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NCERT Solutions · Class 11 Maths · Chapter 3

Chapter 3: Trigonometric Functions (Sets and Functions)

Q: cot x = 3/4, x in quadrant III.

4/3; −5/4; −4/5; −3/5; −5/3

Step-by-step solutions to all exercises of Chapter 3, Trigonometric Functions (NCERT Class 11 Maths, 2026-27): radian measure, signs and values of trigonometric functions, compound and multiple angle formulas, and the Miscellaneous Exercise. All 52 questions are answered, with the key answer highlighted.

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$π$ rad $= 18 0^{\circ}$ , arc $l = r θ$ . Signs: All, Sin, Tan, Cos in quadrants I–IV. $sin (x \pm y)$ , $cos (x \pm y)$ , $tan (x \pm y)$ ; $sin 2 x = 2 sin x cos x$ , $cos 2 x = cos^{2} x - sin^{2} x$ ; sum-to-product: $sin A + sin B = 2 sin \frac{A + B}{2} cos \frac{A - B}{2}$ .

Exercise 3.1

Find the radian measures of (i) 25° (ii) −47°30′ (iii) 240° (iv) 520°.

Solution

Multiply by π/180. (i) 25π/180 = 5π/36. (ii) −47½° = −(95/2)(π/180) = −19π/72. (iii) 240π/180 = 4π/3. (iv) 520π/180 = 26π/9.

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

Find the degree measures (π = 22/7) of (i) 11/16 (ii) −4 (iii) 5π/3 (iv) 7π/6.

Solution

Multiply by 180/π = 180 × 7/22. (i) (11/16)(180 × 7/22) = 315/8 = 39.375° = 39°22′30″. (ii) −4 × 1260/22 = −229.0909° = −229°5′27″ (0.0909° = 5.45′ and 0.45′ = 27″). (iii) (5/3) × 180 = 300°. (iv) (7/6) × 180 = 210°.

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

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

Solution

360 rev/min = 6 rev/s; each revolution is 2π rad: 6 × 2π = 12π radians.

12π radians

Find the degree measure of the angle subtended at the centre of a circle of radius 100 cm by an arc of 22 cm (π = 22/7).

Solution

θ = l/r = 22/100 rad = (22/100)(180 × 7/22) = 12.6° = 12°36′.

12°36′

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.

Solution

r = 20 cm = chord, so the chord and two radii form an equilateral triangle: θ = 60° = π/3. Arc = rθ = 20 × π/3 = 20π/3 cm.

20π/3 cm

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

Solution

r₁θ₁ = r₂θ₂ ⇒ r₁/r₂ = θ₂/θ₁ = 75/60 = 5 : 4.

5 : 4

Find the angle in radians through which a pendulum swings if its length is 75 cm and the tip describes an arc of (i) 10 cm (ii) 15 cm (iii) 21 cm.

Solution

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

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

Exercise 3.2

cos x = −½, x in quadrant III. Find the other five functions.

Solution

sin x = −√(1 − ¼) = −√3/2 (negative in III); cosec x = −2/√3; sec x = −2; tan x = sin/cos = √3; cot x = 1/√3.

−√3/2; −2/√3; −2; √3; 1/√3

sin x = 3/5, x in quadrant II.

Solution

cos x = −√(1 − 9/25) = −4/5; cosec x = 5/3; sec x = −5/4; tan x = −3/4; cot x = −4/3.

−4/5; 5/3; −5/4; −3/4; −4/3

cot x = 3/4, x in quadrant III.

Solution

tan x = 4/3; cosec²x = 1 + 9/16 = 25/16, cosec x = −5/4 (sin < 0 in III); sin x = −4/5; cos x = sin x cot x = −3/5; sec x = −5/3.

4/3; −5/4; −4/5; −3/5; −5/3

sec x = 13/5, x in quadrant IV.

Solution

cos x = 5/13; sin x = −√(1 − 25/169) = −12/13; cosec x = −13/12; tan x = −12/5; cot x = −5/12.

5/13; −12/13; −13/12; −12/5; −5/12

tan x = −5/12, x in quadrant II.

Solution

sec²x = 1 + 25/144 = 169/144, sec x = −13/12 (cos < 0 in II); cos x = −12/13; sin x = tan x cos x = 5/13; cosec x = 13/5; cot x = −12/5.

−13/12; −12/13; 5/13; 13/5; −12/5

Find: sin 765°

Solution

sin(2 × 360° + 45°) = sin 45° = 1/√2.

1/√2

Find: cosec(−1410°)

Solution

cosec(−1410° + 4 × 360°) = cosec 30° = 2.

Find: tan(19π/3)

Solution

tan(6π + π/3) = tan(π/3) = √3.

√3

Find: sin(−11π/3)

Solution

sin(−11π/3 + 4π) = sin(π/3) = √3/2.

√3/2

Find: cot(−15π/4).

Solution

cot(−15π/4 + 4π) = cot(π/4) = 1.

Exercise 3.3

Prove that: sin²(π/6) + cos²(π/3) − tan²(π/4) = −½.

Solution

¼ + ¼ − 1 = −½.

−½

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

Solution

cosec(7π/6) = −cosec(π/6) = −2: 2(¼) + 4(¼) = ½ + 1 = 3/2.

3/2

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

Solution

cosec(5π/6) = cosec(π/6) = 2: 3 + 2 + 3(⅓) = 6.

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

Solution

sin(3π/4) = 1/√2: 2(½) + 2(½) + 2(4) = 1 + 1 + 8 = 10.

Find (i) sin 75° (ii) tan 15°.

Solution

(i) sin(45° + 30°) = (1/√2)(√3/2) + (1/√2)(½) = (√3 + 1)/(2√2) = (√6 + √2)/4. (ii) tan(45° − 30°) = (1 − 1/√3)/(1 + 1/√3) = (√3 − 1)/(√3 + 1) = (√3 − 1)²/2 = 2 − √3.

(√3 + 1)/(2√2); 2 − √3

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

Solution

LHS = cos[(π/4 − x) + (π/4 − y)] = cos[π/2 − (x + y)] = sin(x + y).

sin(x + y)

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); tan(π/4 − x) = (1 − tan x)/(1 + tan x). The quotient is $(\frac{1 + tan x}{1 - tan x})^{2}$ = RHS.

See the working above.

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

Solution

= (−cos x)(cos x)/[(sin x)(−sin x)] = cos²x/sin²x = cot²x.

cot²x

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. LHS = sin x cos x (tan x + cot x) = sin x cos x × (sin²x + cos²x)/(sin x cos x) = 1.

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

Solution

This is cos A cos B + sin A sin B = cos(A − B) with A = (n + 2)x, B = (n + 1)x: = cos[(n + 2)x − (n + 1)x] = cos x.

cos x

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

Solution

cos A − cos B = −2 sin((A + B)/2) sin((A − B)/2) = −2 sin(3π/4) sin x = −2(1/√2) sin x = −√2 sin x.

−√2 sin x

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

Solution

= (sin 6x + sin 4x)(sin 6x − sin 4x) = (2 sin 5x cos x)(2 cos 5x sin x) = (2 sin 5x cos 5x)(2 sin x cos x) = sin 10x sin 2x.

sin 10x sin 2x

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

Solution

= (cos 2x + cos 6x)(cos 2x − cos 6x) = (2 cos 4x cos 2x)(2 sin 4x sin 2x) = (2 sin 4x cos 4x)(2 sin 2x cos 2x) = sin 8x sin 4x.

sin 8x sin 4x

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

Solution

= (sin 6x + sin 2x) + 2 sin 4x = 2 sin 4x cos 2x + 2 sin 4x = 2 sin 4x (1 + cos 2x) = 2 sin 4x (2 cos²x) = 4 cos²x sin 4x.

4 cos²x sin 4x

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

Solution

LHS = (cos 4x/sin 4x)(2 sin 4x cos x) = 2 cos 4x cos x. RHS = (cos x/sin x)(2 cos 4x sin x) = 2 cos 4x cos x. Equal.

Equal.

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

Solution

= (−2 sin 7x sin 2x)/(2 cos 10x sin 7x) = −sin 2x/cos 10x.

−sin 2x/cos 10x

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

Solution

= (2 sin 4x cos x)/(2 cos 4x cos x) = tan 4x.

tan 4x

Prove that: (sin x − sin y)/(cos x + cos y) = tan((x − y)/2).

Solution

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

tan((x − y)/2)

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

Solution

= (2 sin 2x cos x)/(2 cos 2x cos x) = tan 2x.

tan 2x

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

Solution

Numerator = 2 cos 2x sin(−x) = −2 cos 2x sin x; denominator = −cos 2x. Quotient = 2 sin x.

2 sin x

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

Solution

Numerator = 2 cos 3x cos x + cos 3x = cos 3x (2 cos x + 1); denominator = 2 sin 3x cos x + sin 3x = sin 3x (2 cos x + 1). Quotient = cot 3x.

cot 3x

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

Solution

cot 3x = cot(2x + x) = (cot 2x cot x − 1)/(cot x + cot 2x). Cross-multiplying: cot 3x cot x + cot 3x cot 2x = cot 2x cot x − 1, i.e. cot x cot 2x − cot 2x cot 3x − cot 3x cot x = 1.

cot x cot 2x − cot 2x cot 3x − cot 3x cot x = 1

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

Solution

Let t = tan x: tan 2x = 2t/(1 − t²). tan 4x = 2 tan 2x/(1 − tan²2x) = [4t/(1 − t²)]/[1 − 4t²/(1 − t²)²] = 4t(1 − t²)/[(1 − t²)² − 4t²] = 4t(1 − t²)/(1 − 6t² + t⁴).

4t(1 − t²)/(1 − 6t² + t⁴)

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

Solution

cos 4x = 1 − 2 sin²2x = 1 − 2(2 sin x cos x)² = 1 − 8 sin²x cos²x.

1 − 8 sin²x cos²x

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

Solution

cos 6x = cos 3(2x) = 4 cos³2x − 3 cos 2x. With c = cos x, cos 2x = 2c² − 1: 4(2c² − 1)³ − 3(2c² − 1) = 4(8c⁶ − 12c⁴ + 6c² − 1) − 6c² + 3 = 32c⁶ − 48c⁴ + 18c² − 1.

32c⁶ − 48c⁴ + 18c² − 1

Miscellaneous Exercise on Chapter 3

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

Solution

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(5π/13). Sum = −cos(3π/13) − cos(5π/13) + cos(3π/13) + cos(5π/13) = 0.

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

Solution

= sin 3x sin x + cos 3x cos x + sin²x − cos²x = cos(3x − x) − cos 2x = cos 2x − cos 2x = 0.

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

Solution

= cos²x + cos²y + 2 cos x cos y + sin²x + sin²y − 2 sin x sin y = 2 + 2 cos(x + y) = 2[1 + cos(x + y)] = 4 cos²((x + y)/2).

4 cos²((x + y)/2)

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

Solution

= 2 − 2(cos x cos y + sin x sin y) = 2 − 2 cos(x − y) = 4 sin²((x − y)/2).

4 sin²((x − y)/2)

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

Solution

(sin 7x + sin x) + (sin 5x + sin 3x) = 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.

4 cos x cos 2x sin 4x

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

Solution

Numerator = 2 sin 6x cos x + 2 sin 6x cos 3x = 2 sin 6x (cos x + cos 3x); denominator = 2 cos 6x cos x + 2 cos 6x cos 3x = 2 cos 6x (cos x + cos 3x). Quotient = tan 6x.

tan 6x

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

Solution

(sin 3x − sin x) + sin 2x = 2 cos 2x sin x + 2 sin x cos x = 2 sin x (cos 2x + cos x) = 2 sin x × 2 cos(3x/2) cos(x/2) = 4 sin x cos(x/2) cos(3x/2).

4 sin x cos(x/2) cos(3x/2)

Find sin(x/2), cos(x/2), tan(x/2) for: tan x = −4/3, x in II

Solution

π/2 < x < π ⇒ π/4 < x/2 < π/2 (all positive). cos x = −3/5. sin(x/2) = √((1 + 3/5)/2) = 2/√5; cos(x/2) = √((1 − 3/5)/2) = 1/√5; tan(x/2) = 2.

2/√5; 1/√5; 2

Find sin(x/2), cos(x/2), tan(x/2) for: cos x = −⅓, x in III

Solution

π < x < 3π/2 ⇒ π/2 < x/2 < 3π/4 (sin +, cos −). sin(x/2) = √((1 + ⅓)/2) = √(2/3) = √6/3; cos(x/2) = −√((1 − ⅓)/2) = −1/√3 = −√3/3; tan(x/2) = −√2.

√6/3; −1/√3; −√2

Find sin(x/2), cos(x/2), tan(x/2) for: sin x = ¼, x in II.

Solution

cos x = −√15/4; π/4 < x/2 < π/2. sin(x/2) = √((1 + √15/4)/2) = √(8 + 2√15)/4; cos(x/2) = √((1 − √15/4)/2) = √(8 − 2√15)/4; tan(x/2) = √((8 + 2√15)/(8 − 2√15)) = (8 + 2√15)/√(64 − 60) = 4 + √15.

√(8 + 2√15)/4; √(8 − 2√15)/4; 4 + √15

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