the NCERT Mathematics Part I, Class 11, Chapter 3.
| Detail | Information |
|---|---|
| Chapter | 3 — Trigonometric Functions |
| Textbook | NCERT Mathematics Part I, Class 11 |
| Subject | Mathematics |
| CBSE Unit | Algebra |
| Approximate Marks Weightage | 13 marks (Algebra unit) |
| Difficulty Level | Medium to Hard |
| Academic Year | 2026-27 |
Key Concepts and Formulas — Trigonometric Functions Class 11
Radian and Degree Conversion
An angle (कोण) is measured in degrees or radians. The relation between them is:
\[ \pi \text{ radians} = 180° \]
To convert degrees to radians, multiply by \( \frac{\pi}{180} \). To convert radians to degrees, multiply by \( \frac{180}{\pi} \).
Arc Length and Central Angle
In a circle of radius \( r \), if an arc of length \( l \) subtends a central angle \( \theta \) (in radians), then:
\[ l = r\theta \quad \Rightarrow \quad \theta = \frac{l}{r} \]
This is the most-tested formula in Exercise 3.1.
Quadrant Sign Rules (ASTC)
The signs of trigonometric functions depend on the quadrant in which the angle lies. Use the mnemonic ASTC — All Students Take Calculus: All positive in Q1, Sin in Q2, Tan in Q3, Cos in Q4.
Pythagorean Identities (मूलभूत सर्वसमिकाएँ)
\[ \sin^2 x + \cos^2 x = 1 \]
\[ 1 + \tan^2 x = \sec^2 x \]
\[ 1 + \cot^2 x = \csc^2 x \]
Sum and Difference Formulas
\[ \sin(A \pm B) = \sin A \cos B \pm \cos A \sin B \]
\[ \cos(A \pm B) = \cos A \cos B \mp \sin A \sin B \]
\[ \tan(A + B) = \frac{\tan A + \tan B}{1 – \tan A \tan B} \]
\[ \tan(A – B) = \frac{\tan A – \tan B}{1 + \tan A \tan B} \]
Product-to-Sum and Sum-to-Product Formulas
\[ \sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right) \]
\[ \sin A – \sin B = 2 \cos\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right) \]
\[ \cos A + \cos B = 2 \cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right) \]
\[ \cos A – \cos B = -2 \sin\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right) \]
NCERT Solutions Exercise 3.1 — Radian and Degree Measure
Exercise 3.1 tests your ability to convert between degrees and radians, apply the arc length formula, and solve problems involving circles. These are 2-3 mark questions in board exams.
Question 1
Easy
Find the radian measures corresponding to the following degree measures: (i) 25° (ii) –47°30′ (iii) 240° (iv) 520°
Step 1: Use the conversion formula: multiply degrees by \( \frac{\pi}{180} \).
\[ 25° = 25 \times \frac{\pi}{180} = \frac{25\pi}{180} = \frac{5\pi}{36} \text{ radians} \]
\( \therefore \) 25° = \( \dfrac{5\pi}{36} \) radians
Step 1: Convert 30 minutes to degrees: \( 30′ = \frac{30}{60}° = \frac{1}{2}° \). So –47°30′ = \( -47\frac{1}{2}° = -\frac{95}{2}° \).
Step 2: Convert to radians:
\[ -\frac{95}{2} \times \frac{\pi}{180} = -\frac{95\pi}{360} = -\frac{19\pi}{72} \text{ radians} \]
\( \therefore \) –47°30′ = \( -\dfrac{19\pi}{72} \) radians
Step 1: Multiply by \( \frac{\pi}{180} \):
\[ 240 \times \frac{\pi}{180} = \frac{240\pi}{180} = \frac{4\pi}{3} \text{ radians} \]
\( \therefore \) 240° = \( \dfrac{4\pi}{3} \) radians
\[ 520 \times \frac{\pi}{180} = \frac{520\pi}{180} = \frac{26\pi}{9} \text{ radians} \]
\( \therefore \) 520° = \( \dfrac{26\pi}{9} \) radians
Question 2
Easy
Find the degree measures corresponding to the following radian measures (Use \( \pi = \frac{22}{7} \)): (i) \( \frac{11}{16} \) (ii) \( -4 \) (iii) \( \frac{5\pi}{3} \) (iv) \( \frac{7\pi}{6} \)
Step 1: Multiply by \( \frac{180}{\pi} \):
\[ \frac{11}{16} \times \frac{180}{\pi} = \frac{11 \times 180}{16\pi} = \frac{11 \times 180 \times 7}{16 \times 22} = \frac{315}{8} = 39.375° \]
Step 2: Convert 0.375° to minutes: \( 0.375 \times 60 = 22.5′ \), i.e., 22′30″.
\( \therefore \) \( \frac{11}{16} \) rad = 39°22′30″
Step 1: \( -4 \times \frac{180}{\pi} = -4 \times \frac{180 \times 7}{22} = -\frac{5040}{22} = -229\frac{1}{11}° \)
Step 2: \( \frac{1}{11}° = \frac{60}{11}’ = 5\frac{5}{11}’ \); \( \frac{5}{11}’ = \frac{300}{11}” \approx 27” \)
\( \therefore \) –4 rad ≈ –229°5′27″
Step 1: \( \frac{5\pi}{3} \times \frac{180}{\pi} = \frac{5 \times 180}{3} = 300° \)
\( \therefore \) \( \frac{5\pi}{3} \) rad = 300°
Step 1: \( \frac{7\pi}{6} \times \frac{180}{\pi} = \frac{7 \times 180}{6} = 210° \)
\( \therefore \) \( \frac{7\pi}{6} \) rad = 210°
Question 3
Medium
A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second?
Key Concept: One complete revolution = \( 2\pi \) radians.
Step 1: Revolutions per second = \( \frac{360}{60} = 6 \) revolutions per second.
Step 2: Radians per second = \( 6 \times 2\pi = 12\pi \) radians.
\( \therefore \) The wheel turns \( 12\pi \) radians in one second.
Question 4
Medium
Find the degree measure of the angle subtended at the centre of a circle of radius 100 cm by an arc of length 22 cm. (Use \( \pi = \frac{22}{7} \))
Step 1: Use \( \theta = \frac{l}{r} \): \( \theta = \frac{22}{100} = \frac{11}{50} \) radians.
Step 2: Convert to degrees: \( \frac{11}{50} \times \frac{180}{\pi} = \frac{11 \times 180 \times 7}{50 \times 22} = \frac{13860}{1100} = \frac{63}{5} = 12.6° \)
Step 3: \( 0.6° = 0.6 \times 60′ = 36′ \)
\( \therefore \) The angle is 12°36′.
Question 5
Medium
In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the length of the minor arc of the chord.
Step 1: Radius \( r = 20 \) cm. Chord AB = 20 cm = radius. So triangle OAB is equilateral.
Step 2: Each angle of an equilateral triangle = 60° = \( \frac{\pi}{3} \) radians.
Step 3: Arc length \( = r\theta = 20 \times \frac{\pi}{3} = \frac{20\pi}{3} \) cm.
\( \therefore \) Length of minor arc = \( \dfrac{20\pi}{3} \) cm
Question 6
Medium
If in two circles, arcs of the same length subtend angles 60° and 75° at the centre, find the ratio of their radii.
Step 1: Convert: \( 60° = \frac{\pi}{3} \) rad, \( 75° = \frac{5\pi}{12} \) rad.
Step 2: Since arc length \( l \) is the same: \( l = r_1 \cdot \frac{\pi}{3} = r_2 \cdot \frac{5\pi}{12} \).
Step 3: \( \frac{r_1}{r_2} = \frac{5\pi/12}{\pi/3} = \frac{5\pi}{12} \times \frac{3}{\pi} = \frac{5}{4} \).
\( \therefore \) \( r_1 : r_2 = 5 : 4 \)
Question 7
Easy
Find the angle in radian through which a pendulum swings if its length is 75 cm and the tip describes an arc of length (i) 10 cm (ii) 15 cm (iii) 21 cm.
Why does this work? The pendulum tip traces an arc of a circle with radius = pendulum length.
\( \therefore \) \( \theta = \dfrac{2}{15} \) radians
\( \therefore \) \( \theta = \dfrac{1}{5} \) radians
\( \therefore \) \( \theta = \dfrac{7}{25} \) radians
NCERT Solutions Exercise 3.2 — Values of Trigonometric Functions
Exercise 3.2 asks you to find all six trigonometric functions when one is given, and to evaluate trig functions of large or negative angles using periodicity. These are important for class 11 maths ncert solutions and frequently appear in board exams.
Question 1
Medium
Find the values of other five trigonometric functions if \( \cos x = -\frac{1}{2} \), \( x \) lies in the third quadrant.
Step 1: \( \sec x = \frac{1}{\cos x} = -2 \)
Step 2: \( \sin^2 x = 1 – \cos^2 x = 1 – \frac{1}{4} = \frac{3}{4} \Rightarrow \sin x = -\frac{\sqrt{3}}{2} \) (negative in Q3)
Step 3: \( \csc x = \frac{1}{\sin x} = -\frac{2}{\sqrt{3}} = -\frac{2\sqrt{3}}{3} \)
Step 4: \( \tan x = \frac{\sin x}{\cos x} = \frac{-\sqrt{3}/2}{-1/2} = \sqrt{3} \) (positive in Q3 ✓)
Step 5: \( \cot x = \frac{1}{\tan x} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3} \)
\( \therefore \) \( \sin x = -\frac{\sqrt{3}}{2},\ \csc x = -\frac{2\sqrt{3}}{3},\ \tan x = \sqrt{3},\ \cot x = \frac{\sqrt{3}}{3},\ \sec x = -2 \)
Question 2
Medium
Find the values of other five trigonometric functions if \( \sin x = \frac{3}{5} \), \( x \) lies in the second quadrant.
Step 1: \( \csc x = \frac{5}{3} \)
Step 2: \( \cos^2 x = 1 – \frac{9}{25} = \frac{16}{25} \Rightarrow \cos x = -\frac{4}{5} \) (negative in Q2)
Step 3: \( \sec x = -\frac{5}{4} \)
Step 4: \( \tan x = \frac{3/5}{-4/5} = -\frac{3}{4} \) (negative in Q2 ✓)
Step 5: \( \cot x = -\frac{4}{3} \)
\( \therefore \) \( \cos x = -\frac{4}{5},\ \tan x = -\frac{3}{4},\ \csc x = \frac{5}{3},\ \sec x = -\frac{5}{4},\ \cot x = -\frac{4}{3} \)
Question 6
Easy
Find the value of the trigonometric function \( \sin 765° \).
Step 1: sin repeats every 360°. \( 765° = 2 \times 360° + 45° \).
Step 2: \( \sin 765° = \sin 45° = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \)
\( \therefore \) \( \sin 765° = \dfrac{\sqrt{2}}{2} \)
Question 7
Easy
Find the value of the trigonometric function \( \csc(-1410°) \).
Step 1: \( -1410° + 4 \times 360° = -1410° + 1440° = 30° \)
Step 2: \( \csc(-1410°) = \csc 30° = 2 \)
\( \therefore \) \( \csc(-1410°) = 2 \)
Question 8
Easy
Find the value of \( \tan\dfrac{19\pi}{3} \).
Step 1: tan repeats every \( \pi \). \( \frac{19\pi}{3} = 6\pi + \frac{\pi}{3} \). Since \( 6\pi \) is a multiple of \( \pi \):
Step 2: \( \tan\frac{19\pi}{3} = \tan\frac{\pi}{3} = \tan 60° = \sqrt{3} \)
\( \therefore \) \( \tan\dfrac{19\pi}{3} = \sqrt{3} \)
NCERT Solutions Exercise 3.3 — Proving Trigonometric Identities
Exercise 3.3 is the most important exercise for board exams. It involves proving identities using sum-and-difference formulas, allied angle identities, and product-to-sum conversions. These are 5-mark long-answer questions in CBSE papers.
Question 1
Easy
Prove that: \( \sin^2\dfrac{\pi}{6} + \cos^2\dfrac{\pi}{3} – \tan^2\dfrac{\pi}{4} = -\dfrac{1}{2} \)
Step 1: Substitute standard values: \( \sin\frac{\pi}{6} = \frac{1}{2},\ \cos\frac{\pi}{3} = \frac{1}{2},\ \tan\frac{\pi}{4} = 1 \).
Step 2:
\[ \text{LHS} = \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2 – (1)^2 = \frac{1}{4} + \frac{1}{4} – 1 = \frac{1}{2} – 1 = -\frac{1}{2} = \text{RHS} \]
\( \therefore \) LHS = RHS. Hence Proved.
Question 4
Easy
Prove that: \( 2\sin^2\dfrac{3\pi}{4} + 2\cos^2\dfrac{\pi}{4} + 2\sec^2\dfrac{\pi}{3} = 10 \)
Step 1: \( \sin\frac{3\pi}{4} = \sin\left(\pi – \frac{\pi}{4}\right) = \sin\frac{\pi}{4} = \frac{1}{\sqrt{2}} \)
Step 2: \( \cos\frac{\pi}{4} = \frac{1}{\sqrt{2}},\ \sec\frac{\pi}{3} = 2 \)
Step 3:
\[ \text{LHS} = 2\cdot\frac{1}{2} + 2\cdot\frac{1}{2} + 2\cdot 4 = 1 + 1 + 8 = 10 = \text{RHS} \]
Question 5
Medium
Find the value of: (i) \( \sin 75° \) (ii) \( \tan 15° \)
Step 1: Write \( 75° = 45° + 30° \) and apply the sum formula.
\[ \sin 75° = \sin(45°+30°) = \sin 45°\cos 30° + \cos 45°\sin 30° \]
\[ = \frac{1}{\sqrt{2}}\cdot\frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}}\cdot\frac{1}{2} = \frac{\sqrt{3}+1}{2\sqrt{2}} = \frac{\sqrt{6}+\sqrt{2}}{4} \]
\( \therefore \) \( \sin 75° = \dfrac{\sqrt{6}+\sqrt{2}}{4} \)
Step 1: Write \( 15° = 45° – 30° \).
\[ \tan 15° = \tan(45°-30°) = \frac{\tan 45° – \tan 30°}{1 + \tan 45°\tan 30°} = \frac{1 – \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}} = \frac{\sqrt{3}-1}{\sqrt{3}+1} \]
Step 2: Rationalise: \( \frac{(\sqrt{3}-1)^2}{(\sqrt{3})^2-1^2} = \frac{3-2\sqrt{3}+1}{2} = \frac{4-2\sqrt{3}}{2} = 2-\sqrt{3} \)
\( \therefore \) \( \tan 15° = 2 – \sqrt{3} \)
Question 8
Medium
Prove that: \( \dfrac{\cos(\pi+x)\cos(-x)}{\sin(\pi-x)\cos\left(\dfrac{\pi}{2}+x\right)} = \cot^2 x \)
Step 1: Apply allied angle identities: \( \cos(\pi+x) = -\cos x,\ \cos(-x) = \cos x,\ \sin(\pi-x) = \sin x,\ \cos\left(\frac{\pi}{2}+x\right) = -\sin x \).
Step 2:
\[ \text{LHS} = \frac{(-\cos x)(\cos x)}{(\sin x)(-\sin x)} = \frac{-\cos^2 x}{-\sin^2 x} = \frac{\cos^2 x}{\sin^2 x} = \cot^2 x = \text{RHS} \]
Question 12
Hard
Prove that: \( \sin^2 6x – \sin^2 4x = \sin 2x \sin 10x \)
Step 1: Use the identity \( \sin^2 A – \sin^2 B = (\sin A + \sin B)(\sin A – \sin B) \).
Step 2: Apply sum-to-product: \( \sin 6x + \sin 4x = 2\sin 5x\cos x \) and \( \sin 6x – \sin 4x = 2\cos 5x\sin x \).
Step 3:
\[ \text{LHS} = (2\sin 5x\cos x)(2\cos 5x\sin x) = (2\sin 5x\cos 5x)(2\sin x\cos x) = \sin 10x\cdot\sin 2x = \text{RHS} \]
Formula Reference Table — Trigonometric Functions Chapter 3
| Formula Name | Formula | Variables / Notes |
|---|---|---|
| Degree to Radian | \( \theta_{\text{rad}} = \theta_{\text{deg}} \times \dfrac{\pi}{180} \) | Multiply degrees by π/180 |
| Radian to Degree | \( \theta_{\text{deg}} = \theta_{\text{rad}} \times \dfrac{180}{\pi} \) | Multiply radians by 180/π |
| Arc Length | \( l = r\theta \) | θ must be in radians |
| Pythagorean Identity 1 | \( \sin^2 x + \cos^2 x = 1 \) | All quadrants |
| Pythagorean Identity 2 | \( 1 + \tan^2 x = \sec^2 x \) | \( x \neq \frac{\pi}{2} + n\pi \) |
| Pythagorean Identity 3 | \( 1 + \cot^2 x = \csc^2 x \) | \( x \neq n\pi \) |
| sin(A+B) | \( \sin A\cos B + \cos A\sin B \) | Sum formula |
| cos(A+B) | \( \cos A\cos B – \sin A\sin B \) | Sum formula |
| tan(A+B) | \( \dfrac{\tan A + \tan B}{1 – \tan A\tan B} \) | Sum formula |
| sin A + sin B | \( 2\sin\dfrac{A+B}{2}\cos\dfrac{A-B}{2} \) | Sum-to-product |
| cos A − cos B | \( -2\sin\dfrac{A+B}{2}\sin\dfrac{A-B}{2} \) | Sum-to-product |
Important Questions for CBSE Board Exam — Chapter 3 Trigonometric Functions
1-Mark Questions (Definition / Standard Value)
- What is the radian measure of 180°? Answer: \( \pi \) radians
- What is the period of \( \tan x \)? Answer: \( \pi \) radians (180°)
- In which quadrant is \( \cos x \) negative and \( \sin x \) positive? Answer: Second quadrant
3-Mark Questions
Q1: Find the value of \( \sin\left(-\frac{11\pi}{3}\right) \).
Solution: \( \sin\left(-\frac{11\pi}{3}\right) = -\sin\frac{11\pi}{3} = -\sin\left(4\pi – \frac{\pi}{3}\right) = -(-\sin\frac{\pi}{3}) = \sin\frac{\pi}{3} = \frac{\sqrt{3}}{2} \). Answer: \( \frac{\sqrt{3}}{2} \)
Q2: If \( \tan x = -\frac{5}{12} \) and \( x \) lies in the second quadrant, find \( \sec x \).
Solution: \( \sec^2 x = 1 + \tan^2 x = 1 + \frac{25}{144} = \frac{169}{144} \Rightarrow \sec x = -\frac{13}{12} \) (negative in Q2). Answer: \( -\frac{13}{12} \)
5-Mark Questions (Proof)
Q1: Prove that \( \sin 2x + 2\sin 4x + \sin 6x = 4\cos^2 x\sin 4x \).
Solution: Group \( \sin 2x + \sin 6x = 2\sin 4x\cos 2x \) (sum-to-product). Then LHS \( = 2\sin 4x\cos 2x + 2\sin 4x = 2\sin 4x(\cos 2x + 1) = 2\sin 4x(2\cos^2 x – 1 + 1) = 2\sin 4x \cdot 2\cos^2 x = 4\cos^2 x\sin 4x \) = RHS. Hence Proved.
Common Mistakes Students Make in Chapter 3
Mistake 1: Using the arc length formula \( l = r\theta \) with \( \theta \) in degrees.
Why it’s wrong: The formula only works when \( \theta \) is in radians. Plugging in degrees gives a completely wrong answer.
Correct approach: Always convert degrees to radians first, then apply \( l = r\theta \).
Mistake 2: Applying the wrong period — using \( 2\pi \) for tan and cot.
Why it’s wrong: tan and cot have period \( \pi \), not \( 2\pi \). Using \( 2\pi \) still gives the correct answer sometimes, but the working is unnecessarily long and may lose marks.
Correct approach: Reduce using \( \pi \) for tan/cot, \( 2\pi \) for sin/cos/sec/csc.
Mistake 3: Forgetting the quadrant sign when finding remaining trig functions.
Why it’s wrong: \( \cos^2 x = \frac{16}{25} \) gives \( \cos x = \pm\frac{4}{5} \). Students often write \( +\frac{4}{5} \) without checking the quadrant.
Correct approach: Always identify the quadrant first and apply ASTC to determine the correct sign.
Mistake 4: In proofs, starting from both sides simultaneously.
Why it’s wrong: CBSE expects you to start from one side (usually LHS) and reach the other side. Working on both sides simultaneously is not accepted.
Correct approach: Start with LHS, apply identities step by step, and write “= RHS. Hence Proved.”
Mistake 5: Confusing \( \sin^2 x \) with \( \sin(2x) \).
Why it’s wrong: \( \sin^2 x = (\sin x)^2 \) while \( \sin 2x = 2\sin x\cos x \). These are completely different expressions.
Correct approach: Read the notation carefully. \( \sin^2 x \) means the square of sin x.
Exam Tips for 2026-27 — Trigonometric Functions CBSE Board
- Memorise the standard values table for sin, cos, tan at 0°, 30°, 45°, 60°, 90°. These appear in almost every question.
- For proof questions (5 marks): Write “LHS =” at the start, show every algebraic step on a new line, and end with “= RHS. Hence Proved.” The CBSE 2026-27 marking scheme awards 1 mark per correct step.
- Ex 3.4 (Trigonometric Equations) is removed from the current CBSE rationalised syllabus. Do not waste time on it for board exams — focus on Ex 3.1, 3.2, 3.3, and Miscellaneous.
- Arc length problems from Ex 3.1 are easy 2-mark scoring questions. Always write the formula \( l = r\theta \) first, then substitute.
- For “find all six trig functions” questions: Use a structured format — write all six values clearly labelled. Missing even one value can cost you a mark.
- Rationalise surds in final answers. Leaving \( \frac{\sqrt{3}-1}{\sqrt{3}+1} \) unsimplified loses marks.
For more chapter-wise solutions, visit our NCERT Solutions for Class 11 page, or explore the full NCERT Solutions library for all classes.
Frequently Asked Questions — Trigonometric Functions Class 11
How do you convert degrees to radians in Class 11 Chapter 3?
To convert degrees to radians, multiply the degree value by \( \frac{\pi}{180} \). For example, \( 90° = 90 \times \frac{\pi}{180} = \frac{\pi}{2} \) radians. This comes from the fundamental relation \( \pi \text{ radians} = 180° \). Always simplify the fraction and keep \( \pi \) in the answer unless a decimal approximation is asked.
How do you find the other five trigonometric functions given one value?
First, identify which quadrant the angle lies in and note the sign rules (ASTC). Then use the reciprocal relations (\( \csc = 1/\sin \), etc.) and Pythagorean identities (\( \sin^2 x + \cos^2 x = 1 \)) to find the remaining values. Always assign the correct sign based on the quadrant before writing the final answer. This is the standard method for all Exercise 3.2 questions.
Which exercises are in NCERT Class 11 Maths Chapter 3 for CBSE 2026-27?
For CBSE 2026-27, the relevant exercises are Ex 3.1 (radian-degree conversion, arc length), Ex 3.2 (finding trig values, periodicity), Ex 3.3 (proving identities using sum-difference formulas), and the Miscellaneous Exercise. Exercise 3.4 on Trigonometric Equations has been removed from the rationalised CBSE syllabus and is not required for board exams.
How many marks does Trigonometric Functions carry in CBSE Class 11 Maths?
Trigonometric Functions is part of the Algebra unit in CBSE Class 11 Maths, which carries approximately 13 marks in the annual examination. Questions from this chapter appear as 2-3 mark short answers (finding trig values, converting measures) and 5-mark long answers (proving identities). It is one of the highest-weightage chapters in Class 11 Maths.
Where can I download NCERT Solutions for Class 11 Maths Chapter 3 PDF for 2026-27?
You can access the complete NCERT solutions for Class 11 Maths Chapter 3 Trigonometric Functions on this page — all exercises are solved step-by-step and updated for 2026-27. For the official textbook PDF, visit the NCERT official website. Our solutions cover Ex 3.1, Ex 3.2, Ex 3.3, and the Miscellaneous Exercise with full working and LaTeX-rendered formulas.