Class 11 Maths Chapter 3: Trigonometric Functions worksheet with answer key

Answer: (a) sin 380°

380° = 360° + 20°, and sin (2nπ + x) = sin x, so sin 380° = sin 20°. sin (−20°) = −sin 20°, sin 200° = sin (180° + 20°) = −sin 20°, and cos 20° = sin 70°.

Answer: (d) π/3

For every real x, −1 ≤ cos x ≤ 1. π/3 ≈ 1.047 is greater than 1; the other three lie in [−1, 1] (cos π = −1).

Answer: (d) 45

sin 2x = 2 tan x/(1 + tan² x) = 1/(1 + 1/4) = 4/5. 4/3 is tan 2x and 3/5 is cos 2x.

Answer: (b) second quadrant

sin x is positive in quadrants I and II; tan x is negative in quadrants II and IV. Both hold only in the second quadrant.

Answer: (b) 3π/4

Radian measure = (π/180) × 135 = 135π/180 = 3π/4. 2π/3 is 120° and 5π/6 is 150°.

Take π ≈ 3.14. sin 2 = sin (π − 2) = sin 1.14 and sin 3 = sin (π − 3) = sin 0.14. The numbers 0.14, 1 and 1.14 all lie between 0 and π/2 ≈ 1.57, where sin x increases as x increases. As 0.14 < 1 < 1.14, sin 3 < sin 1 < sin 2.

• rewrites sin 2 and sin 3 as sin (π − 2) and sin (π − 3), angles in (0, π/2) (1)
• uses that sin x increases on (0, π/2) to give sin 3 < sin 1 < sin 2 (1)

tan x = sin x/cos x needs cos x ≠ 0: domain {x : x ∈ R, x ≠ (2n + 1)π/2, n ∈ Z}. cot x = cos x/sin x needs sin x ≠ 0: domain {x : x ∈ R, x ≠ nπ, n ∈ Z}. A number in neither domain must be both an odd multiple of π/2 and a multiple of π, which is impossible: an odd multiple of π/2 has cos x = 0 and so sin x = ±1 ≠ 0. So there is no such number; at every real x at least one of tan x and cot x is defined.

• writes both domains correctly (1)
• answers no, with the reason that cos x and sin x are never both 0 (1)

30° = π/6 radian, so r = l/θ = 11/(π/6) = 66/π = 66 × 7/22 = 21 cm. The second arc is l = rθ = 21 × 1.5 = 31.5 cm.

• converts 30° to π/6 and finds r = 21 cm (1)
• arc = 21 × 1.5 = 31.5 cm (1)

−1 ≤ sin x ≤ 1, so −2 ≤ −2 sin x ≤ 2 (multiplying by −2 reverses the inequalities) and 1 ≤ 3 − 2 sin x ≤ 5. Every value in [1, 5] is taken, as sin x takes every value in [−1, 1]; the range is [1, 5]. The greatest value 5 comes when sin x = −1, at x = 3π/2; the least value 1 when sin x = 1, at x = π/2.

• uses −1 ≤ sin x ≤ 1 and reverses the inequality for −2 (1)
• range [1, 5] (1)
• greatest at x = 3π/2, least at x = π/2 (1)

tan (x + y) = (tan x + tan y)/(1 − tan x tan y) = (1/2 + 1/3)/(1 − 1/6) = (5/6)/(5/6) = 1. As x and y are acute, 0 < x + y < π. In (0, π) the only angle with tangent 1 is π/4 (tan x is negative in (π/2, π) and undefined at π/2). So x + y = π/4.

• applies tan (x + y) correctly (1)
• gets tan (x + y) = 1 (1)
• uses 0 < x + y < π to conclude x + y = π/4 (1)

tan A + tan B = sin A/cos A + sin B/cos B = (sin A cos B + cos A sin B)/(cos A cos B) = sin (A + B)/(cos A cos B). With A = 15° and B = 75°: sin (A + B) = sin 90° = 1, and cos 75° = cos (90° − 15°) = sin 15°. So tan 15° + tan 75° = 1/(cos 15° sin 15°) = 2/(2 sin 15° cos 15°) = 2/sin 30° = 2/(1/2) = 4.

• writes tan A + tan B = sin (A + B)/(cos A cos B) (1)
• uses sin 90° = 1 and cos 75° = sin 15° (1)
• uses 2 sin 15° cos 15° = sin 30° = 1/2 to get 4 (1)