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Class 11 Maths NCERT Solutions

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Trigonometric Functions Class 11 Maths NCERT Solutions

The complete NCERT exercise solutions for Chapter 3, Trigonometric Functions — 51 questions from Ex 3.1 to Ex 3.4, each worked through step by step in the CBSE marking pattern. Radian measure, compound and multiple angle identities, transformations, and the general solutions of trigonometric equations.

Class:11Subject:MathsChapter:3
4 Key Formulas
DWritten byDeep Narayan
Updated
Key Concept Summary

How many questions are in NCERT Class 11 Maths Chapter 3?

Chapter 3 carries 4 exercise questions, numbered Ex 3.1 to Ex 3.4. All of them are solved step by step on this page, along with the chapter's key formulas and exam pointers.

01

Chapter Overview

This chapter converts between degree and radian measure, defines the six trigonometric functions on the unit circle, and develops the compound-angle, double-angle and sum-to-product identities together with general solutions of trigonometric equations. These identities are the workhorses of Class 11 and reappear in Class 12 calculus. Every question below is from the NCERT Class 11 textbook (rationalised edition), worked line by line.

Board pattern

Proof questions earn method marks: state the identity you use, apply it to both sides (or LHS only), and stop when both sides match. For general solutions always include '+ nπ' or '+ 2nπ' with the correct period — forgetting the period loses the mark. Remember π radians = 180° and the sign of each function in the four quadrants.
02

Exercise 3.1 — Angles and their Measures

7Exercise questions

Step-by-step solution

  1. 1Use 180° = π radians, so 1° = π/180.
  2. 2(i) 25° = 25π/180 = 5π/36 rad.
  3. 3(ii) −47°30′ = −(47.5)° = −47.5 × π/180 = −19π/72 rad.
  4. 4(iii) 240° = 240π/180 = 4π/3 rad.
  5. 5(iv) 520° = 520π/180 = 26π/9 rad.

Final answer

(i) 5π/36 (ii) −19π/72 (iii) 4π/3 (iv) 26π/9 radians.

Step-by-step solution

  1. 1(i) (11/16) × 180/π = (11/16) × 180 × 7/22 = 39.375° = 39°22′30″.
  2. 2(ii) −4 rad = −4 × 180 × 7/22 = −2520/11 = −229 1/11° = −229°5′27″.
  3. 3(iii) (5π/3)(180/π) = 300°.
  4. 4(iv) (7π/6)(180/π) = 210°.

Final answer

(i) 39°22′30″ (ii) −229°5′27″ (iii) 300° (iv) 210°.

Step-by-step solution

  1. 11 revolution = 2π radians.
  2. 2360 revolutions per minute = 360/60 = 6 revolutions per second.
  3. 3Angle per second = 6 × 2π = 12π radians.

Final answer

12π radians per second.

Step-by-step solution

  1. 1Arc length l = rθ → θ = l/r = 22/100 rad.
  2. 2θ = 0.22 rad = 0.22 × 180/π = 0.22 × 180 × 7/22 = 12.6°.
  3. 312.6° = 12°36′.

Final answer

12°36′.

Step-by-step solution

  1. 1Radius r = 20 cm. Chord length 20 = 2r sin(θ/2) → 20 = 40 sin(θ/2).
  2. 2sin(θ/2) = 1/2 → θ/2 = π/6 → θ = π/3.
  3. 3Minor arc = rθ = 20 × π/3 = 20π/3 cm.

Final answer

20π/3 cm.

Step-by-step solution

  1. 1l = r₁θ₁ = r₂θ₂ with the same arc length l.
  2. 2θ₁ = 60° = π/3, θ₂ = 75° = 5π/12.
  3. 3r₁/r₂ = θ₂/θ₁ = (5π/12)/(π/3) = 5/4.

Final answer

r₁ : r₂ = 5 : 4.

Step-by-step solution

  1. 1θ = l/r with r = 75 cm.
  2. 2(i) θ = 10/75 = 2/15 rad.
  3. 3(ii) θ = 15/75 = 1/5 rad.
  4. 4(iii) θ = 21/75 = 7/25 rad.

Final answer

(i) 2/15 (ii) 1/5 (iii) 7/25 radians.

03

Exercise 3.2 — Trigonometric Functions of an Angle

10Exercise questions

Step-by-step solution

  1. 1sin²x = 1 − cos²x = 1 − 1/4 = 3/4.
  2. 2Third quadrant → sin x < 0 → sin x = −√3/2.
  3. 3tan x = sin x/cos x = (−√3/2)/(−1/2) = √3.
  4. 4cosec x = −2/√3, sec x = −2, cot x = 1/√3.

Final answer

sin x = −√3/2, tan x = √3, cosec x = −2/√3, sec x = −2, cot x = 1/√3.

Step-by-step solution

  1. 1cos²x = 1 − 9/25 = 16/25 → cos x = ±4/5.
  2. 2Second quadrant → cos x = −4/5.
  3. 3tan x = (3/5)/(−4/5) = −3/4.
  4. 4cosec x = 5/3, sec x = −5/4, cot x = −4/3.

Final answer

cos x = −4/5, tan x = −3/4, cosec x = 5/3, sec x = −5/4, cot x = −4/3.

Step-by-step solution

  1. 1tan x = 1/cot x = 4/3 (cot is positive in the third quadrant).
  2. 2Draw a right triangle with opposite 4, adjacent 3 → hypotenuse 5.
  3. 3Third quadrant → sin x = −4/5, cos x = −3/5.
  4. 4cosec x = −5/4, sec x = −5/3.

Final answer

tan x = 4/3, sin x = −4/5, cos x = −3/5, cosec x = −5/4, sec x = −5/3.

Step-by-step solution

  1. 1cos x = 1/sec x = 5/13.
  2. 2sin²x = 1 − 25/169 = 144/169 → sin x = ±12/13.
  3. 3Fourth quadrant → sin x = −12/13.
  4. 4cosec x = −13/12, tan x = −12/5, cot x = −5/12.

Final answer

cos x = 5/13, sin x = −12/13, cosec x = −13/12, tan x = −12/5, cot x = −5/12.

Step-by-step solution

  1. 1Triangle: opposite 5, adjacent 12 → hypotenuse 13.
  2. 2Second quadrant → sin x = 5/13, cos x = −12/13.
  3. 3cosec x = 13/5, sec x = −13/12, cot x = −12/5.

Final answer

cot x = −12/5, sin x = 5/13, cos x = −12/13, cosec x = 13/5, sec x = −13/12.

Step-by-step solution

  1. 1Subtract full revolutions: 765° − 720° = 45°.
  2. 2sin 765° = sin 45° = √2/2.

Final answer

√2/2.

Step-by-step solution

  1. 1cos(−1710°) = cos(1710°). 1710° = 4 × 360° + 270°.
  2. 2cos 1710° = cos 270° = 0.

Final answer

0.

Step-by-step solution

  1. 1−1410° + 4 × 360° = −1410° + 1440° = 30°.
  2. 2cosec 30° = 2.

Final answer

2.

Step-by-step solution

  1. 119π/3 = 18π/3 + π/3 = 6π + π/3.
  2. 2tan has period π: tan(6π + π/3) = tan(π/3) = √3.

Final answer

√3.

Step-by-step solution

  1. 1−15π/4 = −12π/4 − 3π/4 → cot(−15π/4) = cot(−3π/4) since cot has period π: −12π/4 = −3π is an integer multiple of π.
  2. 2cot(−3π/4) = −cot(3π/4) = −(−1) = 1.

Final answer

1.

04

Exercise 3.3 — Trigonometric Identities

25Exercise questions

Step-by-step solution

  1. 1sin(π/6) = 1/2 → sin²(π/6) = 1/4.
  2. 2cos(π/3) = 1/2 → cos²(π/3) = 1/4. tan(π/4) = 1 → tan²(π/4) = 1.
  3. 3LHS = 1/4 + 1/4 − 1 = −1/2 = RHS.

Final answer

Proved: LHS = −1/2.

Step-by-step solution

  1. 1sin²(π/6) = 1/4 so 2 sin²(π/6) = 1/2.
  2. 2cosec(7π/6) = 1/sin(7π/6) = 1/(−1/2) = −2 → cosec²(7π/6) = 4.
  3. 3cos²(π/3) = 1/4, so 4 × 1/4 = 1.
  4. 4LHS = 1/2 + 1 = 3/2 = RHS.

Final answer

Proved: LHS = 3/2.

Step-by-step solution

  1. 1cot(π/6) = √3 → cot² = 3. tan(π/6) = 1/√3 → 3 tan² = 3 × 1/3 = 1.
  2. 2cosec(5π/6) = 1/sin(5π/6) = 1/(1/2) = 2.
  3. 3LHS = 3 + 2 + 1 = 6 = RHS.

Final answer

Proved: LHS = 6.

Step-by-step solution

  1. 1sin(3π/4) = √2/2 → sin² = 1/2. cos(π/4) = √2/2 → cos² = 1/2.
  2. 2sec(π/3) = 2 → sec² = 4.
  3. 3LHS = 2(1/2) + 2(1/2) + 2(4) = 1 + 1 + 8 = 10 = RHS.

Final answer

Proved: LHS = 10.

Step-by-step solution

  1. 1sin 75° = sin(45° + 30°).
  2. 2= sin 45° cos 30° + cos 45° sin 30° = (√2/2)(√3/2) + (√2/2)(1/2).
  3. 3= (√6 + √2)/4.

Final answer

(√6 + √2)/4.

Step-by-step solution

  1. 1Use cos A cos B − sin A sin B = cos(A + B) with A = π/4 − x, B = π/4 − y.
  2. 2LHS = cos((π/4 − x) + (π/4 − y)) = cos(π/2 − (x + y)) = sin(x + y) = RHS.

Final answer

Proved.

Step-by-step solution

  1. 1Use tan(A + B) = (tan A + tan B)/(1 − tan A tan B) with tan(π/4) = 1.
  2. 2tan(π/4 + x) = (1 + tan x)/(1 − tan x) and tan(π/4 − x) = (1 − tan x)/(1 + tan x).
  3. 3Ratio = [(1 + tan x)/(1 − tan x)] / [(1 − tan x)/(1 + tan x)] = ((1 + tan x)/(1 − tan x))² = RHS.

Final answer

Proved.

Step-by-step solution

  1. 1cos(π + x) = −cos x; cos(−x) = cos x; sin(π − x) = sin x; cos(π/2 + x) = −sin x.
  2. 2Numerator = (−cos x)(cos x) = −cos²x. Denominator = (sin x)(−sin x) = −sin²x.
  3. 3LHS = cos²x/sin²x = cot²x = RHS.

Final answer

Proved.

Step-by-step solution

  1. 1cos(3π/2 + x) = sin x; cos(2π + x) = cos x.
  2. 2cot(3π/2 − x) = tan x; cot(2π + x) = cot x.
  3. 3LHS = sin x cos x (tan x + cot x) = sin x cos x (sin x/cos x + cos x/sin x).
  4. 4= sin x cos x × (sin²x + cos²x)/(sin x cos x) = 1 = RHS.

Final answer

Proved.

Step-by-step solution

  1. 1This is cos A cos B + sin A sin B = cos(A − B) with A = (n + 2)x, B = (n + 1)x.
  2. 2LHS = cos((n + 2)x − (n + 1)x) = cos x = RHS.

Final answer

Proved.

Step-by-step solution

  1. 1Use cos(A + x) − cos(A − x) = −2 sin A sin x.
  2. 2Here A = 3π/4: LHS = −2 sin(3π/4) sin x = −2(√2/2) sin x = −√2 sin x = RHS.

Final answer

Proved.

Step-by-step solution

  1. 1sin²A − sin²B = sin(A + B) sin(A − B).
  2. 2LHS = sin(6x + 4x) sin(6x − 4x) = sin 10x sin 2x = RHS.

Final answer

Proved.

Step-by-step solution

  1. 1Use cos²A − cos²B = −sin(A + B) sin(A − B) (derived from sum-to-product or identities).
  2. 2LHS = −sin(2x + 6x) sin(2x − 6x) = −sin 8x sin(−4x) = sin 8x sin 4x = RHS.

Final answer

Proved.

Step-by-step solution

  1. 1Group sin 2x + sin 6x = 2 sin 4x cos 2x (sum-to-product).
  2. 2LHS = 2 sin 4x cos 2x + 2 sin 4x = 2 sin 4x (cos 2x + 1).
  3. 3cos 2x + 1 = 2 cos²x → LHS = 2 sin 4x × 2 cos²x = 4 cos²x sin 4x = RHS.

Final answer

Proved.

Step-by-step solution

  1. 1sin 5x + sin 3x = 2 sin 4x cos x (sum-to-product).
  2. 2sin 5x − sin 3x = 2 cos 4x sin x.
  3. 3LHS = cot 4x · 2 sin 4x cos x = 2 cos 4x cos x.
  4. 4RHS = cot x · 2 cos 4x sin x = 2 cos x cos 4x × (cos x/sin x × sin x)... = 2 cos 4x cos x. Both sides equal, hence proved (correcting: RHS = (cos x/sin x)·2 cos 4x sin x = 2 cos 4x cos x).

Final answer

Proved — both sides equal 2 cos 4x cos x.

Step-by-step solution

  1. 1cos 9x − cos 5x = −2 sin 7x sin 2x.
  2. 2sin 17x − sin 3x = 2 cos 10x sin 7x.
  3. 3Ratio = (−2 sin 7x sin 2x)/(2 cos 10x sin 7x) = −sin 2x / cos 10x = RHS.

Final answer

Proved.

Step-by-step solution

  1. 1sin 5x + sin 3x = 2 sin 4x cos x.
  2. 2cos 5x + cos 3x = 2 cos 4x cos x.
  3. 3Ratio = (2 sin 4x cos x)/(2 cos 4x cos x) = tan 4x = RHS.

Final answer

Proved.

Step-by-step solution

  1. 1sin x − sin y = 2 cos((x+y)/2) sin((x−y)/2).
  2. 2cos x + cos y = 2 cos((x+y)/2) cos((x−y)/2).
  3. 3Ratio = sin((x−y)/2)/cos((x−y)/2) = tan((x−y)/2) = RHS.

Final answer

Proved.

Step-by-step solution

  1. 1sin x + sin 3x = 2 sin 2x cos x.
  2. 2cos x + cos 3x = 2 cos 2x cos x.
  3. 3Ratio = sin 2x/cos 2x = tan 2x = RHS.

Final answer

Proved.

Step-by-step solution

  1. 1sin x − sin 3x = −2 cos 2x sin x (using sin 3x − sin x = 2 cos 2x sin x).
  2. 2sin²x − cos²x = −cos 2x.
  3. 3Ratio = (−2 cos 2x sin x)/(−cos 2x) = 2 sin x = RHS.

Final answer

Proved.

Step-by-step solution

  1. 1cos 4x + cos 2x = 2 cos 3x cos x, and sin 4x + sin 2x = 2 sin 3x cos x.
  2. 2Numerator = 2 cos 3x cos x + cos 3x = cos 3x(2 cos x + 1).
  3. 3Denominator = 2 sin 3x cos x + sin 3x = sin 3x(2 cos x + 1).
  4. 4Ratio = cos 3x/sin 3x = cot 3x = RHS.

Final answer

Proved.

Step-by-step solution

  1. 1Write 3x = x + 2x: cot 3x = cot(x + 2x) = (cot x cot 2x − 1)/(cot x + cot 2x).
  2. 2Cross-multiply: cot 3x (cot x + cot 2x) = cot x cot 2x − 1.
  3. 3cot x cot 3x + cot 2x cot 3x = cot x cot 2x − 1.
  4. 4Rearrange: cot x cot 2x − cot 2x cot 3x − cot 3x cot x = 1 = RHS.

Final answer

Proved.

Step-by-step solution

  1. 1tan 2x = 2t/(1 − t²) where t = tan x.
  2. 2tan 4x = tan(2x + 2x) = 2 tan 2x/(1 − tan² 2x) = [4t/(1 − t²)] / [1 − 4t²/(1 − t²)²].
  3. 3= 4t(1 − t²)/[(1 − t²)² − 4t²] = 4t(1 − t²)/(1 − 2t² + t⁴ − 4t²).
  4. 4= 4 tan x (1 − tan²x)/(1 − 6 tan²x + tan⁴x) = RHS.

Final answer

Proved.

Step-by-step solution

  1. 1cos 4x = cos(2 · 2x) = 1 − 2 sin²2x.
  2. 2sin 2x = 2 sin x cos x → sin²2x = 4 sin²x cos²x.
  3. 3LHS = 1 − 2(4 sin²x cos²x) = 1 − 8 sin²x cos²x = RHS.

Final answer

Proved.

Step-by-step solution

  1. 1cos 6x = cos(2 · 3x) = 2 cos²3x − 1.
  2. 2cos 3x = 4 cos³x − 3 cos x, so cos²3x = 16 cos⁶x − 24 cos⁴x + 9 cos²x.
  3. 32 cos²3x − 1 = 32 cos⁶x − 48 cos⁴x + 18 cos²x − 1 = RHS.

Final answer

Proved.

05

Exercise 3.4 — Trigonometric Equations

9Exercise questions

Step-by-step solution

  1. 1tan(π/3) = √3 and tan(4π/3) = √3; principal solutions are the ones in [0, 2π): π/3 and 4π/3.
  2. 2General solution: tan x = tan α with α = π/3 gives x = nπ + π/3, n ∈ Z.

Final answer

Principal: x = π/3, 4π/3. General: x = nπ + π/3, n ∈ Z.

Step-by-step solution

  1. 1sec x = 2 → cos x = 1/2.
  2. 2cos(π/3) = 1/2 and cos(5π/3) = 1/2 → principal solutions π/3 and 5π/3.
  3. 3General: x = 2nπ ± π/3, n ∈ Z.

Final answer

Principal: π/3, 5π/3. General: x = 2nπ ± π/3.

Step-by-step solution

  1. 1cot x = −√3 → tan x = −1/√3 = tan(−π/6).
  2. 2In [0, 2π): tan x = −1/√3 at x = 5π/6 and 11π/6 → principal solutions.
  3. 3General: x = nπ − π/6, n ∈ Z.

Final answer

Principal: 5π/6, 11π/6. General: x = nπ − π/6.

Step-by-step solution

  1. 1cosec x = −2 → sin x = −1/2.
  2. 2sin(7π/6) = −1/2 and sin(11π/6) = −1/2 → principal solutions 7π/6, 11π/6.
  3. 3General: x = nπ + (−1)ⁿ(7π/6), n ∈ Z.

Final answer

Principal: 7π/6, 11π/6. General: x = nπ + (−1)ⁿ(7π/6).

Step-by-step solution

  1. 1cos 4x − cos 2x = 0 → −2 sin 3x sin x = 0.
  2. 2sin 3x = 0 → 3x = nπ → x = nπ/3.
  3. 3sin x = 0 → x = nπ, already included in nπ/3 for n multiple of 3.
  4. 4General solution: x = nπ/3, n ∈ Z.

Final answer

x = nπ/3, n ∈ Z.

Step-by-step solution

  1. 1cos 3x + cos x = 2 cos 2x cos x.
  2. 2Equation: 2 cos 2x cos x − cos 2x = 0 → cos 2x (2 cos x − 1) = 0.
  3. 3cos 2x = 0 → 2x = (2n + 1)π/2 → x = (2n + 1)π/4.
  4. 42 cos x − 1 = 0 → cos x = 1/2 → x = 2nπ ± π/3.

Final answer

x = (2n+1)π/4 or x = 2nπ ± π/3, n ∈ Z.

Step-by-step solution

  1. 1sin 2x = 2 sin x cos x → cos x (2 sin x + 1) = 0.
  2. 2cos x = 0 → x = (2n + 1)π/2.
  3. 3sin x = −1/2 → x = nπ + (−1)ⁿ(7π/6).

Final answer

x = (2n+1)π/2 or x = nπ + (−1)ⁿ(7π/6), n ∈ Z.

Step-by-step solution

  1. 1sec²2x = 1 + tan²2x, so 1 + tan²2x = 1 − tan 2x.
  2. 2tan²2x + tan 2x = 0 → tan 2x (tan 2x + 1) = 0.
  3. 3tan 2x = 0 → 2x = nπ → x = nπ/2.
  4. 4tan 2x = −1 → 2x = nπ − π/4 → x = nπ/2 − π/8.

Final answer

x = nπ/2 or x = nπ/2 − π/8, n ∈ Z.

Step-by-step solution

  1. 1sin x + sin 5x = 2 sin 3x cos 2x.
  2. 22 sin 3x cos 2x + sin 3x = 0 → sin 3x (2 cos 2x + 1) = 0.
  3. 3sin 3x = 0 → 3x = nπ → x = nπ/3.
  4. 4cos 2x = −1/2 → 2x = 2nπ ± 2π/3 → x = nπ ± π/3.

Final answer

x = nπ/3 or x = nπ ± π/3, n ∈ Z.

Quick Revision

Key formulas at a glance

Memorise these equations — direct application numericals and derivations in CBSE & JEE frequently hinge on these.

Pythagorean identity

Double angle

General solution of sin x = k

General solution of cos x = k

Exam Strategy

How this chapter is asked

High-yield question patterns observed across CBSE boards, JEE Main & Advanced, and NEET.

  • General solutions come in only two forms — nπ ± or 2nπ ± — so get the form right before worrying about the principal value.
  • The ambiguous case tan x = k needs both the positive and negative branches, and dropping one loses half the solutions.

FAQ

Frequently asked questions

How many questions are in NCERT Class 11 Maths Chapter 3 (Trigonometric Functions)?

There are 4 exercise questions in this chapter, numbered Ex 3.1 to Ex 3.4. Every one is solved step by step on this page in the official NCERT numbering.

Which formulas come up in Trigonometric Functions Class 11 Maths?

The formulas this chapter's questions actually turn on are: Pythagorean identity, Double angle, General solution of sin x = k, General solution of cos x = k. They are listed with their expressions in the key formulas section below, and the solved questions show where each one is used.

Is Trigonometric Functions important for JEE Main?

Very important — identities and general solutions are tested directly in boards and JEE Main, and they underpin every trigonometric equation in Class 12.

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Previous: Relations and FunctionsNext: Complex Numbers and Quadratic EquationsTrigonometric Functions revision notesAll Class 11 Maths NCERT Solutions

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