Trigonometric Functions — Previous Year Questions (Class 11 Mathematics)
Trigonometric functions relate angles to ratios of sides in triangles. Master angle measures, identities, and function graphs.
TL;DR: Trigonometric functions relate angles to ratios of sides in triangles. Master angle measures, identities, and function graphs.
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Trigonometric functions relate angles to ratios of sides in triangles. Master angle measures, identities, and function graphs.
Trigonometric Functions — Previous Year Questions with Solutions
Q (2023, 2 marks): Convert 45° to radians and 3π/4 radians to degrees.
Answer: 45° to radians:
45° × (π / 180°) = 45π / 180 = π / 4 radians
3π/4 radians to degrees:
(3π/4) × (180° / π) = 3 × 180° / 4 = 540° / 4 = 135°
Q (2022, 3 marks): If sin θ = 3/5 and θ is in first quadrant, find cos θ, tan θ, and cot θ.
Answer: sin θ = 3/5, θ in first quadrant (all ratios positive)
Using sin²θ + cos²θ = 1:
(3/5)² + cos²θ = 1
9/25 + cos²θ = 1
cos²θ = 16/25
cos θ = 4/5 (positive in Q1)
tan θ = sin θ / cos θ = (3/5) / (4/5) = 3/4
cot θ = cos θ / sin θ = (4/5) / (3/5) = 4/3
Q (2023, 2 marks): Prove the identity: (1 - cos²θ) / sin²θ = tan²θ
Answer: Left side: (1 - cos²θ) / sin²θ
= sin²θ / sin²θ [using sin²θ + cos²θ = 1, so 1 - cos²θ = sin²θ]
= 1
Wait, let me recalculate.
Actually: (1 - cos²θ) / sin²θ = sin²θ / sin²θ = 1, not tan²θ
Let me prove the correct identity:
Prove: sin²θ / cos²θ = tan²θ
Left side: sin²θ / cos²θ = (sin θ / cos θ)² = tan²θ = Right side ✓
Or prove: 1 - sin²θ / cos²θ = 1, which gives (sin²θ + cos²θ) / cos²θ = sec²θ
Q (2021, 3 marks): Find the value of sin(A + B) if sin A = 1/2, cos B = √3/2, where A and B are acute angles.
Answer: sin A = 1/2, cos B = √3/2, A and B acute
From sin A = 1/2: A = 30°
From cos B = √3/2: B = 30°
Finding missing values:
cos A = √(1 - sin²A) = √(1 - 1/4) = √(3/4) = √3/2
sin B = √(1 - cos²B) = √(1 - 3/4) = √(1/4) = 1/2
Using sin(A + B) = sin A cos B + cos A sin B:
= (1/2)(√3/2) + (√3/2)(1/2)
= √3/4 + √3/4
= 2√3/4 = √3/2
Q (2022, 3 marks): Find the period of f(x) = tan(2x) and sketch one complete cycle.
Answer: Period of tan(x) = π
Period of tan(2x) = π / 2
One complete cycle occurs from x = -π/4 to x = π/4
Key points:
At x = -π/4: tan(-π/2) undefined (vertical asymptote)
At x = 0: tan(0) = 0
At x = π/4: tan(π/2) undefined (vertical asymptote)
The function:
- Has vertical asymptotes at x = ±π/4, ±3π/4, ...
- Passes through origin (0, 0)
- Increases from -∞ to +∞ in each period
- Period = π/2
Q (2023, 3 marks): Solve sin x = 1/2 for x ∈ [0, 2π]. Also give general solution.
Answer: sin x = 1/2
Basic angle: sin θ = 1/2 → θ = π/6 or 30°
For x ∈ [0, 2π]:
sin x = sin(π/6) gives:
x = π/6 (in Q1) or x = π - π/6 = 5π/6 (in Q2)
Solutions in [0, 2π]: x = π/6, 5π/6
General solution:
x = 2nπ + π/6 or x = 2nπ + 5π/6, where n ∈ Z
Or: x = nπ + (-1)^n(π/6), where n ∈ Z
Frequently Asked Questions
What is the relationship between degree and radian measure?
180° = π radians. Conversion: degrees × (π/180) = radians, radians × (180/π) = degrees. Radians are preferred in calculus because they make derivatives and integrals simpler (derivative of sin x is cos x only when x is in radians).
What are the ASTC rule and its application?
ASTC rule tells which trigonometric ratios are positive in each quadrant: All positive (Q1), Sin positive (Q2), Tan positive (Q3), Cos positive (Q4). Use this to determine signs when finding equivalent angles or solving trigonometric equations.
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