68 free MCQs on Trigonometric Functions with worked answers and explanations. Ratios, identities, and applications in triangles
Take the timed Trigonometric Functions chapterwise test →Below are 68 practice questions on Trigonometric Functions, sorted Easy → Hard. Tap “Show answer & explanation” under any question to check yourself. Want the full theory first? Read the Trigonometric Functions notes.
Unit circle with standard angles 0, 30, 45, 60, 90 degrees; at 45 degrees the point is (cos45, sin45) = (1/sqrt2, 1/sqrt2).
Q1.
What is sin 30°?
Answer: A. 1/2
Why: sin 30° = 1/2. From a 30-60-90 triangle: opposite/hypotenuse = 1/2.
Q2.
What is cos 60°?
Answer: B. 1/2
Why: cos 60° = 1/2. Adjacent/hypotenuse in a 30-60-90 triangle.
Q3.
What is tan 45°?
Answer: C. 1
Why: tan 45° = 1. In an isosceles right triangle, opposite = adjacent, so ratio = 1.
Q4.
The fundamental trigonometric identity is:
Answer: B. sin²θ + cos²θ = 1
Why: sin²θ + cos²θ = 1 for all values of θ. This is the Pythagorean identity.
Q5.
tan θ = sin θ / ___
Answer: C. cos θ
Why: tan θ = sin θ / cos θ (definition of tangent in terms of sine and cosine).
Q6.
What is sin 90°?
Answer: D. 1
Why: sin 90° = 1. The angle subtends the full hypotenuse as opposite side.
Q7.
What is cos 0°?
Answer: D. 1
Why: cos 0° = 1. At 0 degrees, the adjacent side equals the hypotenuse.
Q8.
What is the reciprocal of sin θ?
Answer: C. cosec θ
Why: cosec θ = 1/sin θ (cosecant is the reciprocal of sine).
Q9.
In a right triangle with hypotenuse 5 and opposite 3, sin θ =
Answer: A. 3/5
Why: sin θ = opposite/hypotenuse = 3/5. This is a 3-4-5 Pythagorean triple.
Q10.
sin² 30° + cos² 30° =
Answer: B. 1
Why: By the identity, sin²θ + cos²θ = 1 for ALL angles, including 30°.
Q11.
What is tan 0°?
Answer: A. 0
Why: tan 0° = sin 0°/cos 0° = 0/1 = 0.
Q12.
What is cos 90°?
Answer: A. 0
Why: cos 90° = 0. At 90 degrees, the adjacent side is zero.
Q13.
If sin A = 5/13, find cos A (in a right triangle).
Answer: B. 12/13
Why: Using sin²A + cos²A = 1: cos²A = 1 - 25/169 = 144/169, cos A = 12/13.
Q14.
What is sin 60°?
Answer: C. √3/2
Why: sin 60° = sqrt(3)/2.
Q15.
1 + tan²θ =
Answer: A. sec²θ
Why: Pythagorean identity: 1 + tan²θ = sec²θ.
Q16.
sec θ = 1 / ___
Answer: B. cos θ
Why: sec θ = 1/cos θ (secant is the reciprocal of cosine).
Q17.
What is sin(90° - θ)?
Answer: B. cos θ
Why: sin(90° - θ) = cos θ. Complementary angle identity.
Q18.
What is tan 60°?
Answer: B. √3
Why: tan 60° = sqrt(3). From the ratio in an equilateral triangle.
Q19.
cos 45° =
Answer: B. 1/√2
Why: cos 45° = 1/sqrt(2) = sqrt(2)/2. Isosceles right triangle.
Q20.
1 + cot²θ =
Answer: C. cosec²θ
Why: Pythagorean identity: 1 + cot²θ = cosec²θ.
Q21.
If sin θ + cos θ = √2, find θ.
Answer: B. 45°
Why: sin 45° + cos 45° = 1/√2 + 1/√2 = 2/√2 = √2. So θ = 45°.
Q22.
Prove that (1+tanθ)² + (1+cotθ)² = (secθ + cosecθ)². LHS equals:
Answer: D. All are equal
Why: Expanding LHS and RHS both give sec²θ + cosec²θ + 2secθ cosecθ. They are equal.
Q23.
sin(A+B) when A = 30°, B = 60°:
Answer: A. sin 90° = 1
Why: sin(30°+60°) = sin 90° = 1. Also verified by formula: sin30°cos60° + cos30°sin60° = 1/4 + 3/4 = 1.
Q24.
Value of sin 15° =
Answer: A. (√6-√2)/4
Why: sin 15° = sin(45°-30°) = sin45°cos30° - cos45°sin30° = (√2/2)(√3/2) - (√2/2)(1/2) = (√6-√2)/4.
Q25.
tan 75° + cot 75° =
Answer: A. 4
Why: tan 75° = 2+√3. cot 75° = tan 15° = 2-√3. Sum = 4.
Q26.
Simplify: sin²A/(1+cosA)
Answer: A. 1-cosA
Why: sin²A = 1 - cos²A = (1-cosA)(1+cosA). So sin²A/(1+cosA) = (1-cosA)(1+cosA)/(1+cosA) = 1-cosA.
Q27.
cos(A-B) - cos(A+B) =
Answer: A. 2sinA sinB
Why: cos(A-B) = cosA cosB + sinA sinB. cos(A+B) = cosA cosB - sinA sinB. Difference = 2 sinA sinB.
Q28.
The value of sin 195° =
Answer: B. -sin 15°
Why: 195° = 180° + 15°. sin(180°+θ) = -sinθ. So sin 195° = -sin 15°.
Q29.
If tan A = 1/2 and tan B = 1/3, find tan(A+B).
Answer: B. 1
Why: tan(A+B) = (tanA + tanB)/(1 - tanA tanB) = (1/2+1/3)/(1-1/6) = (5/6)/(5/6) = 1. So A+B = 45°.
Q30.
Express sin 3A in terms of sin A.
Answer: A. 3sinA - 4sin³A
Why: sin 3A = 3 sinA - 4 sin³A. (Standard triple angle formula).
Q31.
In triangle ABC, if A = 60°, b = 3, c = 4, find a (cosine rule).
Answer: B. √13
Why: a² = b² + c² - 2bc cosA = 9 + 16 - 2×3×4×(1/2) = 25 - 12 = 13. a = √13.
Q32.
Height of a tower from a point 50m away at elevation 30° is:
Answer: B. 50/√3 m
Why: tan 30° = h/50. h = 50 × tan 30° = 50/√3 m.
Q33.
Find the general solution of sin θ = 1/2.
Answer: B. θ = nπ + (-1)ⁿ π/6
Why: General solution: θ = nπ + (-1)ⁿ α, where sin α = 1/2, α = π/6.
Q34.
cos 2A = 1 - 2sin²A. Verify for A = 30°:
Answer: A. cos 60° = 1/2 and 1-2sin²30° = 1/2 (true)
Why: cos 60° = 1/2. 1 - 2sin²30° = 1 - 2(1/4) = 1/2. Equal. Identity verified.
Q35.
sin(90°+θ) =
Answer: C. cosθ
Why: sin(90°+θ) = cosθ (complementary angle identity).
Q36.
cos²θ - sin²θ =
Answer: D. Both B and C
Why: cos²θ - sin²θ = cos 2θ. Also cos 2θ = 2cos²θ - 1. So B and C are equal expressions.
Q37.
If sin A = 3/5 and A is in second quadrant, cos A =
Answer: B. -4/5
Why: sin²A + cos²A = 1. cosA = ±4/5. In Q2, cosine is negative. cosA = -4/5.
Q38.
The range of sin x for all x is:
Answer: B. [-1,1]
Why: sin x ranges from -1 to 1 inclusive for all real x.
Q39.
sin 2A =
Answer: C. 2sinA cosA
Why: Double angle formula: sin 2A = 2 sinA cosA.
Q40.
In a triangle, if a = 5, b = 6, c = 7, use the cosine rule to find angle A:
Answer: A. cos A = 60/84
Why: cos A = (b² + c² - a²)/2bc = (36+49-25)/(2×6×7) = 60/84 = 5/7. A = cos⁻¹(5/7).
Q41.
Solve: 2cos²θ - 3cosθ + 1 = 0. Find all solutions in [0, 2π].
Answer: A. 0, π/3, 5π/3, 2π
Why: Quadratic factoring: (2cosθ−1)(cosθ−1)=0. cosθ=1/2 → Q<sub>1</sub>: θ=π/3, Q4: θ=5π/3. cosθ=1 → θ=0 and θ=2π (boundary). All four solutions in [0,2π]: 0, π/3, 5π/3, 2π.
Q42.
In a triangle, prove R = a/(2sinA). If a = 8, A = 30°, find R.
Answer: B. 8
Why: Sine rule states a/sinA=b/sinB=c/sinC=2R (R=circumradius). Substituting a=8, A=30°: 2R=8/sin30°=8/(1/2)=16, so R=8. The circumscribed circle of this triangle has radius 8.
Q43.
If tan α = 1/7 and sin β = 1/√10, find value of α + 2β (in first quadrant).
Answer: A. π/4
Why: tan(2β): sin β = 1/√10, cos β = 3/√10. tan β = 1/3. tan(2β) = 2(1/3)/(1-1/9) = (2/3)/(8/9) = 3/4. tan(α+2β) = (1/7 + 3/4)/(1 - 3/28) = (25/28)/(25/28) = 1. So α+2β = π/4.
Q44.
General solution of 2cos²x + 3sinx = 0 is:
Answer: C. x = 7π/6 + 2nπ or x = 11π/6 + 2nπ
Why: 2(1-sin²x) + 3sinx = 0. 2sin²x - 3sinx - 2 = 0. (2sinx+1)(sinx-2)=0. sinx = -1/2 (sinx=2 impossible). x = 7π/6 + 2nπ or x = 11π/6 + 2nπ.
Q45.
tan(A+B+C) formula: if A+B+C = π, then tanA + tanB + tanC =
Answer: C. tanA tanB tanC
Why: If A+B+C = π, then tan(A+B+C) is undefined (tan π = 0 but formula gives tanA+tanB+tanC - tanAtanBtanC = 0 at numerator when sum is 0... Actually the identity: tanA+tanB+tanC = tanA tanB tanC when A+B+C=nπ.
Q46.
Value of cos 36° - cos 72° =
Answer: A. 1/2
Why: Exact values from pentagon: cos36°=(1+√5)/4 and cos72°=(√5−1)/4. Difference=(1+√5)/4−(√5−1)/4=(1+√5−√5+1)/4=2/4=1/2. Verify: cos36°≈0.809, cos72°≈0.309, diff≈0.5 ✓.
Q47.
In triangle ABC, if b + c = 2a, then tanB/2 × tanC/2 =
Answer: A. 1/3
Why: tan(B/2)tan(C/2) = r/(s-a) × ... Using formulas: tan(B/2)tan(C/2) = [s-b][s-c]/s(s-a). Since b+c=2a and a+b+c=2s, b+c=2s-2a=2a so s=2a. s-a=a, s-b=2a-b, s-c=2a-c. tan(B/2)tan(C/2) = (2a-b)(2a-c)/(2a×a). Also (b+c=2a): (2a-b)(2a-c) = 4a²-2a(b+c)+bc = 4a²-4a²+bc = bc. So = bc/(2a²). Using law of cosines... Actually standard result: tan(B/2)tan(C/2) = (s-b)(s-c) / [s(s-a)]. With s=2a, s-a=a: = (2a-b)(2a-c)/(2a²) = bc/(2a²). Need more info. Common result for b+c=2a is 1/3.
Q48.
Sum of cosines: cosθ + cos(θ+2π/n) + cos(θ+4π/n) + ... (n terms) =
Answer: B. 0
Why: Phasor method: each term is Re(e^(i(θ+2πk/n))). Sum = Re(e<sup>iθ</sup>·Σe<sup>2πik/n</sup>) for k=0..n−1. The inner sum is the sum of all n-th roots of unity = 0. Therefore total sum = 0 for all θ.
Q49.
If sin θ = 3/5 and θ is in second quadrant, find tan 2θ.
Answer: A. -24/7
Why: Q2: sinθ=3/5, cosθ=−4/5 (cosine is negative in Q2). tanθ=sinθ/cosθ=−3/4. Double-angle formula: tan2θ=2tanθ/(1−tan²θ)=2(−3/4)/(1−9/16)=(−3/2)/(7/16)=−24/7.
Q50.
cos 20° × cos 40° × cos 80° =
Answer: A. 1/8
Why: Using product formula: cos 20° cos 40° cos 80° = (1/4)sin(2<sup>3</sup> × 20°)/sin(20°) ... = 1/8. Standard product identity result.
Q51.
In triangle ABC, a/(sinA) = 2R. This is the:
Answer: B. Sine rule
Why: This is the Sine Rule (Law of Sines): a/sinA=b/sinB=c/sinC=2R, where R is the circumradius. It connects each side to the sine of its opposite angle. Used to solve triangles when two angles and a side are known.
Q52.
Value of sin 18° =
Answer: A. (√5-1)/4
Why: Let θ=18°, so 5θ=90° → sin3θ=cos2θ. Expanding: 3sinθ−4sin³θ=1−2sin²θ. Set x=sinθ: 4x²+3x−1=0 (after dividing by sinθ≠0), giving x=(√5−1)/4. Answer: sin18°=(√5−1)/4.
Q53.
If in triangle ABC, a = 3, b = 4, C = 90°, find area using trigonometry.
Answer: A. 6
Why: Trig area formula: Δ=(1/2)ab sinC. Given a=3, b=4, C=90°: sin90°=1. Area=(1/2)(3)(4)(1)=6. The right angle simplifies sinC to 1, so area is simply half the product of the two enclosing sides.
Q54.
Solve tan²θ - 3 = 0 for θ in (0, 2π).
Answer: A. π/3, 2π/3, 4π/3, 5π/3
Why: tan²θ=3 → tanθ=±√3. Reference angle α=π/3. tanθ=+√3 in Q<sub>1</sub>, Q<sub>3</sub>: θ=π/3, 4π/3. tanθ=−√3 in Q2, Q4: θ=2π/3, 5π/3. Four solutions: π/3, 2π/3, 4π/3, 5π/3. General form: nπ±π/3.
Q55.
tan⁻¹(1) + tan⁻¹(2) + tan⁻¹(3) =
Answer: A. π
Why: tan⁻¹1=π/4. For tan⁻¹2+tan⁻¹3: since xy=6>1 and both positive, formula gives π+tan⁻¹((2+3)/(1−6))=π+tan⁻¹(−1)=π−π/4=3π/4. Total: π/4+3π/4=π. Answer: π.
Q56.
The minimum value of 3 sin²θ + 4 cos²θ is:
Answer: A. 3
Why: Rewrite using cos²θ=1−sin²θ: 3sin²θ+4(1−sin²θ)=4−sin²θ. Since 0≤sin²θ≤1, the minimum value occurs at sin²θ=1 (maximum subtraction): 4−1=3. The maximum is 4 (at sinθ=0). Min=3.
Q57.
In triangle, sin(A/2) = √[(s-b)(s-c)/bc]. This is:
Answer: B. Half-angle sine formula using s (semi-perimeter)
Why: Half-angle sine formula using semi-perimeter s=(a+b+c)/2. Derived by combining cosine rule cosA=(b²+c²−a²)/2bc with cos A=1−2sin²(A/2): substituting a²=(b+c)²−4bc·sin²(A/2) and simplifying yields sin(A/2)=√[(s−b)(s−c)/bc].
Q58.
If tan(x+y) = 33 and tan(x) = 3, find tan(y).
Answer: B. 30/99
Why: tan(x+y) = (tanx + tany)/(1 - tanx tany) = 33. (3+tany)/(1-3tany) = 33. 3+tany = 33-99tany. 100tany = 30. tany = 30/100 = 3/10. Wait that is same as option A. But 3/10 = 30/100. So both A and C say same thing differently.
Q59.
The maximum value of 3 sinθ + 4 cosθ is:
Answer: A. 5
Why: The maximum of a sinθ + b cosθ is √(a² + b²) = √(9 + 16) = 5.
Q60.
The exact value of sin 75° is:
Answer: A. (√6 + √2)/4
Why: sin 75° = sin(45° + 30°) = (√6 + √2)/4.
Q61.
If sinθ + cosθ = 1/2, then sin2θ equals:
Answer: B. −3/4
Why: Squaring: 1 + sin2θ = 1/4, so sin2θ = −3/4.
Q62.
The value of cos(π/7)·cos(2π/7)·cos(4π/7) is:
Answer: B. −1/8
Why: Using cosA·cos2A·cos4A = sin8A/(8 sinA) with A = π/7 gives sin(8π/7)/(8 sin(π/7)) = −1/8.
Q63.
The number of solutions of tanx + secx = 2cosx in [0, 2π) is:
Answer: B. 2
Why: Rearranging gives 2sin²x + sinx − 1 = 0, so sinx = 1/2 or sinx = −1. sinx = −1 makes secx undefined, leaving x = π/6, 5π/6: 2 solutions.
Q64.
In a triangle ABC, tanA + tanB + tanC = 6 and tanA·tanB = 2. Then tanC equals:
Answer: B. 3
Why: In a triangle tanA + tanB + tanC = tanA·tanB·tanC. So 6 = 2·tanC, giving tanC = 3.
Q65.
The range of f(x) = 3sinx + 4cosx + 5 is:
Answer: B. [0, 10]
Why: 3sinx + 4cosx lies in [−5, 5] since the amplitude is √(9+16) = 5. Adding 5 gives [0, 10].
Q66.
If tanθ = 3/4 and θ lies in the third quadrant, then sinθ + cosθ equals:
Answer: B. −7/5
Why: In the third quadrant both are negative: sinθ = −3/5, cosθ = −4/5, so the sum is −7/5.
Q67.
The value of sin²(π/8) + sin²(3π/8) + sin²(5π/8) + sin²(7π/8) is:
Answer: C. 2
Why: By symmetry sin(5π/8) = sin(3π/8) and sin(7π/8) = sin(π/8), and sin(3π/8) = cos(π/8). The sum is 2[sin²(π/8) + cos²(π/8)] = 2.
Q68.
The maximum value of sin⁴θ + cos⁴θ is:
Answer: C. 1
Why: sin⁴θ + cos⁴θ = 1 − (1/2)sin²2θ, which is largest when sin2θ = 0, giving 1.