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Admissions tests / TMUA / Paper 1 / Trigonometry

Test standard. 12 questions, 12 marks, about 45 minutes.

TMUA Paper 1: Trigonometry, set 1

Sine and cosine rules, exact values, graphs and transformations of trigonometric functions, identities and equations over a stated interval.

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  • Answer all questions. No calculator.
  • Each question has exactly one correct answer.
  1. 11 mark

    In triangle ABC, angle BAC = 30 degrees and side AC (= b) = 10 cm. Side BC (= a) is opposite the given angle. For which value of a below are there exactly two different triangles ABC satisfying these measurements (the 'ambiguous case')?

    1. A a = 4 cm
    2. B a = 5 cm
    3. C a = 6 cm
    4. D a = 12 cm
  2. 21 mark

    Triangle PQR has PQ = 7 cm, PR = 8 cm and angle QPR = 60 degrees. Find the exact area of triangle PQR.

    1. A 28 sqrt(3) cm^2
    2. B 14 sqrt(3) cm^2
    3. C 7 sqrt(3) cm^2
    4. D 14 cm^2
  3. 31 mark

    A sector of a circle has radius 9 cm, and its arc subtends an angle of 2pi/3 radians at the centre. Find the exact length of the arc.

    1. A 6 pi cm
    2. B 12 pi cm
    3. C 27 pi cm
    4. D 3 pi cm
  4. 41 mark

    A sector of a circle has radius 6 cm and the angle at the centre is pi/3 radians. Find the exact area of the corresponding minor segment (the region between the chord and the arc).

    1. A 6 pi cm^2
    2. B (12 pi - 18 sqrt(3)) cm^2
    3. C (9 sqrt(3) - 6 pi) cm^2
    4. D (6 pi - 9 sqrt(3)) cm^2
  5. 51 mark

    Find the exact value of 3 tan^2(30) + 2.

    1. A 11
    2. B 3
    3. C sqrt(3) + 2
    4. D 7/3
  6. 61 mark

    Find the exact value of cos^2(45) - sin^2(30).

    1. A (sqrt(2) - 1)/2
    2. B -1/4
    3. C 0
    4. D 1/4
  7. 71 mark

    The graph of y = cos(x) (x in degrees) is transformed to give the graph of y = cos(2x) + 3. Which of the following correctly states the period and the range of y = cos(2x) + 3?

    1. A Period 180 degrees, range [2, 4]
    2. B Period 720 degrees, range [2, 4]
    3. C Period 360 degrees, range [2, 4]
    4. D Period 180 degrees, range [-1, 1]
  8. 81 mark

    The graph of y = sin(x) for 0 <= x <= 180 (degrees) is symmetric about the line x = 90. Given that sin(50) = k, use this symmetry to write down the exact value of sin(130) in terms of k.

    1. A -k
    2. B k
    3. C 1 - k
    4. D sqrt(1 - k^2)
  9. 91 mark

    Simplify fully: sin(theta)/(1 - cos(theta)) + (1 - cos(theta))/sin(theta), for 0 < theta < 180 degrees.

    1. A 2(1 + cos(theta))/sin(theta)
    2. B 1/sin(theta)
    3. C 2/sin(theta)
    4. D 2 sin(theta)
  10. 101 mark

    Given that sin(theta) = 3/5 and 90 < theta < 180 (degrees), find the exact value of tan(theta).

    1. A -3/4
    2. B 3/4
    3. C -4/3
    4. D 4/3
  11. 111 mark

    Solve 2 sin(x) + 1 = 0 for 0 <= x <= 360 (degrees). Which of the following gives the complete solution set?

    1. A {30, 150}
    2. B {150, 330}
    3. C {210, 330}
    4. D {240, 300}
  12. 121 mark

    Solve 2 cos^2(x) + cos(x) - 1 = 0 for 0 <= x <= 360 (degrees). How many solutions does the equation have in this interval?

    1. A 3
    2. B 2
    3. C 1
    4. D 4

Worked solutions

Every question below carries the reasoning, not just the answer. The official material for this test publishes a correct option letter and nothing else.

  1. Question 1Answer: C

    1. The ambiguous (angle-side-side) case arises when angle A, the adjacent side b, and the opposite side a are known: two triangles are possible exactly when b sin A < a < b.
    2. Here b sin A = 10 x sin(30) = 10 x (1/2) = 5, so two triangles occur precisely when 5 < a < 10.
    3. Checking each option: a = 6 lies strictly between 5 and 10, so it gives two triangles. a = 4 is below 5 (no triangle), a = 5 is exactly the boundary (one right-angled triangle), and a = 12 is at least 10 (one triangle only).
    4. Therefore the answer is C: a = 6 cm.
    • Why not A: Assumes any a less than b gives two triangles without checking a is large enough to reach the opposite ray: here a = 4 is less than b sin A = 5, so no triangle exists at all.
    • Why not B: Treats a = 5, which equals b sin A exactly, as still giving two triangles; this boundary value in fact gives exactly one right-angled triangle (angle B = 90 degrees), not two.
    • Why not D: Assumes the ambiguous case applies whenever angle A, b and a are known, regardless of relative side length; once a is at least as long as b (here a = 12 > b = 10), there is only ever one triangle.

  2. Question 2Answer: B

    1. The area of a triangle given two sides and the included angle is Area = (1/2) ab sin(C).
    2. Here a = 7, b = 8 and C = 60 degrees, so Area = (1/2)(7)(8) sin(60).
    3. sin(60) = sqrt(3)/2, an exact standard value, so Area = (1/2)(56)(sqrt(3)/2) = 14 sqrt(3).
    4. Therefore the answer is B: 14 sqrt(3) cm^2.
    • Why not A: Uses Area = ab sin(C) instead of (1/2) ab sin(C), omitting the 1/2 factor and so doubling the correct area to 28 sqrt(3).
    • Why not C: Applies the 1/2 factor an extra time, computing (1/2) x (1/2) x 56 x sin(60), which halves the correct area to 7 sqrt(3).
    • Why not D: Uses cos(60) = 1/2 in place of sin(60) in the area formula, giving (1/2)(7)(8)(1/2) = 14 instead of the correct value.

  3. Question 3Answer: A

    1. Arc length is given by s = r theta, where theta is measured in radians.
    2. Here r = 9 and theta = 2pi/3, so s = 9 x (2pi/3).
    3. 9 x (2pi/3) = (9 x 2 / 3) pi = 6 pi.
    4. Therefore the answer is A: 6 pi cm.
    • Why not B: Uses the diameter (18 cm) instead of the radius in s = r theta, doubling the correct arc length to 12 pi.
    • Why not C: Confuses the arc length formula r theta with the sector area formula (1/2) r^2 theta, computing (1/2)(81)(2pi/3) = 27 pi instead.
    • Why not D: Carries the 1/2 factor from the area formula across into the arc length calculation, halving the correct answer of 6 pi to 3 pi.

  4. Question 4Answer: D

    1. The area of a segment is the sector area minus the triangle area: Area = (1/2) r^2 theta - (1/2) r^2 sin(theta) = (1/2) r^2 (theta - sin(theta)).
    2. Here r = 6 and theta = pi/3, so Area = (1/2)(36)(pi/3 - sin(pi/3)).
    3. sin(pi/3) = sqrt(3)/2, an exact standard value, so Area = 18 x (pi/3 - sqrt(3)/2) = 18 x pi/3 - 18 x sqrt(3)/2 = 6 pi - 9 sqrt(3).
    4. Therefore the answer is D: (6 pi - 9 sqrt(3)) cm^2.
    • Why not A: Gives the area of the whole sector, (1/2) r^2 theta = 6 pi, forgetting that a segment excludes the triangular region cut off by the chord.
    • Why not B: Omits the 1/2 factor from the segment formula, using r^2(theta - sin(theta)) instead of (1/2) r^2 (theta - sin(theta)), doubling the correct area.
    • Why not C: Subtracts the sector area from the triangle area instead of the other way round, giving the negative of the correct expression.

  5. Question 5Answer: B

    1. The exact value tan(30) = 1/sqrt(3), which rationalises to sqrt(3)/3; squaring gives tan^2(30) = 1/3.
    2. So 3 tan^2(30) + 2 = 3 x (1/3) + 2 = 1 + 2.
    3. 1 + 2 = 3.
    4. Therefore the answer is B: 3.
    • Why not A: Uses tan(60) = sqrt(3) in place of tan(30), so tan^2(60) = 3 and 3(3) + 2 = 11 - the two standard angles 30 and 60 have been swapped.
    • Why not C: Uses tan(30) itself rather than tan^2(30), giving 3 x (1/sqrt(3)) + 2 = sqrt(3) + 2 - the squaring in the expression has been dropped.
    • Why not D: Misremembers the exact value as tan(30) = 1/3 (rather than 1/sqrt(3)), so tan^2(30) is taken as 1/9, giving 3(1/9) + 2 = 7/3.

  6. Question 6Answer: D

    1. cos(45) = sqrt(2)/2, so cos^2(45) = (sqrt(2)/2)^2 = 2/4 = 1/2.
    2. sin(30) = 1/2, so sin^2(30) = 1/4.
    3. cos^2(45) - sin^2(30) = 1/2 - 1/4 = 1/4.
    4. Therefore the answer is D: 1/4.
    • Why not A: Uses cos(45) - sin(30) instead of the squared values, giving (sqrt(2)/2) - (1/2) = (sqrt(2) - 1)/2 - the squaring in the expression has been ignored.
    • Why not B: Computes sin^2(30) - cos^2(45) instead, reversing the order of subtraction and giving the negative of the correct value.
    • Why not C: Misremembers cos(45) as 1/2 (the value of cos(60), not cos(45)), so cos^2(45) is taken as 1/4, which then cancels sin^2(30) to give 0.

  7. Question 7Answer: A

    1. For y = cos(x), the period is 360 degrees and the range is [-1, 1].
    2. Replacing x with 2x compresses the graph horizontally by a factor of 2, so the period of cos(2x) is 360/2 = 180 degrees; a horizontal transformation does not change the output values, so the range of cos(2x) is still [-1, 1].
    3. Adding 3 shifts the whole graph up by 3, so the range becomes [-1 + 3, 1 + 3] = [2, 4], while a vertical shift does not affect the period.
    4. Therefore y = cos(2x) + 3 has period 180 degrees and range [2, 4], which is answer A.
    • Why not B: Treats the coefficient 2 as stretching rather than compressing the period, doubling 360 to 720 instead of halving it.
    • Why not C: Assumes multiplying x by 2 only changes the graph's shape and not its period, incorrectly leaving the period at 360 degrees (confusing y = cos(2x) with y = 2cos(x), which does have period 360).
    • Why not D: Correctly compresses the period to 180 degrees but forgets to apply the +3 vertical shift to the range, leaving it as [-1, 1].

  8. Question 8Answer: B

    1. 130 and 50 are symmetric about x = 90, since (50 + 130)/2 = 90.
    2. By the graph's symmetry about x = 90 (equivalently, the identity sin(180 - theta) = sin(theta)), sin(130) = sin(180 - 130) = sin(50).
    3. Since sin(50) = k, it follows that sin(130) = k.
    4. Therefore the answer is B: k.
    • Why not A: Assumes sin(180 - theta) = -sin(theta), confusing this reflection about x = 90 with the different identity sin(180 + theta) = -sin(theta); the actual symmetry about x = 90 introduces no sign change.
    • Why not C: Invents a linear relationship between sin(130) and sin(50), such as subtracting k from 1, with no valid identity behind it.
    • Why not D: Uses the Pythagorean identity to find cos(50) = sqrt(1 - k^2) and gives this as the answer, confusing sin(130) with a value belonging to a different trigonometric ratio.

  9. Question 9Answer: C

    1. Combine the two fractions over a common denominator: [sin^2(theta) + (1 - cos(theta))^2] / [sin(theta)(1 - cos(theta))].
    2. Expand (1 - cos(theta))^2 = 1 - 2cos(theta) + cos^2(theta), so the numerator is sin^2(theta) + cos^2(theta) + 1 - 2cos(theta).
    3. Using sin^2(theta) + cos^2(theta) = 1, the numerator becomes 1 + 1 - 2cos(theta) = 2 - 2cos(theta) = 2(1 - cos(theta)).
    4. The (1 - cos(theta)) factor cancels between the numerator and the denominator (it is nonzero for 0 < theta < 180), leaving 2/sin(theta).
    5. Therefore the answer is C: 2/sin(theta).
    • Why not A: Expands (1 - cos(theta))^2 as 1 - cos^2(theta), dropping the middle cross term -2cos(theta); this squaring error leads to 2(1 + cos(theta))/sin(theta) after the rest of the algebra is carried through correctly.
    • Why not B: Simplifies 2 - 2cos(theta) to 1 - cos(theta) by losing the factor of 2, then cancels (1 - cos(theta)) to leave 1/sin(theta) instead of 2/sin(theta).
    • Why not D: Reaches the correct simplified fraction 2/sin(theta) but then inverts it, a reciprocal slip that turns division by sin(theta) into multiplication.

  10. Question 10Answer: A

    1. Use sin^2(theta) + cos^2(theta) = 1: cos^2(theta) = 1 - (3/5)^2 = 1 - 9/25 = 16/25, so cos(theta) = 4/5 or -4/5.
    2. Since 90 < theta < 180, theta is in the second quadrant, where cosine is negative, so cos(theta) = -4/5.
    3. tan(theta) = sin(theta)/cos(theta) = (3/5) / (-4/5) = -3/4.
    4. Therefore the answer is A: -3/4.
    • Why not B: Takes cos(theta) = 4/5 (the positive root), forgetting that the second quadrant makes cosine negative, giving tan(theta) = 3/4.
    • Why not C: Uses cos(theta)/sin(theta) instead of sin(theta)/cos(theta) (the tan identity written upside down), but does correctly keep cosine negative, giving (-4/5)/(3/5) = -4/3.
    • Why not D: Combines both errors above: takes cosine as the positive root 4/5 and also inverts the tan identity, giving (4/5)/(3/5) = 4/3.

  11. Question 11Answer: C

    1. 2 sin(x) + 1 = 0 rearranges to sin(x) = -1/2.
    2. The reference angle is the acute angle with sin(reference) = 1/2, which is 30 degrees (an exact standard value).
    3. sin(x) is negative in the third and fourth quadrants, giving x = 180 + 30 = 210 and x = 360 - 30 = 330.
    4. Both lie in the given interval 0 <= x <= 360, so the full solution set is {210, 330}, answer C.
    • Why not A: Solves sin(x) = 1/2 instead of -1/2 (dropping the negative sign), giving the first- and second-quadrant solutions 30 and 150 instead of the correct third- and fourth-quadrant ones.
    • Why not B: Uses the second-quadrant formula 180 - 30 = 150 for one solution instead of the third-quadrant formula 180 + 30 = 210, mixing up which quadrants give a negative sine value.
    • Why not D: Correctly identifies that sin is negative in the third and fourth quadrants but misremembers the reference angle as 60 degrees instead of 30, giving 180 + 60 = 240 and 360 - 60 = 300.

  12. Question 12Answer: A

    1. Let c = cos(x). The equation becomes 2c^2 + c - 1 = 0, which factorises as (2c - 1)(c + 1) = 0.
    2. So cos(x) = 1/2 or cos(x) = -1.
    3. cos(x) = 1/2 gives x = 60 or x = 300 (using the exact value cos(60) = 1/2) within 0 <= x <= 360.
    4. cos(x) = -1 gives exactly one solution in this interval, x = 180.
    5. In total there are three solutions: 60, 180 and 300, so the answer is A: 3.
    • Why not B: Rejects cos(x) = -1 as if it were not a valid solution, keeping only the two solutions from cos(x) = 1/2.
    • Why not C: Finds only the cos(x) = -1 solution (x = 180) and stops, missing that the quadratic has a second factor giving two more solutions.
    • Why not D: Factorises the quadratic with a sign error as (2c - 1)(c - 1) instead of (2c - 1)(c + 1), giving cos(x) = 1/2 or cos(x) = 1; cos(x) = 1 is then double-counted at both x = 0 and x = 360, giving four solutions instead of three.

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