- Radian-Degree Relation: \( 180° = \pi \) radians; to convert degrees → radians multiply by \( \frac{\pi}{180} \)
- Arc Length Formula: \( l = r\theta \) where \( \theta \) is in radians
- ASTC Rule (Quadrant Signs): Q1 — All positive; Q2 — Sin/Cosec positive; Q3 — Tan/Cot positive; Q4 — Cos/Sec positive
- Pythagorean Identities: \( \sin^2 x + \cos^2 x = 1 \), \( 1 + \tan^2 x = \sec^2 x \), \( 1 + \cot^2 x = \cosec^2 x \)
- Sum Formula: \( \sin(A+B) = \sin A \cos B + \cos A \sin B \)
- Periodicity: \( \sin \) and \( \cos \) have period \( 2\pi \); \( \tan \) and \( \cot \) have period \( \pi \)
- Syllabus Note (2026-27): Exercise 3.4 (Trigonometric Equations) is removed from CBSE rationalised syllabus
The NCERT Solutions CBSE free for Class 11th Maths Chapter 3 Trigonometric Functions on this page cover every question from Exercises 3.1, 3.2, 3.3, and the Miscellaneous Exercise — fully updated for the 2026-27 academic year. Whether you need to convert angles between degrees and radians, find all six trigonometric values from one given value, or prove identities using sum and difference formulas, you will find clear step-by-step solutions here. This page is part of our complete NCERT Solutions for Class 11 series. You can also browse all subjects at NCERT Solutions. The official textbook is available on the NCERT official website.
Table of Contents
- Quick Revision Box
- Chapter Overview — Class 11 Maths Chapter 3
- Key Concepts and Theorems
- Formula Reference Table
- NCERT Solutions — All Exercises Step by Step
- Solved Examples Beyond NCERT
- Topic-Wise Important Questions for Board Exam
- Common Mistakes Students Make
- Exam Tips for 2026-27
- Frequently Asked Questions
Chapter Overview — Class 11 Maths Chapter 3 Trigonometric Functions
Chapter 3 of the NCERT Class 11 Mathematics textbook extends trigonometry from right-triangle ratios (studied in Class 9–10) to the full real-number domain using the unit circle. You will learn to measure angles in radians, define all six trigonometric functions for any real number, and use compound-angle identities to evaluate and prove expressions.
This chapter carries significant weight in the CBSE board exam — questions from this chapter appear in 2-mark, 3-mark, and 5-mark sections. Proof-based questions from Exercise 3.3 are especially popular in board papers. Mastering this chapter also directly supports JEE Main preparation.
| Detail | Information |
|---|---|
| Chapter | 3 — Trigonometric Functions |
| Textbook | NCERT Mathematics Part I, Class 11 |
| Class | 11 (Grade 11) |
| Subject | Mathematics |
| Academic Year | 2026-27 |
| Exercises | 3.1, 3.2, 3.3, Miscellaneous (Ex 3.4 removed from CBSE) |
| Prerequisites | Basic trigonometry (Class 10), coordinate geometry basics |
Key Concepts and Theorems — Trigonometric Functions Class 11
Radian Measure and Angle Conversion
An angle is measured in radians when the arc length equals the radius. The fundamental relation is \( \pi \text{ radians} = 180° \). To convert degrees to radians, use \( \theta_{\text{rad}} = \theta_{\text{deg}} \times \frac{\pi}{180} \). To convert radians to degrees, use \( \theta_{\text{deg}} = \theta_{\text{rad}} \times \frac{180}{\pi} \).
The arc length formula \( l = r\theta \) (where \( \theta \) is in radians) connects geometry to trigonometry and appears directly in Exercise 3.1.
Unit Circle Definition of Trigonometric Functions
For any real number \( x \), if \( P(a, b) \) is the point on the unit circle corresponding to arc length \( x \), then \( \cos x = a \) and \( \sin x = b \). The other four functions are defined as reciprocals and ratios: \( \tan x = \frac{\sin x}{\cos x} \), \( \cot x = \frac{\cos x}{\sin x} \), \( \sec x = \frac{1}{\cos x} \), \( \cosec x = \frac{1}{\sin x} \).
Quadrant Sign Rule (ASTC)
The sign of each trigonometric function depends on the quadrant of the terminal ray. Remember ASTC: All positive (Q1), Sin positive (Q2), Tan positive (Q3), Cos positive (Q4). This is the most frequently tested concept in Exercise 3.2.
Pythagorean Identities
Three fundamental identities follow directly from \( \sin^2 x + \cos^2 x = 1 \):
- \( \sin^2 x + \cos^2 x = 1 \)
- \( 1 + \tan^2 x = \sec^2 x \)
- \( 1 + \cot^2 x = \cosec^2 x \)
Sum and Difference Formulas
These are the backbone of Exercise 3.3 and Miscellaneous Exercise:
- \( \sin(A \pm B) = \sin A \cos B \pm \cos A \sin B \)
- \( \cos(A \pm B) = \cos A \cos B \mp \sin A \sin B \)
- \( \tan(A \pm B) = \frac{\tan A \pm \tan B}{1 \mp \tan A \tan B} \)
Formula Reference Table — Class 11 Maths Chapter 3
| Formula Name | Formula | Variables |
|---|---|---|
| Degree to Radian | \( \theta_r = \theta_d \times \frac{\pi}{180} \) | \( \theta_d \) = degrees, \( \theta_r \) = radians |
| Arc Length | \( l = r\theta \) | \( l \) = arc length, \( r \) = radius, \( \theta \) = radians |
| Pythagorean Identity 1 | \( \sin^2 x + \cos^2 x = 1 \) | \( x \) = any real number |
| Pythagorean Identity 2 | \( 1 + \tan^2 x = \sec^2 x \) | \( x \neq \frac{\pi}{2} + n\pi \) |
| Pythagorean Identity 3 | \( 1 + \cot^2 x = \cosec^2 x \) | \( x \neq n\pi \) |
| sin(A+B) | \( \sin A \cos B + \cos A \sin B \) | A, B real numbers |
| sin(A−B) | \( \sin A \cos B – \cos A \sin B \) | A, B real numbers |
| cos(A+B) | \( \cos A \cos B – \sin A \sin B \) | A, B real numbers |
| cos(A−B) | \( \cos A \cos B + \sin A \sin B \) | A, B real numbers |
| tan(A+B) | \( \frac{\tan A + \tan B}{1 – \tan A \tan B} \) | \( \tan A \tan B \neq 1 \) |
| Sum-to-Product (sin) | \( \sin A + \sin B = 2\sin\frac{A+B}{2}\cos\frac{A-B}{2} \) | A, B real numbers |
| Product-to-Sum (cos) | \( \cos A – \cos B = -2\sin\frac{A+B}{2}\sin\frac{A-B}{2} \) | A, B real numbers |
NCERT Solutions CBSE Free for Class 11th Maths Chapter 3 — All Exercises Step by Step
Below are complete, step-by-step solutions for all questions in Chapter 3. These solutions match the official NCERT answer key and are structured for maximum marks in your 2026-27 board exam. Each solution shows full working — never skip steps in your exam answer.
Exercise 3.1 — Angles and Radian Measure
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 \( \theta_{\text{rad}} = \theta_{\text{deg}} \times \frac{\pi}{180} \)
\[ 25° = 25 \times \frac{\pi}{180} = \frac{25\pi}{180} = \frac{5\pi}{36} \text{ radians} \]
\( \therefore \) 25° = \( \frac{5\pi}{36} \) radians
Step 1: Convert minutes to degrees. \( 30′ = \frac{30}{60}° = \frac{1}{2}° \)
Step 2: So \( -47°30′ = -47\frac{1}{2}° = -\frac{95}{2}° \)
Step 3: Convert to radians:
\[ -\frac{95}{2} \times \frac{\pi}{180} = -\frac{95\pi}{360} = -\frac{19\pi}{72} \text{ radians} \]
\( \therefore \) −47°30′ = \( -\frac{19\pi}{72} \) radians
Step 1: Apply the conversion formula:
\[ 240 \times \frac{\pi}{180} = \frac{240\pi}{180} = \frac{4\pi}{3} \text{ radians} \]
\( \therefore \) 240° = \( \frac{4\pi}{3} \) radians
Step 1: Apply the conversion formula:
\[ 520 \times \frac{\pi}{180} = \frac{520\pi}{180} = \frac{26\pi}{9} \text{ radians} \]
\( \therefore \) 520° = \( \frac{26\pi}{9} \) 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: Use \( \theta_{\text{deg}} = \theta_{\text{rad}} \times \frac{180}{\pi} \)
\[ \frac{11}{16} \times \frac{180}{\pi} = \frac{11 \times 180}{16\pi} = \frac{11 \times 180 \times 7}{16 \times 22} = \frac{11 \times 45 \times 7}{4 \times 22} = \frac{315}{8} \]
Step 2: Convert \( \frac{315}{8} = 39\frac{3}{8}° = 39° + \frac{3 \times 60}{8}’ = 39°22’30” \)
\( \therefore \) \( \frac{11}{16} \) rad = 39°22′30″
Step 1: \( -4 \times \frac{180}{\pi} = \frac{-720}{\pi} = \frac{-720 \times 7}{22} = \frac{-5040}{22} = -229\frac{1}{11}° \)
Step 2: \( \frac{1}{11}° = \frac{60}{11}’ = 5\frac{5}{11}’ \); \( \frac{5}{11}’ = \frac{300}{11}” = 27\frac{3}{11}” \)
\( \therefore \) −4 rad = −229°5′27″ (approx)
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?
Step 1: Revolutions per second \( = \frac{360}{60} = 6 \) revolutions/second.
Step 2: One complete revolution = \( 2\pi \) radians.
Step 3: Radians in 6 revolutions \( = 6 \times 2\pi = 12\pi \) 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 \( \theta = \frac{l}{r} \), where \( l = 22 \) cm, \( r = 100 \) cm.
\[ \theta = \frac{22}{100} = \frac{11}{50} \text{ radians} \]
Step 2: Convert to degrees: \( \frac{11}{50} \times \frac{180}{\pi} = \frac{11 \times 180 \times 7}{50 \times 22} = \frac{11 \times 180 \times 7}{1100} = \frac{1386}{110} = 12.6° \)
Step 3: \( 12.6° = 12° + 0.6 \times 60′ = 12°36′ \)
\( \therefore \) 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 \( r = \frac{40}{2} = 20 \) cm. Chord AB = 20 cm.
Step 2: Since OA = OB = AB = 20 cm, triangle OAB is equilateral. So the central angle \( \theta = 60° = \frac{\pi}{3} \) radians.
Step 3: Arc length \( = r\theta = 20 \times \frac{\pi}{3} = \frac{20\pi}{3} \) cm.
\( \therefore \) Length of minor arc \( = \frac{20\pi}{3} \) 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 angles: \( 60° = \frac{\pi}{3} \) rad, \( 75° = \frac{5\pi}{12} \) rad.
Step 2: Since arc length \( l \) is the same: \( l = r_1 \theta_1 = r_2 \theta_2 \)
\[ r_1 \cdot \frac{\pi}{3} = r_2 \cdot \frac{5\pi}{12} \]
Step 3: Solve for the ratio:
\[ \frac{r_1}{r_2} = \frac{5\pi/12}{\pi/3} = \frac{5\pi}{12} \times \frac{3}{\pi} = \frac{5}{4} \]
\( \therefore \) \( r_1 : r_2 = 5 : 4 \)
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.
\( \therefore \) \( \theta = \frac{2}{15} \) radians
\( \therefore \) \( \theta = \frac{1}{5} \) radians
\( \therefore \) \( \theta = \frac{7}{25} \) radians
Exercise 3.2 — Trigonometric Functions and Quadrants
Question 1
Medium
Find the values of other five trigonometric functions if \( \cos x = -\frac{1}{2} \), x lies in third quadrant.
Step 1: \( \sec x = \frac{1}{\cos x} = -2 \)
Step 2: Use \( \sin^2 x + \cos^2 x = 1 \): \( \sin^2 x = 1 – \frac{1}{4} = \frac{3}{4} \), so \( \sin x = \pm\frac{\sqrt{3}}{2} \).
Step 3: In Q3, sin is negative: \( \sin x = -\frac{\sqrt{3}}{2} \), \( \cosec x = -\frac{2}{\sqrt{3}} = -\frac{2\sqrt{3}}{3} \)
Step 4: \( \tan x = \frac{\sin x}{\cos x} = \frac{-\sqrt{3}/2}{-1/2} = \sqrt{3} \), \( \cot x = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3} \)
\( \therefore \) \( \sin x = -\frac{\sqrt{3}}{2},\ \cosec x = -\frac{2\sqrt{3}}{3},\ \tan x = \sqrt{3},\ \cot x = \frac{\sqrt{3}}{3},\ \sec x = -2 \)
Question 2
Medium
Find the values of other five trigonometric functions if \( \sin x = \frac{3}{5} \), x lies in second quadrant.
Step 1: \( \cosec x = \frac{5}{3} \)
Step 2: \( \cos^2 x = 1 – \frac{9}{25} = \frac{16}{25} \). In Q2, \( \cos x = -\frac{4}{5} \), \( \sec x = -\frac{5}{4} \)
Step 3: \( \tan x = \frac{3/5}{-4/5} = -\frac{3}{4} \), \( \cot x = -\frac{4}{3} \)
\( \therefore \) \( \cos x = -\frac{4}{5},\ \sec x = -\frac{5}{4},\ \tan x = -\frac{3}{4},\ \cot x = -\frac{4}{3},\ \cosec x = \frac{5}{3} \)
Question 3
Medium
Find the values of other five trigonometric functions if \( \cot x = \frac{3}{4} \), x lies in third quadrant.
Step 1: \( \tan x = \frac{4}{3} \) (In Q3, tan is positive — consistent.)
Step 2: \( \sec^2 x = 1 + \tan^2 x = 1 + \frac{16}{9} = \frac{25}{9} \). In Q3, \( \cos x < 0 \), so \( \cos x = -\frac{3}{5} \), \( \sec x = -\frac{5}{3} \).
Step 3: \( \sin x = \tan x \cdot \cos x = \frac{4}{3} \times \left(-\frac{3}{5}\right) = -\frac{4}{5} \), \( \cosec x = -\frac{5}{4} \)
\( \therefore \) \( \sin x = -\frac{4}{5},\ \cos x = -\frac{3}{5},\ \tan x = \frac{4}{3},\ \sec x = -\frac{5}{3},\ \cosec x = -\frac{5}{4} \)
Question 4
Medium
Find the values of other five trigonometric functions if \( \sec x = \frac{13}{5} \), x lies in fourth quadrant.
Step 1: \( \cos x = \frac{5}{13} \) (In Q4, cos is positive — consistent.)
Step 2: \( \sin^2 x = 1 – \frac{25}{169} = \frac{144}{169} \). In Q4, \( \sin x = -\frac{12}{13} \), \( \cosec x = -\frac{13}{12} \)
Step 3: \( \tan x = \frac{-12/13}{5/13} = -\frac{12}{5} \), \( \cot x = -\frac{5}{12} \)
\( \therefore \) \( \sin x = -\frac{12}{13},\ \cos x = \frac{5}{13},\ \tan x = -\frac{12}{5},\ \cot x = -\frac{5}{12},\ \cosec x = -\frac{13}{12} \)
Question 5
Medium
Find the values of other five trigonometric functions if \( \tan x = -\frac{5}{12} \), x lies in second quadrant.
Step 1: \( \cot x = -\frac{12}{5} \)
Step 2: \( \sec^2 x = 1 + \tan^2 x = 1 + \frac{25}{144} = \frac{169}{144} \). In Q2, \( \cos x < 0 \): \( \cos x = -\frac{12}{13} \), \( \sec x = -\frac{13}{12} \)
Step 3: \( \sin x = \tan x \cdot \cos x = \left(-\frac{5}{12}\right)\left(-\frac{12}{13}\right) = \frac{5}{13} \), \( \cosec x = \frac{13}{5} \)
\( \therefore \) \( \sin x = \frac{5}{13},\ \cos x = -\frac{12}{13},\ \cot x = -\frac{12}{5},\ \sec x = -\frac{13}{12},\ \cosec x = \frac{13}{5} \)
Question 6
Easy
Find the value of the trigonometric function \( \sin 765° \).
Key Concept: \( \sin \) has period \( 360° \), so \( \sin(\theta + 360°n) = \sin\theta \).
\[ \sin 765° = \sin(2 \times 360° + 45°) = \sin 45° = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \]
\( \therefore \) \( \sin 765° = \frac{\sqrt{2}}{2} \)
Question 7
Easy
Find the value of the trigonometric function \( \cosec(-1410°) \).
Step 1: Add multiples of 360° to get a standard angle:
\[ -1410° + 4 \times 360° = -1410° + 1440° = 30° \]
Step 2: \( \cosec(-1410°) = \cosec 30° = 2 \)
\( \therefore \) \( \cosec(-1410°) = 2 \)
Question 8
Easy
Find the value of the trigonometric function \( \tan\frac{19\pi}{3} \).
Key Concept: \( \tan \) has period \( \pi \).
\[ \tan\frac{19\pi}{3} = \tan\left(6\pi + \frac{\pi}{3}\right) = \tan\frac{\pi}{3} = \sqrt{3} \]
\( \therefore \) \( \tan\frac{19\pi}{3} = \sqrt{3} \)
Question 9
Easy
Find the value of the trigonometric function \( \sin\left(-\frac{11\pi}{3}\right) \).
Step 1: Add \( 2 \times 2\pi = 4\pi \):
\[ -\frac{11\pi}{3} + 4\pi = -\frac{11\pi}{3} + \frac{12\pi}{3} = \frac{\pi}{3} \]
\[ \sin\left(-\frac{11\pi}{3}\right) = \sin\frac{\pi}{3} = \frac{\sqrt{3}}{2} \]
\( \therefore \) \( \sin\left(-\frac{11\pi}{3}\right) = \frac{\sqrt{3}}{2} \)
Question 10
Easy
Find the value of the trigonometric function \( \cot\left(-\frac{15\pi}{4}\right) \).
Step 1: \( \cot \) has period \( \pi \). Add \( 4\pi \):
\[ -\frac{15\pi}{4} + 4\pi = -\frac{15\pi}{4} + \frac{16\pi}{4} = \frac{\pi}{4} \]
\[ \cot\left(-\frac{15\pi}{4}\right) = \cot\frac{\pi}{4} = 1 \]
\( \therefore \) \( \cot\left(-\frac{15\pi}{4}\right) = 1 \)
Exercise 3.3 — Sum and Difference Formulas (Proofs)
Question 1
Easy
Prove that: \( \sin^2\frac{\pi}{6} + \cos^2\frac{\pi}{3} – \tan^2\frac{\pi}{4} = -\frac{1}{2} \)
Step 1: Substitute standard values: \( \sin\frac{\pi}{6} = \frac{1}{2},\ \cos\frac{\pi}{3} = \frac{1}{2},\ \tan\frac{\pi}{4} = 1 \)
\[ \text{LHS} = \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2 – (1)^2 = \frac{1}{4} + \frac{1}{4} – 1 = \frac{1}{2} – 1 = -\frac{1}{2} \]
\( \therefore \) LHS = RHS = \( -\frac{1}{2} \). Hence proved.
Question 4
Easy
Prove that: \( 2\sin^2\frac{3\pi}{4} + 2\cos^2\frac{\pi}{4} + 2\sec^2\frac{\pi}{3} = 10 \)
Step 1: \( \sin\frac{3\pi}{4} = \sin\left(\pi – \frac{\pi}{4}\right) = \sin\frac{\pi}{4} = \frac{1}{\sqrt{2}} \)
Step 2: \( \cos\frac{\pi}{4} = \frac{1}{\sqrt{2}} \), \( \sec\frac{\pi}{3} = 2 \)
\[ \text{LHS} = 2\left(\frac{1}{\sqrt{2}}\right)^2 + 2\left(\frac{1}{\sqrt{2}}\right)^2 + 2(2)^2 = 2 \cdot \frac{1}{2} + 2 \cdot \frac{1}{2} + 8 = 1 + 1 + 8 = 10 \]
\( \therefore \) LHS = 10 = RHS. Hence proved.
Question 5
Medium
Find the value of: (i) \( \sin 75° \) (ii) \( \tan 15° \)
Step 1: Write \( 75° = 45° + 30° \) and apply \( \sin(A+B) \) formula:
\[ \sin 75° = \sin 45° \cos 30° + \cos 45° \sin 30° \]
\[ = \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{2} = \frac{\sqrt{3} + 1}{2\sqrt{2}} = \frac{\sqrt{6}+\sqrt{2}}{4} \]
\( \therefore \) \( \sin 75° = \frac{\sqrt{6}+\sqrt{2}}{4} \)
Step 1: Write \( 15° = 45° – 30° \) and apply \( \tan(A-B) \) formula:
\[ \tan 15° = \frac{\tan 45° – \tan 30°}{1 + \tan 45° \tan 30°} = \frac{1 – \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}} = \frac{\sqrt{3}-1}{\sqrt{3}+1} \]
Step 2: Rationalise: \( \frac{(\sqrt{3}-1)^2}{(\sqrt{3}+1)(\sqrt{3}-1)} = \frac{3 – 2\sqrt{3} + 1}{2} = 2 – \sqrt{3} \)
\( \therefore \) \( \tan 15° = 2 – \sqrt{3} \)
Question 8
Medium
Prove that: \( \frac{\cos(\pi+x)\cos(-x)}{\sin(\pi-x)\cos\left(\frac{\pi}{2}+x\right)} = \cot^2 x \)
Step 1: Apply allied angle identities: \( \cos(\pi+x) = -\cos x \), \( \cos(-x) = \cos x \), \( \sin(\pi-x) = \sin x \), \( \cos\left(\frac{\pi}{2}+x\right) = -\sin x \)
\[ \text{LHS} = \frac{(-\cos x)(\cos x)}{(\sin x)(-\sin x)} = \frac{-\cos^2 x}{-\sin^2 x} = \frac{\cos^2 x}{\sin^2 x} = \cot^2 x \]
\( \therefore \) LHS = \( \cot^2 x \) = RHS. Hence proved.
Question 12
Hard
Prove that: \( \sin^2 6x – \sin^2 4x = \sin 2x \sin 10x \)
Step 1: Use the identity \( \sin^2 A – \sin^2 B = (\sin A + \sin B)(\sin A – \sin B) \):
\[ \text{LHS} = (\sin 6x + \sin 4x)(\sin 6x – \sin 4x) \]
Step 2: Apply sum-to-product: \( \sin 6x + \sin 4x = 2\sin 5x \cos x \) and \( \sin 6x – \sin 4x = 2\cos 5x \sin x \)
\[ = (2\sin 5x \cos x)(2\cos 5x \sin x) = (2\sin 5x \cos 5x)(2\sin x \cos x) = \sin 10x \cdot \sin 2x \]
\( \therefore \) LHS = \( \sin 2x \sin 10x \) = RHS. Hence proved.
Solved Examples Beyond NCERT — Trigonometric Functions Class 11
Extra Example 1
Medium
If \( \sin A = \frac{4}{5} \) and \( \cos B = \frac{5}{13} \), where A and B are in the first quadrant, find \( \sin(A+B) \).
Step 1: Find \( \cos A \): \( \cos A = \sqrt{1 – \frac{16}{25}} = \frac{3}{5} \)
Step 2: Find \( \sin B \): \( \sin B = \sqrt{1 – \frac{25}{169}} = \frac{12}{13} \)
Step 3: Apply \( \sin(A+B) = \sin A \cos B + \cos A \sin B \):
\[ = \frac{4}{5} \cdot \frac{5}{13} + \frac{3}{5} \cdot \frac{12}{13} = \frac{20}{65} + \frac{36}{65} = \frac{56}{65} \]
\( \therefore \) \( \sin(A+B) = \frac{56}{65} \)
Extra Example 2
Hard
Prove that \( \cos 20° \cos 40° \cos 80° = \frac{1}{8} \).
Step 1: Multiply and divide by \( 2\sin 20° \):
\[ = \frac{2\sin 20° \cos 20°}{2\sin 20°} \cdot \cos 40° \cos 80° = \frac{\sin 40°}{2\sin 20°} \cdot \cos 40° \cos 80° \]
Step 2: Multiply and divide by 2 again: \( = \frac{\sin 80°}{4\sin 20°} \cdot \cos 80° = \frac{\sin 160°}{8\sin 20°} \)
Step 3: \( \sin 160° = \sin(180°-20°) = \sin 20° \), so the expression \( = \frac{\sin 20°}{8\sin 20°} = \frac{1}{8} \)
\( \therefore \) \( \cos 20° \cos 40° \cos 80° = \frac{1}{8} \). Hence proved.
Topic-Wise Important Questions for Board Exam — Class 11 Maths Chapter 3
1-Mark Questions
- Convert \( 240° \) to radians. Answer: \( \frac{4\pi}{3} \)
- What is the period of \( \tan x \)? Answer: \( \pi \)
- Find \( \sin(-30°) \). Answer: \( -\frac{1}{2} \)
3-Mark Questions
- If \( \cos x = -\frac{3}{5} \) and x is in Q3, find all six trigonometric values.
- Prove: \( \sin 2x + 2\sin 4x + \sin 6x = 4\cos^2 x \sin 4x \)
5-Mark Questions
- Prove that \( \cos^2 2x – \cos^2 6x = \sin 4x \sin 8x \) using sum-to-product identities. Show all steps clearly.
Common Mistakes Students Make — Class 11 Maths Chapter 3
Mistake 1: Forgetting to check the quadrant when finding \( \sin x \) from \( \cos x \).
Why it’s wrong: \( \sin x = \pm\sqrt{1-\cos^2 x} \) — both values are algebraically valid but only one is correct for the given quadrant.
Correct approach: Always state the quadrant rule (ASTC) before choosing the sign.
Mistake 2: Writing \( \cos(A+B) = \cos A + \cos B \).
Why it’s wrong: The cosine of a sum is NOT the sum of cosines. This is one of the most common errors in Exercise 3.3.
Correct approach: Use \( \cos(A+B) = \cos A \cos B – \sin A \sin B \).
Mistake 3: Not converting angles to radians before using \( l = r\theta \).
Why it’s wrong: The arc length formula requires \( \theta \) in radians. Using degrees gives a wrong answer.
Correct approach: Always convert to radians first: \( \theta_{\text{rad}} = \theta_{\text{deg}} \times \frac{\pi}{180} \).
Mistake 4: Confusing the period of \( \sin/\cos \) (which is \( 2\pi \)) with the period of \( \tan/\cot \) (which is \( \pi \)).
Correct approach: Memorise: \( \sin, \cos, \sec, \cosec \) → period \( 2\pi \); \( \tan, \cot \) → period \( \pi \).
Mistake 5: Leaving \( \frac{\sqrt{3}-1}{\sqrt{3}+1} \) without rationalising in the final answer.
Correct approach: Always rationalise surds in the denominator. The answer \( \tan 15° = 2-\sqrt{3} \) is the expected form.
Exam Tips for 2026-27 — CBSE Class 11 Maths Chapter 3
- Write the formula before substituting: The 2026-27 CBSE marking scheme awards 1 mark for writing the correct formula, even if the calculation has a minor error.
- Proof questions — always start from LHS: Unless the question says otherwise, always start from the more complex side. Examiners deduct marks for starting from RHS without justification.
- State the quadrant rule explicitly: In Exercise 3.2-type questions, write “Since x lies in Q__, the sign of __ is __” before assigning signs. This earns method marks.
- Radian measure questions are easy marks: Exercise 3.1 questions are straightforward formula applications — practise these for guaranteed marks in the 2026-27 board exam.
- Note on Ex 3.4: Trigonometric Equations (Exercise 3.4) is removed from the CBSE 2026-27 rationalised syllabus. Do not spend time on it for board exam preparation.
- Last-minute checklist: (1) All standard values (sin/cos/tan of 0°, 30°, 45°, 60°, 90°) ✓ (2) ASTC rule ✓ (3) Sum-difference formulas ✓ (4) Sum-to-product formulas ✓ (5) Periodicity of all six functions ✓
For more solutions across all chapters, visit our complete NCERT Solutions for Class 11 hub, or explore all subjects at NCERT Solutions.