IOE entrance Mathematics · Chapter 3
Trigonometry
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50 questions in 3 syllabus topics · 20 are 2-mark questions.
3.1 Trigonometric equations and general values
15 questions
1. The general solution of sinθ = 0 is:
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Where does the sine curve cut the θ-axis?
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Answer: A. θ = nπ
sinθ = 0 at θ = 0, ±π, ±2π, …, i.e. at every integer multiple of π, so θ = nπ, n ∈ Z.
2. The general solution of cosθ = 0 is:
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Cosine vanishes at odd multiples of a certain angle.
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Answer: A. θ = (2n + 1)π/2
cosθ = 0 at θ = ±π/2, ±3π/2, …, i.e. at odd multiples of π/2: θ = (2n + 1)π/2.
3. The general solution of tanθ = √3 is:
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tanθ = tanα gives θ = nπ + α.
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Answer: D. θ = nπ + π/3
tan(π/3) = √3 and tan has period π, so θ = nπ + π/3.
4. The general solution of cosθ = −½ is:
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Find α in [0, π] with cosα = −½, then use the cosine general form.
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Answer: D. θ = 2nπ ± 2π/3
cos(2π/3) = −½, and cosθ = cosα ⇒ θ = 2nπ ± α, so θ = 2nπ ± 2π/3.
5. If sinθ = sinα, then the general value of θ is:
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sin(π − α) = sinα; combine the two families.
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Answer: B. nπ + (−1)ⁿα
Equal sines occur at α and π − α plus multiples of 2π; both families combine as θ = nπ + (−1)ⁿα.
6. The general solution of tan²θ = 1/3 is:
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tan²θ = tan²α ⇒ θ = nπ ± α.
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Answer: D. θ = nπ ± π/6
tan²θ = tan²(π/6) ⇒ θ = nπ ± π/6 (this covers both tanθ = 1/√3 and tanθ = −1/√3).
7. How many solutions does sin x = ½ have in the interval [0, 2π]?
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Sine is positive in the first and second quadrants only.
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Answer: B. 2
In [0, 2π], sin x = ½ at x = π/6 and x = 5π/6 only (sine is negative in the third and fourth quadrants).
8. The general solution of sin2x = 1 is:
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Solve for 2x first, then divide the whole general value by 2.
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Answer: B. x = nπ + π/4
sin2x = 1 ⇒ 2x = 2nπ + π/2 ⇒ x = nπ + π/4.
9. The values of x in [0, 2π) satisfying tan x = −1 are:
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Tangent is negative in the second and fourth quadrants.
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Answer: D. 3π/4 and 7π/4
tan is negative in Q2 and Q4 with reference angle π/4: x = π − π/4 = 3π/4 and x = 2π − π/4 = 7π/4.
10. The general solution of 2cos²θ + 3sinθ = 0 is:
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Write cos²θ = 1 − sin²θ and solve the quadratic in sinθ.
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Answer: D. θ = nπ + (−1)ⁿ(−π/6)
2(1 − sin²θ) + 3sinθ = 0 ⇒ 2sin²θ − 3sinθ − 2 = 0 ⇒ (2sinθ + 1)(sinθ − 2) = 0. sinθ = 2 is impossible, so sinθ = −½ = sin(−π/6), giving θ = nπ + (−1)ⁿ(−π/6).
11. The general solution of sin x + cos x = √2 is:
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Write sin x + cos x as √2 sin(x + π/4).
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Answer: B. x = 2nπ + π/4
sin x + cos x = √2 sin(x + π/4) = √2 ⇒ sin(x + π/4) = 1 ⇒ x + π/4 = 2nπ + π/2 ⇒ x = 2nπ + π/4.
12. The general solution of √3 cosθ + sinθ = 1 is:
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Divide by √(3 + 1) = 2 and write the left side as cos(θ − α).
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Answer: C. θ = 2nπ + π/6 ± π/3
Divide by 2: cos(θ − π/6) = ½ = cos(π/3) ⇒ θ − π/6 = 2nπ ± π/3 ⇒ θ = 2nπ + π/6 ± π/3 (e.g. θ = π/2 or −π/6; both check).
13. The number of solutions of tan x + sec x = 2cos x in [0, 2π] is:
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Convert to sin and cos, but remember tan x and sec x must be defined.
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Answer: B. 2
Multiply by cos x (cos x ≠ 0): sin x + 1 = 2cos²x = 2 − 2sin²x ⇒ 2sin²x + sin x − 1 = 0 ⇒ sin x = ½ or −1. sin x = −1 makes cos x = 0 (tan, sec undefined), so only sin x = ½: x = π/6, 5π/6 → 2 solutions.
14. The general solution of cos3θ = cosθ is:
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cosA = cosB ⇒ A = 2nπ ± B; check whether one family contains the other.
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Answer: A. θ = nπ/2
cos3θ = cosθ ⇒ 3θ = 2nπ ± θ. With +: 2θ = 2nπ ⇒ θ = nπ. With −: 4θ = 2nπ ⇒ θ = nπ/2. The second family contains the first, so θ = nπ/2.
15. Every solution of tanθ + tan2θ + tanθ·tan2θ = 1 is of the form:
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Rearrange to recognise the tan(A + B) formula.
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Answer: B. θ = nπ/3 + π/12
tanθ + tan2θ = 1 − tanθ tan2θ ⇒ (tanθ + tan2θ)/(1 − tanθ tan2θ) = 1 ⇒ tan3θ = 1 ⇒ 3θ = nπ + π/4 ⇒ θ = nπ/3 + π/12. (Not every value of this form works: e.g. θ = 3π/4 makes tan2θ undefined, so such values must be excluded; but every actual solution has this form.)
3.2 Inverse trigonometric functions
15 questions
16. The principal value of sin⁻¹(−½) is:
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Principal values of sin⁻¹ lie in [−π/2, π/2].
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Answer: C. −π/6
The principal range of sin⁻¹ is [−π/2, π/2]; the angle there with sine −½ is −π/6.
17. The principal value of cos⁻¹(−½) is:
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Principal values of cos⁻¹ lie in [0, π].
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Answer: D. 2π/3
cos⁻¹(−x) = π − cos⁻¹x, so cos⁻¹(−½) = π − π/3 = 2π/3, which lies in [0, π].
18. The principal value branch (range) of tan⁻¹x is:
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Can tan⁻¹x ever equal π/2?
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Answer: A. (−π/2, π/2)
tan⁻¹x takes values strictly between −π/2 and π/2; the endpoints are excluded because tan(±π/2) is undefined.
19. The domain of the function f(x) = sin⁻¹(2x) is:
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The argument of sin⁻¹ must lie in [−1, 1].
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Answer: D. [−½, ½]
We need −1 ≤ 2x ≤ 1, i.e. −½ ≤ x ≤ ½.
20. The value of sin(cos⁻¹(3/5)) is:
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Draw a right triangle with adjacent 3 and hypotenuse 5.
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Answer: C. 4/5
Let θ = cos⁻¹(3/5) ∈ [0, π], so sinθ ≥ 0 and sinθ = √(1 − 9/25) = 4/5.
21. The value of sin⁻¹(sin(5π/6)) is:
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sin⁻¹(sin x) = x only when x lies in [−π/2, π/2].
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Answer: C. π/6
sin(5π/6) = ½, and the principal value of sin⁻¹(½) is π/6. (5π/6 is outside [−π/2, π/2], so the answer is not 5π/6.)
22. For −1 ≤ x ≤ 1, cos⁻¹(−x) is equal to:
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The range of cos⁻¹ is [0, π], so it cannot simply change sign.
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Answer: C. π − cos⁻¹x
If cos⁻¹x = θ ∈ [0, π], then cos(π − θ) = −x and π − θ ∈ [0, π], so cos⁻¹(−x) = π − cos⁻¹x.
23. For |x| < 1, 2tan⁻¹x is equal to:
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Use the double-angle formula for tangent.
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Answer: A. tan⁻¹(2x/(1 − x²))
From tan2θ = 2tanθ/(1 − tan²θ) with tanθ = x: 2tan⁻¹x = tan⁻¹(2x/(1 − x²)) for |x| < 1.
24. If sin⁻¹x + sin⁻¹y = 2π/3, then cos⁻¹x + cos⁻¹y equals:
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sin⁻¹t + cos⁻¹t = π/2 for each variable.
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Answer: C. π/3
cos⁻¹x = π/2 − sin⁻¹x and similarly for y, so cos⁻¹x + cos⁻¹y = π − (sin⁻¹x + sin⁻¹y) = π − 2π/3 = π/3.
25. The value of tan⁻¹1 + tan⁻¹2 + tan⁻¹3 is:
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When xy > 1 with x, y > 0, tan⁻¹x + tan⁻¹y = π + tan⁻¹((x + y)/(1 − xy)).
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Answer: A. π
Since 2·3 > 1 (both positive), tan⁻¹2 + tan⁻¹3 = π + tan⁻¹((2 + 3)/(1 − 6)) = π + tan⁻¹(−1) = 3π/4. Adding tan⁻¹1 = π/4 gives π.
26. tan⁻¹(1/4) + tan⁻¹(2/9) is equal to:
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Add with the tan⁻¹ sum formula, then express 2θ through cos2θ = (1 − tan²θ)/(1 + tan²θ).
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Answer: B. ½ cos⁻¹(3/5)
Sum = tan⁻¹[(1/4 + 2/9)/(1 − 2/36)] = tan⁻¹[(17/36)/(34/36)] = tan⁻¹(½). With θ = tan⁻¹(½), cos2θ = (1 − ¼)/(1 + ¼) = 3/5, so tan⁻¹(½) = ½ cos⁻¹(3/5).
27. The solution of tan⁻¹(2x) + tan⁻¹(3x) = π/4 is:
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Combine with the tan⁻¹ addition formula, then check each root in the original equation.
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Answer: A. x = 1/6
5x/(1 − 6x²) = 1 ⇒ 6x² + 5x − 1 = 0 ⇒ (6x − 1)(x + 1) = 0. x = −1 makes both terms negative, so the sum cannot be π/4; hence only x = 1/6.
28. For real x, sin(cot⁻¹(cos(tan⁻¹x))) equals:
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Work from the inside out using right triangles: cosec²φ = 1 + cot²φ.
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Answer: D. √((x² + 1)/(x² + 2))
cos(tan⁻¹x) = 1/√(1 + x²). If cotφ = 1/√(1 + x²), then sinφ = 1/√(1 + cot²φ) = 1/√(1 + 1/(1 + x²)) = √((1 + x²)/(2 + x²)).
29. The value of cos(2 sin⁻¹(1/3)) is:
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Use cos2θ = 1 − 2sin²θ.
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Answer: D. 7/9
Let θ = sin⁻¹(1/3). cos2θ = 1 − 2sin²θ = 1 − 2/9 = 7/9.
30. If cos⁻¹x + cos⁻¹y + cos⁻¹z = 3π, then xy + yz + zx equals:
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What is the largest possible value of cos⁻¹t?
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Answer: A. 3
Each cos⁻¹ value is at most π, so the sum 3π forces cos⁻¹x = cos⁻¹y = cos⁻¹z = π, i.e. x = y = z = −1. Then xy + yz + zx = 1 + 1 + 1 = 3.
3.3 Properties and solution of triangles
20 questions
31. In any triangle ABC with circumradius R, the ratio a/sinA equals:
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Recall the extended sine rule.
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Answer: A. 2R
Sine rule: a/sinA = b/sinB = c/sinC = 2R.
32. In any triangle ABC, which of the following is a correct projection formula?
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Each side equals the sum of projections of the other two sides on it.
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Answer: B. a = b cosC + c cosB
Dropping the altitude from A, side a splits into the projections of b and c on it: a = b cosC + c cosB.
33. In triangle ABC, cosA is equal to:
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Start from a² = b² + c² − 2bc cosA.
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Answer: C. (b² + c² − a²)/(2bc)
Cosine rule: a² = b² + c² − 2bc cosA ⇒ cosA = (b² + c² − a²)/(2bc).
34. If Δ is the area of triangle ABC, then its circumradius R is:
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Combine Δ = ½ bc sinA with the sine rule.
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Answer: D. abc/(4Δ)
Δ = ½ bc sinA and sinA = a/(2R) give Δ = abc/(4R), so R = abc/(4Δ).
35. If Δ is the area and s the semi-perimeter of triangle ABC, the in-radius r is:
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Split the triangle into three parts from the in-centre.
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Answer: C. Δ/s
Joining the in-centre to the vertices splits the triangle into three triangles of heights r: Δ = ½r(a + b + c) = rs, so r = Δ/s. (Δ/(s − a) is the ex-radius.)
36. The circumradius of a triangle with sides 3 cm, 4 cm and 5 cm is:
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Is the triangle right-angled?
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Answer: C. 2.5 cm
3² + 4² = 5², so the triangle is right-angled and the hypotenuse is a diameter of the circumcircle: R = 5/2 = 2.5 cm.
37. The in-radius of a right-angled triangle with sides 5, 12 and 13 is:
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Use r = Δ/s.
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Answer: A. 2
s = 15, Δ = ½·5·12 = 30, r = Δ/s = 30/15 = 2. (Also r = (5 + 12 − 13)/2 = 2.)
38. In triangle ABC, a = 2, b = 3 and C = 60°. Then c equals:
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Apply the cosine rule for c.
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Answer: B. √7
c² = a² + b² − 2ab cosC = 4 + 9 − 2·2·3·½ = 7, so c = √7.
39. In triangle ABC with semi-perimeter s, tan(A/2) equals:
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tan(A/2) = sin(A/2)/cos(A/2); use the half-angle forms for each.
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Answer: C. √[(s − b)(s − c)/(s(s − a))]
Half-angle formula: tan(A/2) = √[(s − b)(s − c)/(s(s − a))]. (The second option is sin(A/2) and the fourth is cos(A/2).)
40. For an equilateral triangle, the ratio of circumradius to in-radius (R : r) is:
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In an equilateral triangle all the centres coincide with the centroid.
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Answer: B. 2 : 1
For side a: R = a/√3 and r = a/(2√3), so R : r = 2 : 1. (The centroid divides each median 2 : 1.)
41. In solving triangle ABC with a, b and acute angle A given, if a < b sinA, then the number of possible triangles is:
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Compare a with the altitude b sinA.
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Answer: D. 0
b sinA is the perpendicular distance from C to the line AB; if a is shorter than this, side a cannot reach the line, so no triangle exists.
42. The ortho-centre of a right-angled triangle lies:
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Which sides of a right triangle are already altitudes?
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Answer: C. at the vertex of the right angle
The two legs are themselves altitudes and they meet at the right-angled vertex, so that vertex is the ortho-centre. (The mid-point of the hypotenuse is the circum-centre.)
43. If in triangle ABC, a cosA = b cosB, then the triangle is:
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Use the sine rule to change sides to sines, then double-angle.
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Answer: A. isosceles or right-angled
With a = 2R sinA, b = 2R sinB: sinA cosA = sinB cosB ⇒ sin2A = sin2B ⇒ 2A = 2B or 2A = π − 2B ⇒ A = B or A + B = π/2.
44. The circumradius of a triangle with sides 13, 14 and 15 is:
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Find the area by Heron's formula, then use R = abc/(4Δ).
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Answer: B. 65/8
s = 21, Δ = √(21·8·7·6) = √7056 = 84. R = abc/(4Δ) = (13·14·15)/(4·84) = 2730/336 = 65/8.
45. In triangle ABC with a = 13, b = 14 and c = 15, the value of tan(A/2) is:
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tan(A/2) = r/(s − a).
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Answer: C. ½
s = 21, Δ = 84, r = Δ/s = 4. tan(A/2) = r/(s − a) = 4/8 = ½. (Check: √[(s − b)(s − c)/(s(s − a))] = √(7·6/(21·8)) = √(¼) = ½.)
46. In triangle ABC, b = 2, c = √3 + 1 and A = 60°. Then angle B is:
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Find a by the cosine rule, then B by the sine rule.
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Answer: C. 45°
a² = 4 + (4 + 2√3) − 2·2(√3 + 1)·½ = 6, so a = √6. sinB = b sinA/a = 2(√3/2)/√6 = 1/√2. Since b < a, B < A, so B = 45° (and then C = 75°).
47. In any triangle ABC, (b + c)cosA + (c + a)cosB + (a + b)cosC equals:
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Regroup the terms in pairs to form projection formulas.
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Answer: D. a + b + c
Regroup: (b cosC + c cosB) + (c cosA + a cosC) + (a cosB + b cosA) = a + b + c by the projection formulas.
48. A triangle has sides 6, 8 and 10. The distance between its in-centre and circum-centre is:
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Place the right angle at the origin and find both centres.
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Answer: C. √5
Right angle at the origin, legs along the axes: in-centre is (r, r) with r = (6 + 8 − 10)/2 = 2, i.e. (2, 2). Circum-centre is the mid-point of the hypotenuse joining (6, 0) and (0, 8), i.e. (3, 4). Distance = √(1² + 2²) = √5. (Also OI² = R² − 2Rr = 25 − 20 = 5.)
49. If the circumradius and in-radius of a triangle are R = 5 and r = 2, then cosA + cosB + cosC equals:
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Recall the identity that relates cosA + cosB + cosC to r/R.
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Answer: C. 7/5
Identity: cosA + cosB + cosC = 1 + r/R = 1 + 2/5 = 7/5. (Check with the 6-8-10 triangle: 0 + 0.8 + 0.6 = 1.4.)
50. The largest angle of the triangle with sides 3, 5 and 7 is:
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The largest angle is opposite the longest side; use the cosine rule.
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Answer: D. 120°
The largest angle is opposite the side 7: cosθ = (9 + 25 − 49)/(2·3·5) = −15/30 = −½, so θ = 120°.