NCERT Solutions for Class 11 Maths Chapter 3: Trigonometric Functions (NCERT 2026–27)

These Class 11 Maths Chapter 3 solutions cover Trigonometric Functions from the NCERT textbook (Reprint 2026–27). Every question of Exercise 3.1, Exercise 3.2, Exercise 3.3 and the Miscellaneous Exercise is reproduced verbatim and solved step by step — radian–degree conversion, signs and values of trigonometric functions, sum-and-difference identities and half-angle results — with every answer cross-checked against the NCERT answer key.

Class: 11 Subject: Mathematics Chapter: 3 Name: Trigonometric Functions Exercises: 3.1, 3.2, 3.3 + Miscellaneous Session: 2026–27
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Chapter 3 Overview

Chapter 3 of Class 11 Maths, Trigonometric Functions, generalises the trigonometric ratios of acute angles studied earlier into trigonometric functions of any real angle. It begins with the two units of angle measure — degree and radian — and the relation π radian = 180° together with the arc-length formula l = rθ. It then defines sine and cosine using a unit circle, fixes the signs of all six functions in the four quadrants, lists their domains and ranges, and develops the sum and difference identities, double- and triple-angle formulas and the product–to–sum transformations. The Class 11 Maths Chapter 3 solutions below work through every numbered question of each exercise and the Miscellaneous Exercise.

Key Concepts & Definitions

Radian measure: the angle subtended at the centre by an arc equal in length to the radius. π radian = 180°, so 1 radian = 180°/π ≈ 57°16′.

Arc-length relation: if an arc of length l subtends an angle θ (in radians) at the centre of a circle of radius r, then θ = l/r, i.e. l = rθ.

Trigonometric functions: for a point P(a, b) on the unit circle with arc x, cos x = a and sin x = b; the other four are defined as cosec x = 1/sin x, sec x = 1/cos x, tan x = sin x/cos x, cot x = cos x/sin x.

Domain & range: sin and cos are defined for all real x with range [−1, 1]; tan and cot take all real values; sec and cosec have range (−∞, −1] ∪ [1, ∞).

Signs by quadrant (“All–Silver–Tea–Cups”): all positive in I; only sin (& cosec) in II; only tan (& cot) in III; only cos (& sec) in IV.

Periodicity: sin and cos repeat after 2π; tan and cot repeat after π.

Important Formulas (Chapter 3)

Conversions: Radian = (π/180) × Degree  •  Degree = (180/π) × Radian  •  l = rθ.

Pythagorean: sin2x + cos2x = 1  •  1 + tan2x = sec2x  •  1 + cot2x = cosec2x.

Sum/difference: cos(x ± y) = cos x cos y ∓ sin x sin y  •  sin(x ± y) = sin x cos y ± cos x sin y  •  tan(x ± y) = (tan x ± tan y)/(1 ∓ tan x tan y).

Double angle: cos 2x = cos2x − sin2x = 2cos2x − 1 = 1 − 2sin2x = (1 − tan2x)/(1 + tan2x); sin 2x = 2 sin x cos x = 2tan x/(1 + tan2x); tan 2x = 2tan x/(1 − tan2x).

Triple angle: sin 3x = 3 sin x − 4 sin3x; cos 3x = 4 cos3x − 3 cos x; tan 3x = (3 tan x − tan3x)/(1 − 3 tan2x).

Sum–to–product: cos x + cos y = 2 cos½(x+y) cos½(x−y); cos x − cos y = −2 sin½(x+y) sin½(x−y); sin x + sin y = 2 sin½(x+y) cos½(x−y); sin x − sin y = 2 cos½(x+y) sin½(x−y).

Exercise 3.1 Solutions

Questions are reproduced verbatim from the NCERT textbook; the worked solutions are original and verified against the answers at the back of the book.

1. Find the radian measures corresponding to the following degree measures: (i) 25°   (ii) −47°30′   (iii) 240°   (iv) 520°

SOLUTION Radian = (π/180) × degree. (i) 25° = (π/180) × 25 = 5π/36 radian. (ii) 47°30′ = 47½° = 95/2 degree; so −47°30′ = (π/180) × (−95/2) = −19π/72 radian. (iii) 240° = (π/180) × 240 = 4π/3 radian. (iv) 520° = (π/180) × 520 = 26π/9 radian.

2. Find the degree measures corresponding to the following radian measures (Use π = 22/7). (i) 11/16   (ii) −4   (iii) 5π/3   (iv) 7π/6

SOLUTION Degree = (180/π) × radian. (i) (11/16)(180/π) = (11/16) × (180 × 7/22) = (11/16) × (630/11) = 630/16 = 39.375° = 39° + 0.375 × 60′ = 39°22′30″ → 39°22′30″. (ii) −4 × (180 × 7)/(22) = −4 × 1260/22 = −5040/22 = −229.0909°. Now 0.0909° × 60 = 5.454′ → 5′ and 0.454′ × 60 = 27″ → −229°5′27″. (iii) (5π/3)(180/π) = 5 × 60 = 300°. (iv) (7π/6)(180/π) = 7 × 30 = 210°.

3. A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second?

SOLUTION 360 revolutions in 60 seconds → 360/60 = 6 revolutions per second. One revolution = 2π radian, so in one second the wheel turns 6 × 2π = 12π radian.

4. 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 π = 22/7).

SOLUTION θ = l/r = 22/100 = 0.22 radian. Degree = (180/π) × 0.22 = (180 × 7/22) × 0.22 = (1260/22) × 0.22 = 1260 × 0.01 = 12.6°. 0.6° × 60 = 36′, so θ = 12°36′.

5. In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the length of minor arc of the chord.

SOLUTION Radius r = 40/2 = 20 cm; the chord = 20 cm = r. A chord equal to the radius makes an equilateral triangle with the two radii, so the central angle θ = 60° = π/3 radian. Length of minor arc l = rθ = 20 × π/3 = 20π/3 cm.

6. If in two circles, arcs of the same length subtend angles 60° and 75° at the centre, find the ratio of their radii.

SOLUTION Let radii be r1, r2. θ1 = 60° = π/3, θ2 = 75° = 5π/12. Equal arcs: l = r1θ1 = r2θ2, so r1/r2 = θ2/θ1 = (5π/12)/(π/3) = (5/12) × 3 = 5/4. ∴ r1 : r2 = 5 : 4.

7. 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

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

Exercise 3.2 Solutions

In Exercises 1–5, find the values of the other five trigonometric functions; in Exercises 6–10, find the value of the trigonometric function.

1. cos x = −1/2, x lies in third quadrant.

SOLUTION sec x = 1/cos x = −2. sin2x = 1 − cos2x = 1 − 1/4 = 3/4, so sin x = ±√3/2. In Q3 sin is negative → sin x = −√3/2. cosec x = 1/sin x = −2/√3. tan x = sin x/cos x = (−√3/2)/(−1/2) = √3; cot x = 1/√3.

2. sin x = 3/5, x lies in second quadrant.

SOLUTION cosec x = 1/sin x = 5/3. cos2x = 1 − 9/25 = 16/25, cos x = ±4/5; in Q2 cos is negative → cos x = −4/5, sec x = −5/4. tan x = (3/5)/(−4/5) = −3/4; cot x = −4/3.

3. cot x = 3/4, x lies in third quadrant.

SOLUTION tan x = 1/cot x = 4/3. cosec2x = 1 + cot2x = 1 + 9/16 = 25/16, cosec x = ±5/4; in Q3 cosec (sin) is negative → cosec x = −5/4, so sin x = −4/5. cos x = cot x × sin x = (3/4)(−4/5) = −3/5, sec x = −5/3. tan x = 4/3.

4. sec x = 13/5, x lies in fourth quadrant.

SOLUTION cos x = 1/sec x = 5/13. sin2x = 1 − 25/169 = 144/169, sin x = ±12/13; in Q4 sin is negative → sin x = −12/13, cosec x = −13/12. tan x = sin x/cos x = (−12/13)/(5/13) = −12/5; cot x = −5/12.

5. tan x = −5/12, x lies in second quadrant.

SOLUTION sec2x = 1 + tan2x = 1 + 25/144 = 169/144, sec x = ±13/12; in Q2 sec (cos) is negative → sec x = −13/12, cos x = −12/13. sin x = tan x × cos x = (−5/12)(−12/13) = 5/13, cosec x = 13/5. cot x = 1/tan x = −12/5.

6. Find the value of the trigonometric function sin 765°.

SOLUTION sin repeats after 360°: 765° = 720° + 45°. sin 765° = sin(2 × 360° + 45°) = sin 45° = 1/√2.

7. Find the value of the trigonometric function cosec (−1410°).

SOLUTION cosec(−1410°) = −cosec 1410° (cosec is odd). 1410° = 4 × 360° − 30°, but adding 1440°: −1410° + 4 × 360° = −1410° + 1440° = 30°. So cosec(−1410°) = cosec 30° = 2.

8. Find the value of the trigonometric function tan (19π/3).

SOLUTION tan repeats after π. 19π/3 = 6π + π/3 = 3(2π) + π/3. tan(19π/3) = tan(π/3) = √3.

9. Find the value of the trigonometric function sin (−11π/3).

SOLUTION sin(−11π/3) = −sin(11π/3) (sin is odd). 11π/3 = 4π − π/3, so sin(11π/3) = sin(−π/3) = −sin(π/3) = −√3/2. ∴ sin(−11π/3) = −(−√3/2) = √3/2.

10. Find the value of the trigonometric function cot (−15π/4).

SOLUTION cot(−15π/4) = −cot(15π/4) (cot is odd). 15π/4 = 4π − π/4, so cot(15π/4) = cot(−π/4) = −cot(π/4) = −1. ∴ cot(−15π/4) = −(−1) = 1.

Exercise 3.3 Solutions

1. Prove that sin2(π/6) + cos2(π/3) − tan2(π/4) = −1/2.

SOLUTION L.H.S. = (sin π/6)2 + (cos π/3)2 − (tan π/4)2 = (1/2)2 + (1/2)2 − (1)2. = 1/4 + 1/4 − 1 = 1/2 − 1 = −1/2 = R.H.S. ✓

2. Prove that 2sin2(π/6) + cosec2(7π/6) cos2(π/3) = 3/2.

SOLUTION cosec(7π/6) = cosec(π + π/6) = −cosec(π/6) = −2, so cosec2(7π/6) = 4. L.H.S. = 2(1/2)2 + 4 × (1/2)2 = 2(1/4) + 4(1/4) = 1/2 + 1 = 3/2 = R.H.S. ✓

3. Prove that cot2(π/6) + cosec(5π/6) + 3tan2(π/6) = 6.

SOLUTION cot(π/6) = √3 → cot2 = 3. cosec(5π/6) = cosec(π − π/6) = cosec(π/6) = 2. tan(π/6) = 1/√3 → tan2 = 1/3. L.H.S. = 3 + 2 + 3(1/3) = 3 + 2 + 1 = 6 = R.H.S. ✓

4. Prove that 2sin2(3π/4) + 2cos2(π/4) + 2sec2(π/3) = 10.

SOLUTION sin(3π/4) = sin(π − π/4) = 1/√2 → sin2 = 1/2. cos(π/4) = 1/√2 → cos2 = 1/2. sec(π/3) = 2 → sec2 = 4. L.H.S. = 2(1/2) + 2(1/2) + 2(4) = 1 + 1 + 8 = 10 = R.H.S. ✓

5. Find the value of: (i) sin 75°   (ii) tan 15°

SOLUTION (i) sin 75° = sin(45° + 30°) = sin 45° cos 30° + cos 45° sin 30° = (1/√2)(√3/2) + (1/√2)(1/2) = (√3 + 1)/(2√2) → (√3 + 1)/(2√2). (ii) tan 15° = tan(45° − 30°) = (tan 45° − tan 30°)/(1 + tan 45° tan 30°) = (1 − 1/√3)/(1 + 1/√3) = (√3 − 1)/(√3 + 1) = 2 − √3 (after rationalising).

6. Prove that cos(π/4 − x) cos(π/4 − y) − sin(π/4 − x) sin(π/4 − y) = sin(x + y).

SOLUTION Using cos A cos B − sin A sin B = cos(A + B) with A = π/4 − x, B = π/4 − y: L.H.S. = cos[(π/4 − x) + (π/4 − y)] = cos(π/2 − (x + y)) = sin(x + y) = R.H.S. ✓

7. Prove that [tan(π/4 + x)] / [tan(π/4 − x)] = [(1 + tan x)/(1 − tan x)]2.

SOLUTION tan(π/4 + x) = (1 + tan x)/(1 − tan x) and tan(π/4 − x) = (1 − tan x)/(1 + tan x). Dividing: L.H.S. = [(1 + tan x)/(1 − tan x)] ÷ [(1 − tan x)/(1 + tan x)] = [(1 + tan x)/(1 − tan x)]2 = R.H.S. ✓

8. Prove that [cos(π + x) cos(−x)] / [sin(π − x) cos(π/2 + x)] = cot2x.

SOLUTION cos(π + x) = −cos x; cos(−x) = cos x; sin(π − x) = sin x; cos(π/2 + x) = −sin x. L.H.S. = (−cos x)(cos x) / [(sin x)(−sin x)] = (−cos2x)/(−sin2x) = cos2x/sin2x = cot2x = R.H.S. ✓

9. Prove that cos(3π/2 + x) cos(2π + x) [cot(3π/2 − x) + cot(2π + x)] = 1.

SOLUTION cos(3π/2 + x) = sin x; cos(2π + x) = cos x; cot(3π/2 − x) = tan x; cot(2π + x) = cot x. L.H.S. = sin x · cos x · (tan x + cot x) = sin x cos x · (sin x/cos x + cos x/sin x) = sin x cos x · (sin2x + cos2x)/(sin x cos x) = sin2x + cos2x = 1 = R.H.S. ✓

10. Prove that sin(n + 1)x sin(n + 2)x + cos(n + 1)x cos(n + 2)x = cos x.

SOLUTION Using cos A cos B + sin A sin B = cos(A − B) with A = (n + 1)x, B = (n + 2)x: L.H.S. = cos[(n + 1)x − (n + 2)x] = cos(−x) = cos x = R.H.S. ✓

11. Prove that cos(3π/4 + x) − cos(3π/4 − x) = −√2 sin x.

SOLUTION cos C − cos D = −2 sin½(C + D) sin½(C − D), with C = 3π/4 + x, D = 3π/4 − x. ½(C + D) = 3π/4, ½(C − D) = x. L.H.S. = −2 sin(3π/4) sin x = −2 × (1/√2) × sin x = −√2 sin x = R.H.S. ✓

12. Prove that sin26x − sin24x = sin 2x sin 10x.

SOLUTION Use sin2A − sin2B = sin(A + B) sin(A − B), with A = 6x, B = 4x. L.H.S. = sin(10x) sin(2x) = sin 2x sin 10x = R.H.S. ✓

13. Prove that cos22x − cos26x = sin 4x sin 8x.

SOLUTION Use cos2A − cos2B = −sin(A + B) sin(A − B) = sin(A + B) sin(B − A), with A = 2x, B = 6x. L.H.S. = sin(2x + 6x) sin(6x − 2x) = sin 8x sin 4x = sin 4x sin 8x = R.H.S. ✓

14. Prove that sin 2x + 2 sin 4x + sin 6x = 4 cos2x sin 4x.

SOLUTION Group sin 2x + sin 6x = 2 sin 4x cos 2x. L.H.S. = 2 sin 4x cos 2x + 2 sin 4x = 2 sin 4x(cos 2x + 1) = 2 sin 4x(2 cos2x) = 4 cos2x sin 4x = R.H.S. ✓ (using 1 + cos 2x = 2cos2x).

15. Prove that cot 4x (sin 5x + sin 3x) = cot x (sin 5x − sin 3x).

SOLUTION sin 5x + sin 3x = 2 sin 4x cos x and sin 5x − sin 3x = 2 cos 4x sin x. L.H.S. = (cos 4x/sin 4x)(2 sin 4x cos x) = 2 cos 4x cos x. R.H.S. = (cos x/sin x)(2 cos 4x sin x) = 2 cos 4x cos x. L.H.S. = R.H.S. ✓

16. Prove that (cos 9x − cos 5x)/(sin 17x − sin 3x) = −(sin 2x)/(cos 10x).

SOLUTION Numerator: cos 9x − cos 5x = −2 sin 7x sin 2x. Denominator: sin 17x − sin 3x = 2 cos 10x sin 7x. L.H.S. = (−2 sin 7x sin 2x)/(2 cos 10x sin 7x) = −sin 2x/cos 10x = R.H.S. ✓

17. Prove that (sin 5x + sin 3x)/(cos 5x + cos 3x) = tan 4x.

SOLUTION sin 5x + sin 3x = 2 sin 4x cos x; cos 5x + cos 3x = 2 cos 4x cos x. L.H.S. = (2 sin 4x cos x)/(2 cos 4x cos x) = sin 4x/cos 4x = tan 4x = R.H.S. ✓

18. Prove that (sin x − sin y)/(cos x + cos y) = tan[(x − y)/2].

SOLUTION sin x − sin y = 2 cos½(x + y) sin½(x − y); cos x + cos y = 2 cos½(x + y) cos½(x − y). L.H.S. = [2 cos½(x+y) sin½(x−y)]/[2 cos½(x+y) cos½(x−y)] = tan[(x − y)/2] = R.H.S. ✓

19. Prove that (sin x + sin 3x)/(cos x + cos 3x) = tan 2x.

SOLUTION sin x + sin 3x = 2 sin 2x cos x; cos x + cos 3x = 2 cos 2x cos x. L.H.S. = (2 sin 2x cos x)/(2 cos 2x cos x) = tan 2x = R.H.S. ✓

20. Prove that (sin x − sin 3x)/(sin2x − cos2x) = 2 sin x.

SOLUTION sin x − sin 3x = 2 cos 2x sin(−x) = −2 cos 2x sin x. Also sin2x − cos2x = −(cos2x − sin2x) = −cos 2x. L.H.S. = (−2 cos 2x sin x)/(−cos 2x) = 2 sin x = R.H.S. ✓

21. Prove that (cos 4x + cos 3x + cos 2x)/(sin 4x + sin 3x + sin 2x) = cot 3x.

SOLUTION Numerator: (cos 4x + cos 2x) + cos 3x = 2 cos 3x cos x + cos 3x = cos 3x(2 cos x + 1). Denominator: (sin 4x + sin 2x) + sin 3x = 2 sin 3x cos x + sin 3x = sin 3x(2 cos x + 1). L.H.S. = [cos 3x(2 cos x + 1)]/[sin 3x(2 cos x + 1)] = cos 3x/sin 3x = cot 3x = R.H.S. ✓

22. Prove that cot x cot 2x − cot 2x cot 3x − cot 3x cot x = 1.

SOLUTION Since 3x = 2x + x, cot 3x = cot(2x + x) = (cot 2x cot x − 1)/(cot x + cot 2x). So cot 3x(cot x + cot 2x) = cot x cot 2x − 1, i.e. cot 3x cot x + cot 3x cot 2x = cot x cot 2x − 1. Rearranging: cot x cot 2x − cot 2x cot 3x − cot 3x cot x = 1 = R.H.S. ✓

23. Prove that tan 4x = [4 tan x (1 − tan2x)] / (1 − 6 tan2x + tan4x).

SOLUTION Let t = tan x. tan 2x = 2t/(1 − t2). Then tan 4x = tan(2·2x) = 2 tan 2x/(1 − tan22x). Numerator = 2 × 2t/(1 − t2) = 4t/(1 − t2). Denominator = 1 − [2t/(1 − t2)]2 = [(1 − t2)2 − 4t2]/(1 − t2)2 = (1 − 6t2 + t4)/(1 − t2)2. tan 4x = [4t/(1 − t2)] ÷ [(1 − 6t2 + t4)/(1 − t2)2] = 4t(1 − t2)/(1 − 6t2 + t4) = R.H.S. ✓

24. Prove that cos 4x = 1 − 8 sin2x cos2x.

SOLUTION cos 4x = 1 − 2 sin22x = 1 − 2(2 sin x cos x)2. = 1 − 2 × 4 sin2x cos2x = 1 − 8 sin2x cos2x = R.H.S. ✓

25. Prove that cos 6x = 32 cos6x − 48 cos4x + 18 cos2x − 1.

SOLUTION cos 6x = cos 3(2x) = 4 cos3(2x) − 3 cos 2x, with cos 2x = 2cos2x − 1. Write c = cos x and y = 2c2 − 1. 4y3 = 4(2c2 − 1)3 = 4(8c6 − 12c4 + 6c2 − 1) = 32c6 − 48c4 + 24c2 − 4. 3y = 3(2c2 − 1) = 6c2 − 3. cos 6x = 4y3 − 3y = (32c6 − 48c4 + 24c2 − 4) − (6c2 − 3) = 32c6 − 48c4 + 18c2 − 1 = R.H.S. ✓

Miscellaneous Exercise on Chapter 3 Solutions

1. Prove that 2 cos(π/13) cos(9π/13) + cos(3π/13) + cos(5π/13) = 0.

SOLUTION 2 cos A cos B = cos(A + B) + cos(A − B) with A = π/13, B = 9π/13: 2 cos(π/13) cos(9π/13) = cos(10π/13) + cos(8π/13). Now cos(10π/13) = cos(π − 3π/13) = −cos(3π/13) and cos(8π/13) = cos(π − 5π/13) = −cos(5π/13). L.H.S. = −cos(3π/13) − cos(5π/13) + cos(3π/13) + cos(5π/13) = 0 = R.H.S. ✓

2. Prove that (sin 3x + sin x) sin x + (cos 3x − cos x) cos x = 0.

SOLUTION Expand: sin 3x sin x + sin2x + cos 3x cos x − cos2x. Group: (cos 3x cos x + sin 3x sin x) − (cos2x − sin2x) = cos(3x − x) − cos 2x = cos 2x − cos 2x = 0 = R.H.S. ✓

3. Prove that (cos x + cos y)2 + (sin x − sin y)2 = 4 cos2[(x + y)/2].

SOLUTION cos x + cos y = 2 cos½(x+y) cos½(x−y); sin x − sin y = 2 cos½(x+y) sin½(x−y). L.H.S. = 4 cos2½(x+y) cos2½(x−y) + 4 cos2½(x+y) sin2½(x−y). = 4 cos2½(x+y)[cos2½(x−y) + sin2½(x−y)] = 4 cos2[(x + y)/2] = R.H.S. ✓

4. Prove that (cos x − cos y)2 + (sin x − sin y)2 = 4 sin2[(x − y)/2].

SOLUTION cos x − cos y = −2 sin½(x+y) sin½(x−y); sin x − sin y = 2 cos½(x+y) sin½(x−y). L.H.S. = 4 sin2½(x+y) sin2½(x−y) + 4 cos2½(x+y) sin2½(x−y). = 4 sin2½(x−y)[sin2½(x+y) + cos2½(x+y)] = 4 sin2[(x − y)/2] = R.H.S. ✓

5. Prove that sin x + sin 3x + sin 5x + sin 7x = 4 cos x cos 2x sin 4x.

SOLUTION Pair: (sin x + sin 7x) + (sin 3x + sin 5x) = 2 sin 4x cos 3x + 2 sin 4x cos x. = 2 sin 4x(cos 3x + cos x) = 2 sin 4x(2 cos 2x cos x) = 4 cos x cos 2x sin 4x = R.H.S. ✓

6. Prove that [(sin 7x + sin 5x) + (sin 9x + sin 3x)] / [(cos 7x + cos 5x) + (cos 9x + cos 3x)] = tan 6x.

SOLUTION sin 7x + sin 5x = 2 sin 6x cos x; sin 9x + sin 3x = 2 sin 6x cos 3x. Numerator = 2 sin 6x(cos x + cos 3x). cos 7x + cos 5x = 2 cos 6x cos x; cos 9x + cos 3x = 2 cos 6x cos 3x. Denominator = 2 cos 6x(cos x + cos 3x). L.H.S. = [2 sin 6x(cos x + cos 3x)]/[2 cos 6x(cos x + cos 3x)] = sin 6x/cos 6x = tan 6x = R.H.S. ✓

7. Prove that sin 3x + sin 2x − sin x = 4 sin x cos(x/2) cos(3x/2).

SOLUTION sin 3x − sin x = 2 cos 2x sin x. So L.H.S. = 2 cos 2x sin x + sin 2x = 2 cos 2x sin x + 2 sin x cos x = 2 sin x(cos 2x + cos x). cos 2x + cos x = 2 cos(3x/2) cos(x/2). So L.H.S. = 2 sin x × 2 cos(3x/2) cos(x/2) = 4 sin x cos(x/2) cos(3x/2) = R.H.S. ✓

In Questions 8–10, find sin(x/2), cos(x/2) and tan(x/2).

8. tan x = −4/3, x in quadrant II.

SOLUTION In Q2 (π/2 < x < π) cos x is negative. sec2x = 1 + 16/9 = 25/9, so cos x = −3/5 and sin x = tan x · cos x = (−4/3)(−3/5) = 4/5. Quadrant of x/2: π/4 < x/2 < π/2, so x/2 is in Q1 → sin(x/2), cos(x/2), tan(x/2) all positive. 2 sin2(x/2) = 1 − cos x = 1 + 3/5 = 8/5, so sin2(x/2) = 4/5 → sin(x/2) = 2/√5. 2 cos2(x/2) = 1 + cos x = 1 − 3/5 = 2/5, so cos2(x/2) = 1/5 → cos(x/2) = 1/√5. ∴ sin(x/2) = 2/√5, cos(x/2) = 1/√5, tan(x/2) = 2.

9. cos x = −1/3, x in quadrant III.

SOLUTION In Q3 (π < x < 3π/2), x/2 lies in π/2 < x/2 < 3π/4 (Q2): sin(x/2) positive, cos(x/2) and tan(x/2) negative. 2 sin2(x/2) = 1 − cos x = 1 + 1/3 = 4/3, so sin2(x/2) = 2/3 → sin(x/2) = √(2/3) = √6/3. 2 cos2(x/2) = 1 + cos x = 1 − 1/3 = 2/3, so cos2(x/2) = 1/3 → cos(x/2) = −1/√3 = −√3/3. tan(x/2) = sin(x/2)/cos(x/2) = (√6/3)/(−√3/3) = −√2. ∴ sin(x/2) = √6/3, cos(x/2) = −√3/3, tan(x/2) = −√2.

10. sin x = 1/4, x in quadrant II.

SOLUTION In Q2 cos x is negative: cos2x = 1 − 1/16 = 15/16, so cos x = −√15/4. For π/2 < x < π, x/2 lies in Q1 → all of sin(x/2), cos(x/2), tan(x/2) positive. 2 sin2(x/2) = 1 − cos x = 1 + √15/4 = (4 + √15)/4, so sin2(x/2) = (4 + √15)/8 = (8 + 2√15)/16 → sin(x/2) = √(8 + 2√15)/4. 2 cos2(x/2) = 1 + cos x = (4 − √15)/4, so cos2(x/2) = (8 − 2√15)/16 → cos(x/2) = √(8 − 2√15)/4. tan(x/2) = sin(x/2)/cos(x/2) = √[(8 + 2√15)/(8 − 2√15)]. Rationalising the denominator gives tan(x/2) = 4 + √15. ∴ sin(x/2) = √(8 + 2√15)/4, cos(x/2) = √(8 − 2√15)/4, tan(x/2) = 4 + √15.

Common Mistakes to Avoid

Watch out for these

  • Forgetting to convert degrees to radians (or vice-versa) before using l = rθ — the arc-length formula needs θ in radians.
  • Taking the wrong sign for a function: always fix the sign from the quadrant (rule “All–Silver–Tea–Cups”) after taking the square root.
  • Mixing up the period: sin/cos repeat after 2π but tan/cot repeat after π, so reduce large angles correctly.
  • Errors in cos(x + y) — the sign rule is cos cos − sin sin for sum and + for difference (the opposite of sine).
  • For half-angle problems, not checking the quadrant of x/2; the sign of sin(x/2)/cos(x/2) depends on where x/2 lies, not where x lies.
  • Confusing the conversion formulas; remember Radian = (π/180) × Degree and Degree = (180/π) × Radian.

Practice MCQs & Assertion–Reason

1. The radian measure of 240° is:

(a) 3π/4    (b) 4π/3    (c) 5π/3    (d) 2π/3

2. The degree measure of 5π/3 radian is:

(a) 270°    (b) 300°    (c) 330°    (d) 360°

3. A wheel makes 360 revolutions per minute; the angle (in radian) turned in one second is:

(a) 6π    (b) 12π    (c) 24π    (d) 360π

4. If cos x = −1/2 and x lies in the third quadrant, then tan x equals:

(a) −√3    (b) 1/√3    (c) √3    (d) −1/√3

5. The value of sin 765° is:

(a) 1    (b) 1/2    (c) 1/√2    (d) √3/2

6. The value of tan 15° is:

(a) 2 + √3    (b) 2 − √3    (c) √3 − 1    (d) √3 + 1

7. cos 2x in terms of tan x is:

(a) 2tan x/(1 + tan2x)    (b) (1 − tan2x)/(1 + tan2x)    (c) 2tan x/(1 − tan2x)    (d) (1 + tan2x)/(1 − tan2x)

8. The range of the function y = sin x is:

(a) all real numbers    (b) [0, 1]    (c) [−1, 1]    (d) (−∞, −1] ∪ [1, ∞)

9. sin 3x equals:

(a) 3 sin x − 4 sin3x    (b) 4 sin3x − 3 sin x    (c) 4 cos3x − 3 cos x    (d) 3 cos x − 4 cos3x

10. The length of the minor arc of a chord of length 20 cm in a circle of diameter 40 cm is:

(a) 10π/3 cm    (b) 20π/3 cm    (c) 40π/3 cm    (d) 20π cm

Answer key: 1-(b), 2-(b), 3-(b), 4-(c), 5-(c), 6-(b), 7-(b), 8-(c), 9-(a), 10-(b).

For each Assertion–Reason question, choose: (A) Both Assertion and Reason are true and the Reason is the correct explanation of the Assertion; (B) Both are true but the Reason is not the correct explanation; (C) Assertion is true but Reason is false; (D) Assertion is false but Reason is true.

A-R 1. Assertion: 25° equals 5π/36 radian.

Reason: A degree measure is converted to radian by multiplying by π/180.

A-R 2. Assertion: If x lies in the second quadrant, then sin x > 0 while cos x < 0.

Reason: In the second quadrant the x-coordinate of a unit-circle point is negative and the y-coordinate is positive.

A-R 3. Assertion: cos(x + y) = cos x cos y + sin x sin y.

Reason: cos is an even function, so cos(−x) = cos x.

A-R 4. Assertion: tan x has period π, not 2π.

Reason: tan(π + x) = tan x for every x in the domain of tan.

A-R 5. Assertion: For an arc of length l subtending angle θ at the centre of a circle of radius r, l = rθ.

Reason: The angle θ here must be measured in degrees.

Answer key: 1-(A), 2-(A), 3-(C), 4-(B), 5-(C).

Quick Revision Summary

  • π radian = 180°; Radian = (π/180) × Degree, Degree = (180/π) × Radian; arc length l = rθ (θ in radians).
  • On the unit circle cos x = a, sin x = b; sin2x + cos2x = 1, 1 + tan2x = sec2x, 1 + cot2x = cosec2x.
  • Signs by quadrant: I all +, II only sin/cosec +, III only tan/cot +, IV only cos/sec +.
  • sin & cos have range [−1, 1] and period 2π; tan & cot have period π.
  • Sum/difference and double/triple-angle identities convert between products and sums — the backbone of every proof in this chapter.
  • For half-angle values use 2sin2(x/2) = 1 − cos x and 2cos2(x/2) = 1 + cos x, then fix signs from the quadrant of x/2.

How to score full marks in this chapter

Memorise the sum–to–product and product–to–sum formulas cold — almost every “prove that” question is one substitution away from the answer. Always state the identity you are applying before the line that uses it, keep θ in radians for arc-length problems, and decide every sign from the correct quadrant. For half-angle problems, write the quadrant of x/2 first so your final signs are never guesswork.

Frequently Asked Questions

What is Class 11 Maths Chapter 3 about?

Chapter 3, Trigonometric Functions, covers degree and radian measure, the arc-length formula l = rθ, the definition of the six trigonometric functions on the unit circle, their signs, domains, ranges and periods, and the sum, difference, double-, triple-angle and product–to–sum identities.

How many exercises are there in Class 11 Maths Chapter 3?

There are three exercises — Exercise 3.1 (radian/degree and arc length), Exercise 3.2 (values and signs of functions) and Exercise 3.3 (identities) — plus a Miscellaneous Exercise. All questions are solved step by step on this page.

How do you convert degrees to radians?

Multiply the degree measure by π/180. For example, 25° = 25 × π/180 = 5π/36 radian. To go the other way, multiply the radian measure by 180/π.

Are these Class 11 Maths Chapter 3 solutions free?

Yes. All ClearStudy NCERT Solutions for Class 11 Maths Chapter 3 are free and follow the official NCERT textbook for the 2026–27 session, with every answer verified against the book’s answer key.

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