Class 11 Maths Chapter 3 Trigonometric Functions – Extra Questions with Answers

Extra practice questions for Class 11 Maths Chapter 3 (Trigonometric Functions), beyond the textbook. These Class 11 Maths Chapter 3 important questions are handy for last-minute exam practice.

Very Short Answer Questions (1 mark)

Q1. Convert 60° to radians.
Ans: 60 × π/180 = π/3.

Q2. What is the value of sin 90°?
Ans: 1.

Q3. In which quadrant is tanθ positive but sinθ negative?
Ans: Quadrant III.

Q4. State the identity relating sec²θ and tan²θ.
Ans: sec²θ = 1 + tan²θ.

Q5. What is the radian measure equivalent to 180°?
Ans: π.

Short Answer Questions (2–3 marks)

Q6. Prove that sin(90°+θ) = cosθ using the sum formula.
Ans: sin(90+θ) = sin90 cosθ + cos90 sinθ = (1)(cosθ) + (0)(sinθ) = cosθ.

Q7. Find the value of cos 75° using the sum formula, expressing 75° as 45°+30°.
Ans: cos(45+30) = cos45 cos30 − sin45 sin30 = (√2/2)(√3/2) − (√2/2)(1/2) = (√6−√2)/4.

Q8. A wheel of radius 21 cm rotates through an angle of 60°. Find the length of the arc traced (using arc length = radius × angle in radians).
Ans: 60° = π/3 radians. Arc length = 21 × π/3 = 7π ≈ 22 cm.

Higher-Order Thinking / Application Questions

Q9. Explain why radian measure, rather than degree measure, is preferred in advanced mathematics (like calculus), using the arc length formula as part of your reasoning.
Ans: In radian measure, the arc length formula simplifies to s = rθ directly, without any conversion constant, because radians are defined based on the ratio of arc length to radius. This natural relationship makes calculus formulas involving trigonometric functions (like derivatives of sin and cos) much simpler in radians than they would be in degrees, which require extra conversion factors.

Q10. Find the general solution of the equation 2sinθ = 1, and explain why trigonometric equations typically have infinitely many solutions.
Ans: sinθ = 1/2 gives θ = 30° = π/6 as a principal solution. The general solution is θ = nπ + (−1)ⁿ(π/6), for any integer n. Trigonometric equations have infinitely many solutions because trigonometric functions are periodic (sin repeats every 2π), so if one value of θ satisfies the equation, adding any multiple of the period (with the appropriate symmetry adjustments) gives another valid solution.

Written by Satish

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