Trigonometric Functions — JEE Main Questions

33 JEE Main practice questions on Trigonometric Functions, part of Mathematics. Below are 12 of them in full, each with the answer and a written explanation.

Questions & explanations

1. If A + B + C = π, what is the value of cos 2A + cos 2B + cos 2C?

  1. 1 + 4 cos A cos B cos C
  2. −1 − 4 cos A cos B cos C
  3. −1 + 4 cos A cos B cos C
  4. 1 − 4 cos A cos B cos C

Answer: −1 − 4 cos A cos B cos C

Use sum-to-product: cos 2A + cos 2B = 2 cos(A+B) cos(A−B) = −2 cos C cos(A−B). Then cos 2C = 2 cos²C − 1. Adding gives −2 cos C cos(A−B) + 2 cos²C − 1 = −2 cos C[cos(A−B) − cos C] − 1. Since cos C = −cos(A+B), bracket becomes cos(A−B)+cos(A+B)=2 cos A cos B. So expression = −2 cos C × 2 cos A cos B − 1 = −1 − 4 cos A cos B cos C.

2. If A + B + C = π, what is the value of sin A + sin B + sin C?

  1. 2 cos(A/2) cos(B/2) cos(C/2)
  2. 4 sin(A/2) sin(B/2) sin(C/2)
  3. 4 cos(A/2) cos(B/2) cos(C/2)
  4. 4 cos(A/2) cos(B/2) sin(C/2)

Answer: 4 cos(A/2) cos(B/2) cos(C/2)

Use sum-to-product: sin A + sin B = 2 sin((A+B)/2) cos((A−B)/2) = 2 cos(C/2) cos((A−B)/2). Then sin C = 2 sin(C/2) cos(C/2). Adding gives 2 cos(C/2)[cos((A−B)/2) + sin(C/2)]. Since sin(C/2) = cos((A+B)/2), bracket becomes cos((A−B)/2)+cos((A+B)/2)=2 cos(A/2) cos(B/2). Thus sum = 4 cos(A/2) cos(B/2) cos(C/2).

3. For n = 5, evaluate sin(π/5) + sin(2π/5) + sin(3π/5) + sin(4π/5).

  1. cot(π/10)
  2. cot(π/5)
  3. tan(π/10)
  4. tan(π/5)

Answer: cot(π/10)

Using the sum of sines in AP formula: S = sin(nβ/2)/sin(β/2) · sin(α + (n-1)β/2). Here α = π/5, β = π/5, n = 4. Then S = sin(2π/5)/sin(π/10) · sin(π/5 + 3π/10) = sin(2π/5)/sin(π/10) · sin(π/2) = sin(2π/5)/sin(π/10). Since sin(2π/5) = sin(3π/5) = cos(π/10), we get S = cos(π/10)/sin(π/10) = cot(π/10).

4. If A + B + C = π, what is the value of (sin 2A + sin 2B + sin 2C) / (sin A + sin B + sin C)?

  1. 4 cos A cos B cos C
  2. 8 cos(A/2) cos(B/2) cos(C/2)
  3. 4 sin A sin B sin C
  4. 8 sin(A/2) sin(B/2) sin(C/2)

Answer: 8 sin(A/2) sin(B/2) sin(C/2)

Using identities: sin 2A+sin 2B+sin 2C = 4 sin A sin B sin C and sin A+sin B+sin C = 4 cos(A/2) cos(B/2) cos(C/2). Dividing gives (4 sin A sin B sin C)/(4 cos(A/2) cos(B/2) cos(C/2)). Then sin A = 2 sin(A/2) cos(A/2) etc., cancel cos(A/2) cos(B/2) cos(C/2) to get 8 sin(A/2) sin(B/2) sin(C/2).

5. If sin²θ = (1 - cos 2θ)/2, then for which θ is this identity valid?

  1. For all real θ
  2. Only for θ in [0, π]
  3. Only for θ in [0, π/2]
  4. Only for θ where sinθ ≥ 0

Answer: For all real θ

The power-reduction formula sin²θ = (1 - cos 2θ)/2 is derived from cos 2θ = 1 - 2 sin²θ, which holds for all real θ. The derivation involves only algebraic manipulation and the Pythagorean identity, both valid for all θ. Hence the formula is valid for all real θ.

6. If A + B + C = π, what is the value of sin 2A + sin 2B + sin 2C?

  1. 4 sin A sin B sin C
  2. 4 cos A cos B cos C
  3. 2 sin A sin B sin C
  4. 4 sin A sin B cos C

Answer: 4 sin A sin B sin C

Use sum-to-product: sin 2A + sin 2B = 2 sin(A+B) cos(A−B) = 2 sin(π−C) cos(A−B) = 2 sin C cos(A−B). Then sin 2C = 2 sin C cos C = 2 sin C cos(π−(A+B)) = −2 sin C cos(A+B). Adding gives 2 sin C [cos(A−B) − cos(A+B)] = 2 sin C × 2 sin A sin B = 4 sin A sin B sin C.

7. If A + B + C = π, which of the following is true?

  1. tan A + tan B + tan C = tan A + tan B + tan C
  2. tan A + tan B + tan C = 0
  3. tan A + tan B + tan C = 1
  4. tan A + tan B + tan C = tan A tan B tan C

Answer: tan A + tan B + tan C = tan A tan B tan C

Using A + B = π − C, take tan: tan(A+B) = tan(π−C) = −tan C. Expand tan(A+B) = (tan A+tan B)/(1−tan A tan B). Equate and cross-multiply to get tan A+tan B = −tan C + tan A tan B tan C. Rearranging gives tan A+tan B+tan C = tan A tan B tan C.

8. What is the value of cos 20° cos 40° cos 80°?

  1. 1/4
  2. 1/8
  3. 1/2
  4. 1/16

Answer: 1/8

Multiply and divide by 2 sin 20°: P = (2 sin 20° cos 20° cos 40° cos 80°)/(2 sin 20°) = (sin 40° cos 40° cos 80°)/(2 sin 20°). Repeat: = (sin 80° cos 80°)/(4 sin 20°) = (sin 160°)/(8 sin 20°). Since sin 160° = sin 20°, P = 1/8.

9. Which of the following statements is correct?

  1. cos(3π/2 + π/4) = -√2/2
  2. cos(3π/2 + π/4) = 1/2
  3. cos(3π/2 + π/4) = -1/2
  4. cos(3π/2 + π/4) = √2/2

Answer: cos(3π/2 + π/4) = √2/2

3π/2 + π/4 = 7π/4, which lies in the fourth quadrant where cosine is positive. Since 3π/2 is an odd multiple of π/2, cosine changes to sine: cos(3π/2 + θ) = sin θ. So cos(3π/2 + π/4) = sin(π/4) = √2/2.

10. If tan A = 1/2 and tan B = 1/3, find tan(A + B).

  1. 5/7
  2. 1
  3. 1/7
  4. 7/5

Answer: 1

Using tan(A+B) = (tan A + tan B)/(1 - tan A tan B). Substituting tan A = 1/2, tan B = 1/3 gives numerator = 1/2 + 1/3 = 5/6, denominator = 1 - (1/2)(1/3) = 1 - 1/6 = 5/6. So tan(A+B) = (5/6)/(5/6) = 1.

11. If cosθ = 1/3 and θ lies in the fourth quadrant, then sin(θ/2) equals

  1. 1/√3
  2. -1/√3
  3. √(2/3)
  4. -√(2/3)

Answer: 1/√3

Using half-angle formula sin²(θ/2) = (1−cosθ)/2 = (1−1/3)/2 = 1/3, so sin(θ/2) = ±1/√3. Since θ in QIV (270° to 360°), θ/2 in QII (135° to 180°), where sine is positive. Hence sin(θ/2) = 1/√3.

12. If A + B + C = π and A, B, C > 0, what is the maximum value of sin A sin B sin C?

  1. √3/4
  2. 1/2
  3. 3√3/8
  4. 1

Answer: 3√3/8

By AM-GM on sin²A, sin²B, sin²C or using Jensen's inequality on log sin x (concave on (0,π)), the product is maximized when A = B = C = π/3. Then sin(π/3) = √3/2, so product = (√3/2)³ = 3√3/8.

More Mathematics topics

This page shows 12 of 33 questions on this topic. The full set, with progress tracking and five agent perspectives per question, is in the JupiteX app — browse the exam catalogue or browse the Learn library.