Edexcel International A Level Maths · P2 · Pure Mathematics 2
Trigonometric identities and equations
28 free, high-quality past-paper-style questions on Trigonometric identities and equations for Edexcel IAL Maths, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Maths notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.28 new this month · 229 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked
1IALM-P2-6-B-01P2 · ASEasy5 marks
The angle θ satisfies 270° < θ < 360° and cos θ = 2⁄5. Give exact answers to this question.
(a)
Find the value of sin θ.
(2)
(b)
Find the value of tan θ.
(2)
(c)
Find the value of sin2θ − cos2θ.
(1)
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Answer
Mark
Notes
Part (a)
sin2θ = 1 − 4⁄25 = 21⁄25
M1
sin θ < 0 in this quadrant, so sin θ = −√21 ⁄ 5
A1
Part (b)
tan θ = sin θ ÷ cos θ = (−√21 ⁄ 5) ÷ (2⁄5) ft their sin θ
M1
tan θ = −√21 ⁄ 2
A1
Part (c)
21⁄25 − 4⁄25 = 17⁄25
B1
Common error: Taking the positive square root without thinking about the quadrant. Between 270° and 360° the sine is negative, so sin θ = −√21 ⁄ 5.
Prove each of the following identities, showing every step of your working.
(a)
(1 − sin2θ) tan θ ≡ sin θ cos θ
(2)
(b)
(cos θ + 2 sin θ)2 + (2 cos θ − sin θ)2 ≡ 5
(3)
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Answer
Mark
Notes
Part (a)
1 − sin2θ = cos2θ and tan θ = sin θ ⁄ cos θ used
M1
cos2θ × sin θ ⁄ cos θ = sin θ cos θ, conclusion stated
A1
Part (b)
both brackets expanded: cos2θ + 4 sin θ cos θ + 4 sin2θ + 4 cos2θ − 4 sin θ cos θ + sin2θ
M1
5 cos2θ + 5 sin2θ
A1
= 5(sin2θ + cos2θ) = 5, conclusion stated
A1
Exam tip: Start from the more complicated side and end with exactly the other side. Write the identity you use, such as sin2θ + cos2θ = 1, at the step where you use it.
Figure 1 shows a circle with centre O and radius 6, drawn on coordinate axes. The point P(−4, y) lies on the circle below the x-axis. The angle θ is measured anticlockwise from the positive x-axis to OP, as shown. Give exact answers to parts (a) and (b).
(a)
Find the value of y.
(2)
(b)
Write down the values of cos θ, sin θ and tan θ.
(3)
(c)
Find the value of θ, in degrees to 1 decimal place.
(2)
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Answer
Mark
Notes
Part (a)
(−4)2 + y2 = 62, so y2 = 20
M1
P is below the x-axis, so y = −2√5
A1
Part (b)
cos θ = −4 ⁄ 6 = −⅔
B1
sin θ = −2√5 ⁄ 6 = −√5 ⁄ 3 ft their y
B1
tan θ = sin θ ÷ cos θ = (−√5 ⁄ 3) ÷ (−⅔) = √5 ⁄ 2 ft
B1
Part (c)
cos−1(−⅔) = 131.81°, and θ is reflex with P in the third quadrant, so θ = 360° − 131.81°
M1
θ = 228.2°
A1
Common error: Giving cos−1(−⅔) = 131.8° as the answer. The figure puts P in the third quadrant, so θ is between 180° and 270°; a sketch of the quadrant is the quickest check.
Figure 1 shows the graph of y = sin x for 0 ≤ x ≤ 2π, where x is in radians. The horizontal line y = k lies on one of the gridlines and meets the curve at the points A and B.
(a)
Use Figure 1 to write down the value of k.
(1)
(b)
Find the x-coordinates of A and B, giving your answers to 3 significant figures.
(3)
(c)
Using the symmetry of the graph, solve sin x = 0.6 for 0 ≤ x ≤ 2π, giving your answers to 3 significant figures.
(2)
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Answer
Mark
Notes
Part (a)
k = −0.6
B1
Part (b)
sin−1(−0.6) = −0.6435
M1
A: x = π + 0.6435 = 3.79
A1
B: x = 2π − 0.6435 = 5.64
A1
Part (c)
x = 0.644
B1
x = π − 0.6435 = 2.50 ft
B1
Exam tip: Your calculator gives sin−1(−0.6) = −0.644, which is outside 0 ≤ x ≤ 2π. Use the graph: the two solutions are π + 0.644 and 2π − 0.644.
Figure 1 shows the curves y = 5 sin x and y = 2 cos x for 0° ≤ x ≤ 360°. The curves intersect at the points A and B.
(a)
Show that the x-coordinates of A and B satisfy tan x = 0.4.
(2)
(b)
Find the x-coordinates of A and B, giving your answers to 1 decimal place.
(3)
(c)
Find the exact y-coordinate of A, giving your answer in the form p√29 where p is a rational number.
(3)
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Answer
Mark
Notes
Part (a)
5 sin x = 2 cos x; cos x ≠ 0 at an intersection, since then sin x = ±1 and 5 sin x ≠ 0
M1
dividing by 5 cos x: sin x ⁄ cos x = tan x = 0.4 with tan x ≡ sin x ⁄ cos x stated
A1
Part (b)
tan−1 0.4 = 21.80°
M1
A: x = 21.8°
A1
B: x = 21.80° + 180° = 201.8°
A1
Part (c)
sin x = 0.4 cos x substituted in sin2x + cos2x = 1: 1.16 cos2x = 1
M1
A is in the first quadrant, so cos x = 5 ⁄ √29 and sin x = 2 ⁄ √29
A1
y = 5 sin x = 10 ⁄ √29 = 10⁄29√29
A1
Common error: Writing y = 5 × 0.4 = 2 for the y-coordinate of A. The value 0.4 is tan x, not sin x; find sin x from sin²x + cos²x = 1 (or a right-angled triangle with sides 2, 5 and √29).
Figure 1 shows the graph of y = cos(2θ + 50°) for 0° ≤ θ ≤ 360°. The point M is the first minimum point of the curve. The line y = 0.35 meets the curve at A, B, C and D.
(a)
Find the coordinates of M.
(2)
(b)
Write down the interval for 2θ + 50° that corresponds to 0° ≤ θ ≤ 360°.
(1)
(c)
Find the values of θ at A, B, C and D, giving your answers to 1 decimal place.
(4)
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Answer
Mark
Notes
Part (a)
minimum where 2θ + 50° = 180°
M1
M = (65°, −1)
A1
Part (b)
50° ≤ 2θ + 50° ≤ 770°
B1
Part (c)
cos−1 0.35 = 69.51°
M1
2θ + 50° = 69.51°, 290.49°, 429.51°, 650.49°
M1
θ = (value − 50°) ÷ 2 for each
M1
θ = 9.8°, 120.2°, 189.8°, 300.2°
A1
Exam tip: Count the crossings on the figure first. Four points means you need four values of 2θ + 50° inside 50° to 770°, so keep adding 360° until you leave the interval.
Figure 1 shows the graph of y = a sin(x + b) for 0 ≤ x ≤ 2π, where x is in radians and a and b are constants with a > 0 and 0 < b < π ⁄ 2. The curve has a maximum point at (π ⁄ 3, 4) and crosses the x-axis at (5π ⁄ 6, 0), as shown. The line y = 1.5 is also drawn.
(a)
Write down the value of a.
(1)
(b)
Find the exact value of b.
(2)
(c)
Find the x-coordinates of the points where the line y = 1.5 meets the curve for 0 ≤ x ≤ 2π, giving your answers to 3 significant figures.
(5)
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Answer
Mark
Notes
Part (a)
a = 4
B1
Part (b)
at the maximum, π ⁄ 3 + b = π ⁄ 2
M1
b = π ⁄ 6
A1
Part (c)
sin(x + π ⁄ 6) = 0.375 with π ⁄ 6 ≤ x + π ⁄ 6 ≤ 13π ⁄ 6 ft
B1
sin−1 0.375 = 0.3844, which is less than π ⁄ 6
M1
x + π ⁄ 6 = π − 0.3844 = 2.757 and 2π + 0.3844 = 6.668
M1
x = 2.23
A1
x = 6.14
A1
Common error: Stopping at x + π/6 = 0.384. That value is below π/6, so it gives a negative x; the second crossing comes from 2π + 0.384, which the figure shows near the right-hand end.
Figure 1 shows the curves y = 2cos2x and y = 1 + sin x for 0° ≤ x ≤ 360°. The curves meet at the points P, Q and R.
(a)
Show that the x-coordinates of P, Q and R satisfy 2 sin2x + sin x − 1 = 0.
(2)
(b)
Hence find the coordinates of P, Q and R.
(5)
(c)
State the number of solutions of 2cos2x = 1 + sin x in the interval 0° ≤ x < 720°.
(1)
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Answer
Mark
Notes
Part (a)
cos2x = 1 − sin2x used: 2(1 − sin2x) = 1 + sin x
M1
2 − 2 sin2x − 1 − sin x = 0 rearranged to 2 sin2x + sin x − 1 = 0 with no errors seen
A1
Part (b)
(2 sin x − 1)(sin x + 1) = 0
M1
sin x = ½ or sin x = −1
A1
sin x = ½: x = 30°, 150°
A1
sin x = −1: x = 270°
A1
P(30°, 1.5), Q(150°, 1.5), R(270°, 0)
A1
Part (c)
6 ft
B1
Exam tip: When an equation mixes cos²x with sin x, replace cos²x with 1 − sin²x so that only one trig function is left; the result is a quadratic in sin x.
Figure 1 shows the triangle PQR, in which angle PQR = 90°, PQ = 10 cm and angle QPR = θ, where θ is acute. The point S lies on PR and QS is perpendicular to PR.
(a)
Write down expressions, in terms of θ, for the lengths PS and PR.
(2)
(b)
Hence show that SR = 10 sin θ tan θ.
(3)
(c)
Given that SR = 4.5 cm, find the value of θ, in degrees to 1 decimal place.
(4)
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Answer
Mark
Notes
Part (a)
PS = 10 cos θ
B1
PR = 10 ⁄ cos θ
B1
Part (b)
SR = PR − PS = 10 ⁄ cos θ − 10 cos θ = 10(1 − cos2θ) ⁄ cos θ
M1
1 − cos2θ = sin2θ, so SR = 10 sin2θ ⁄ cos θ
M1
= 10 sin θ × (sin θ ⁄ cos θ) = 10 sin θ tan θ, conclusion stated
A1
Part (c)
10(1 − cos2θ) ⁄ cos θ = 4.5, an equation in cos θ only
M1
20 cos2θ + 9 cos θ − 20 = 0
A1
(5 cos θ − 4)(4 cos θ + 5) = 0, so cos θ = 0.8; cos θ = −1.25 is rejected
M1
θ = 36.9°
A1
Exam tip: To prove a length identity, write the length as a difference of two lengths you can find from right-angled triangles, then use sin²θ + cos²θ = 1 to tidy it.
Figure 1 shows the graph of y = tan(2θ − 30°) for 0° ≤ θ ≤ 180°, together with its two vertical asymptotes and the line y = 2.5. The curve meets the y-axis at the point A.
(a)
Find the exact y-coordinate of A.
(2)
(b)
Find the equations of the two asymptotes shown in Figure 1.
(2)
(c)
Solve tan(2θ − 30°) = 2.5 for 0° ≤ θ ≤ 180°, giving your answers to 1 decimal place.
(3)
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Answer
Mark
Notes
Part (a)
θ = 0: y = tan(−30°)
M1
y = −√3 ⁄ 3
A1
Part (b)
2θ − 30° = 90° or 270°
M1
θ = 60° and θ = 150°
A1
Part (c)
−30° ≤ 2θ − 30° ≤ 330°; tan−1 2.5 = 68.20°
M1
2θ − 30° = 68.20° and 248.20°
M1
θ = 49.1° and 139.1°
A1
Common error: Halving first and then subtracting 30°. Undo the steps in reverse order: add 30°, then divide by 2.
f(θ) = cos θ⁄1 − sin θ − tan θ, 0° ≤ θ < 360°, θ ≠ 90°, θ ≠ 270° Figure 1 shows the graph of y = f(θ).
(a)
Prove that f(θ) ≡ 1⁄cos θ.
(4)
(b)
Hence solve f(θ) = 3, giving your answers to 1 decimal place.
(3)
(c)
Using Figure 1 and part (a), find the set of values of k for which the equation f(θ) = k has no solutions.
(2)
(d)
Solve f(θ) = 2 tan θ.
(3)
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Answer
Mark
Notes
Part (a)
tan θ = sin θ ⁄ cos θ and common denominator cos θ(1 − sin θ)
M1
numerator cos2θ − sin θ(1 − sin θ) = cos2θ + sin2θ − sin θ
A1
sin2θ + cos2θ = 1 used, numerator = 1 − sin θ
M1
(1 − sin θ) ⁄ [cos θ(1 − sin θ)] = 1 ⁄ cos θ, conclusion stated
A1
Part (b)
1 ⁄ cos θ = 3, so cos θ = ⅓
M1
θ = 70.5°
A1
θ = 289.5°
A1
Part (c)
the graph never lies strictly between the turning values −1 and 1
B1
−1 < k < 1
B1
Part (d)
1 ⁄ cos θ = 2 sin θ ⁄ cos θ, and multiplying by cos θ (≠ 0)
M1
sin θ = ½
A1
θ = 30° and 150°
A1
Exam tip: Write the identity you are using at the line where you use it. In part (a), sin²θ + cos²θ = 1 is the step that makes the factor 1 − sin θ appear and cancel.
Figure 1 shows a circle with centre O and radius 5 cm. The rectangle PQRS has all four vertices on the circle, with PQ horizontal. The radius OP makes an acute angle θ with the horizontal line through O, as shown. The perimeter of the rectangle is L cm.
(a)
Show that L = 20(sin θ + cos θ).
(2)
(b)
Given that L = 28, show that 25 sin2θ − 35 sin θ + 12 = 0.
(3)
(c)
Hence find the two possible values of θ, giving your answers to 1 decimal place.
(3)
(d)
Show that the area of the rectangle is the same for both values of θ, and state this area.
(2)
(e)
Show that no such rectangle has perimeter 30 cm.
(3)
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Answer
Mark
Notes
Part (a)
PQ = 2 × 5 cos θ = 10 cos θ and PS = 2 × 5 sin θ = 10 sin θ
M1
L = 2(10 cos θ + 10 sin θ) = 20(sin θ + cos θ)
A1
Part (b)
sin θ + cos θ = 1.4, so cos θ = 1.4 − sin θ
M1
substituted in sin2θ + cos2θ = 1: sin2θ + (1.4 − sin θ)2 = 1
M1
2 sin2θ − 2.8 sin θ + 0.96 = 0, multiplied by 12.5 to give 25 sin2θ − 35 sin θ + 12 = 0
A1
Part (c)
(5 sin θ − 3)(5 sin θ − 4) = 0
M1
sin θ = 0.6 or 0.8
A1
θ = 36.9° or 53.1°
A1
Part (d)
area = 10 cos θ × 10 sin θ = 100 × 0.8 × 0.6 for either value
M1
area = 48 cm2 in both cases
A1
Part (e)
sin θ + cos θ = 1.5 leads to 2 sin2θ − 3 sin θ + 1.25 = 0
M1
discriminant = 9 − 4 × 2 × 1.25 = −1
M1
−1 < 0, so there is no real value of sin θ and no such rectangle
A1
Exam tip: Two unknowns sin θ and cos θ need two equations. The perimeter gives one; sin²θ + cos²θ = 1 is always the other.
A seat on an observation wheel moves so that its height above the ground, h metres, t minutes after it passes its lowest point, is modelled by h = p − q cos(rt)°, where p, q and r are positive constants. Figure 1 shows the graph of h against t for 0 ≤ t ≤ 45, together with the line h = 15.
(a)
Use Figure 1 to find the values of p, q and r.
(3)
(b)
Find all the times in the interval 0 ≤ t ≤ 45 at which the seat is 15 m above the ground. Give your answers in minutes to 3 significant figures.
(5)
(c)
Find, in minutes to 3 significant figures, how long the seat spends more than 15 m above the ground during each revolution.
(2)
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Answer
Mark
Notes
Part (a)
p = 10, the mean of the greatest and least heights 18 and 2
B1
q = 8, half the difference
B1
one revolution takes 30 minutes, so 30r = 360 and r = 12
B1
Part (b)
10 − 8 cos(12t)° = 15, so cos(12t)° = −0.625 ft
M1
cos−1(−0.625) = 128.68°
A1
0 ≤ 12t ≤ 540, so 12t = 128.68°, 231.32°, 488.68°
M1
t = 10.7, 19.3
A1
t = 40.7
A1
Part (c)
19.276 − 10.724
M1
8.55 minutes
A1
Common error: Stopping after the first cycle. The interval 0 ≤ t ≤ 45 makes 12t run from 0° to 540°, one and a half cycles, and the figure shows three crossings of h = 15.
f(x) = 6cos2x − sin x − 5, 0 ≤ x < 2π, where x is in radians. Figure 1 shows the graph of y = f(x). The graph crosses the x-axis at A, B, C and D and has minimum points P(π ⁄ 2, −6) and Q(3π ⁄ 2, −4).
(a)
Show that f(x) = 1 − sin x − 6 sin2x.
(2)
(b)
Find the x-coordinates of A, B, C and D, giving exact answers where possible and other answers to 3 significant figures.
(5)
(c)
By completing the square on your answer to part (a), find the greatest value of f(x) and state the number of values of x in the interval at which it occurs.
(3)
(d)
Using Figure 1, state the set of values of k for which the equation f(x) = k has exactly two solutions in the interval 0 ≤ x < 2π.
(2)
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Answer
Mark
Notes
Part (a)
cos2x = 1 − sin2x substituted: 6(1 − sin2x) − sin x − 5
M1
= 1 − sin x − 6 sin2x with no errors seen
A1
Part (b)
6 sin2x + sin x − 1 = 0, so (3 sin x − 1)(2 sin x + 1) = 0
M1
sin x = ⅓ or sin x = −½
A1
sin−1(⅓) = 0.3398 and π − 0.3398
M1
A: x = 0.340, B: x = 2.80
A1
C: x = 7π ⁄ 6, D: x = 11π ⁄ 6
A1
Part (c)
f(x) = 25⁄24 − 6(sin x + 1⁄12)2
M1
greatest value 25⁄24
A1
it occurs when sin x = −1⁄12, which has 2 solutions in the interval
B1
Part (d)
−6 < k < −4
B1
k = 25⁄24 ft their greatest value
B1
Exam tip: A line y = k cuts the graph once for each solution. Slide an imaginary horizontal line down the figure and count crossings; the counts change only at turning-point heights.
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