Edexcel International A Level Maths · P3 · Pure Mathematics 3
Compound angles, double angles and R-form
28 free, high-quality past-paper-style questions on Compound angles, double angles and R-form 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 · 239 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked
1IALM-P3-4-B-01P3 · A2Easy5 marks
Do not use a calculator in this question. Show all your working.
(a)
By writing 165° as 120° + 45°, show that cos 165° = −(√6 + √2)/4.
(2)
(b)
Find the exact value of tan 165°, giving your answer in the form a + b√3, where a and b are integers.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
cos 120° cos 45° − sin 120° sin 45°
M1
(−½)(√2/2) − (√3/2)(√2/2) = −(√2 + √6)/4, as required
A1
Part (b)
(tan 120° + tan 45°)/(1 − tan 120° tan 45°) = (1 − √3)/(1 + √3)
M1
multiplies numerator and denominator by (1 − √3)
M1
(1 − √3)2/(1 − 3) = (4 − 2√3)/(−2) = −2 + √3
A1
Exam tip: tan 120° = −√3, not √3: 120° is in the second quadrant. Rationalise by multiplying top and bottom by the conjugate of the denominator.
Given that tan θ = −5/12 and 90° < θ < 180°, find the exact values of
(a)
sin 2θ,
(3)
(b)
cos 2θ,
(1)
(c)
tan 2θ.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
sin θ = 5/13 and cos θ = −12/13
B1
sin 2θ = 2 sin θ cos θ = 2(5/13)(−12/13)
M1
= −120/169
A1
Part (b)
2 cos2θ − 1 = 288/169 − 1 = 119/169 (or any correct form used)
B1
Part (c)
2(−5/12)/(1 − 25/144) or (−120/169) ÷ (119/169) = −120/119
B1
Exam tip: Double-angle formulae are not in the formulae booklet. Two quick checks: sin2 2θ + cos2 2θ should equal 1, and tan 2θ should equal sin 2θ ÷ cos 2θ.
Figure 1 shows two right-angled triangles, OPQ and OQR. OP = 4, PQ = 3 and QR = 12, with right angles at P and at Q. Angle POQ = α and angle QOR = β. The line PO is extended beyond O, and N is the point on this extension such that RN is perpendicular to it. Give exact answers.
(a)
Find the values of sin α, cos α, sin β and cos β.
(3)
(b)
Find the value of cos(α + β), and explain how your answer shows that angle POR is obtuse.
(2)
(c)
Find the length RN.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
OQ = √(42 + 32) = 5, so sin α = 3/5 and cos α = 4/5
B1
OR = √(52 + 122) = 13
B1
sin β = 12/13 and cos β = 5/13
B1
Part (b)
cos α cos β − sin α sin β = (4/5)(5/13) − (3/5)(12/13)
M1
= 20/65 − 36/65 = −16/65; angle POR = α + β and its cosine is negative, so it is obtuse
Exam tip: When two right-angled triangles share a side, find that shared side first: here OQ = 5 is the hypotenuse of one triangle and the base of the next. Then every ratio of α and β is a fraction you can write down.
Figure 1 shows the lines L1 and L2, which pass through the origin O. The line L1 passes through the point A and the line L2 passes through the point B, where A and B lie on gridline intersections. The lines L1 and L2 make angles β and α respectively with the positive x-axis, and θ is the acute angle between L1 and L2.
(a)
Use Figure 1 to write down the values of tan α and tan β.
(2)
(b)
Show that θ = 45°.
(3)
(c)
The line L3 is obtained by rotating L2 clockwise through 45° about O. Find the exact gradient of L3.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
B is (5, 3), so tan α = 3/5
B1
A is (1, 4), so tan β = 4
B1
Part (b)
θ = β − α, and tan(β − α) = (tan β − tan α)/(1 + tan β tan α)
M1
= (4 − 3/5)/(1 + 12/5) = (17/5)/(17/5) = 1
A1
θ is acute and tan θ = 1, so θ = 45°, as required
A1
Part (c)
gradient = tan(α − 45°) = (3/5 − 1)/(1 + 3/5)
M1
= (−2/5)/(8/5) = −1/4
A1
Exam tip: The gradient of a line through the origin is the tangent of the angle it makes with the positive x-axis. Rotating a line through an angle adds or subtracts that angle, so the tan(A ± B) formula gives the new gradient directly.
Figure 1 shows the graph of y = p sin x + q cos x for 0 ≤ x ≤ 360°, where p and q are positive constants. The point M is the maximum point of the curve in this interval. The expression p sin x + q cos x can be written in the form R sin(x + α), where R > 0 and 0 < α < 90°.
(a)
Use Figure 1 to write down the value of R and the value of α.
(2)
(b)
Find the exact values of p and q.
(3)
(c)
Solve p sin x + q cos x = −2 for 0 ≤ x < 360°.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
R = 4, the maximum value
B1
the maximum occurs where x + α = 90°, and x = 60°, so α = 30°
B1
Part (b)
4 sin(x + 30°) = 4 sin x cos 30° + 4 cos x sin 30°
M1
p = 4 cos 30° = 2√3
A1
q = 4 sin 30° = 2 (consistent with the y-intercept 2 in Figure 1)
A1
Part (c)
sin(x + 30°) = −½
M1
x + 30° = 210° or 330°, the values with 30° ≤ x + 30° < 390°
M1
x = 180° and x = 300°
A1
Exam tip: R sin(x + α) is greatest when x + α = 90°. Reading the x-coordinate of the maximum from a graph therefore gives α straight away, and the y-intercept R sin α is a quick check on your values.
y = sin 22.5° = 0.383 at P and y = sin 202.5° = −0.383 at Q
B1
Common error: Writing cos(x + 45°) as cos x + cos 45°. Use the compound-angle formula, then collect the sin x terms on one side and the cos x terms on the other before dividing by cos x.
f(x) = 3 sin x + 2 cos x, where x is measured in degrees.
(a)
Express f(x) in the form R sin(x + α), where R > 0 and 0 < α < 90°. Give the exact value of R and the value of α to 2 decimal places.
(3)
(b)
Write down the maximum value of f(x) and find the smallest positive value of x at which it occurs.
(2)
(c)
Solve f(x) = 1 for 0 ≤ x < 360°, giving your answers to 1 decimal place.
(4)
Show mark scheme
Answer
Mark
Notes
Part (a)
R cos α = 3 and R sin α = 2
M1
R = √13
A1
tan α = 2/3, α = 33.69°
A1
Part (b)
maximum value √13
B1
x + 33.69° = 90° ⇒ x = 56.31° (ft their α)
B1
Part (c)
sin(x + α) = 1/√13
M1
x + α = 163.90° or 376.10°, the values in 33.69° ≤ x + α < 393.69°
M1
x = 130.2°
A1
x = 342.4°
A1
Model sketch
Figure: f(x) = √13 sin(x + 33.69°) and the line y = 1
Exam tip: Shift the interval first: if 0 ≤ x < 360°, then x + α runs from α to 360° + α. The principal value 16.10° is below α, so it is not used, but 360° + 16.10° is.
An angle θ, measured in degrees, satisfies the equation sin(θ + 60°) = 3 sin θ.
(a)
Show that tan θ = √3/5.
(3)
(b)
Hence find the values of θ in the interval 0 ≤ θ < 360°, giving your answers to 1 decimal place.
(2)
(c)
Without finding θ, find the exact value of tan(θ + 60°).
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
sin θ cos 60° + cos θ sin 60° = ½ sin θ + (√3/2) cos θ
M1
(√3/2) cos θ = (5/2) sin θ
M1
tan θ = √3/5, as required
A1
Part (b)
θ = 19.1°
B1
θ = 199.1° (ft their first value + 180°)
B1
Part (c)
(tan θ + √3)/(1 − √3 tan θ) with tan θ = √3/5
M1
(6√3/5)/(1 − 3/5) = (6√3/5)/(2/5)
M1
= 3√3
A1
Common error: Dividing by cos θ before collecting terms, or losing the 3 on the right. Collect all sin θ terms on one side and all cos θ terms on the other, then divide.
Exam tip: The half-angle results come from cos θ = 1 − 2 sin²(½θ) = 2 cos²(½θ) − 1. The sign of the square root depends on the quadrant of ½θ, not of θ.
The temperature, T °C, inside a greenhouse t hours after midnight is modelled by T = 20 − 2.1 cos(πt/12) − 2 sin(πt/12), 0 ≤ t < 24, where the angle πt/12 is measured in radians.
(a)
Express 2.1 cos(πt/12) + 2 sin(πt/12) in the form R cos(πt/12 − α), where R > 0 and 0 < α < π/2. Give the value of α to 3 decimal places.
(3)
(b)
Find the maximum temperature predicted by the model and the value of t at which it occurs, giving t to 1 decimal place.
(3)
(c)
Find the values of t for which the model predicts a temperature of 21 °C, giving your answers to 1 decimal place.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
R cos α = 2.1 and R sin α = 2
M1
R = √(4.41 + 4) = 2.9
A1
tan α = 2/2.1, α = 0.761
A1
Part (b)
T = 20 − 2.9 cos(πt/12 − α), greatest when cos(πt/12 − α) = −1
M1
Tmax = 22.9 °C
A1
πt/12 − 0.761 = π ⇒ t = 14.9
A1
Part (c)
cos(πt/12 − α) = −1/2.9
M1
πt/12 − α = 1.923 or 2π − 1.923 = 4.360
M1
t = 10.3 and t = 19.6
A1
Model sketch
Figure: the temperature model; vertical axis starts at 16.5 °C
Exam tip: T = k − R cos(…) is greatest when the cosine is −1, not +1. Keep your calculator in radians throughout this question.
A rectangular panel ABCD has AB = 6 m and BC = 4 m. It is held in a vertical plane with corner A on horizontal ground, and AB makes an acute angle θ with the ground, as shown in Figure 1. The horizontal width of the panel, measured between the vertical lines through D and through B, is W m.
(a)
Show that W = 6 cos θ + 4 sin θ.
(2)
(b)
Express W in the form R cos(θ − α), where R > 0 and 0 < α < 90°. Give the exact value of R and the value of α to 2 decimal places.
(3)
(c)
Write down the greatest possible width of the panel and find the value of θ at which it occurs.
(2)
(d)
The panel has to fit exactly between two vertical walls that are 7 m apart. Find the two possible values of θ, giving your answers to 1 decimal place.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
horizontal extent of AB is 6 cos θ; AD makes angle 90° − θ with the ground, so its horizontal extent is 4 cos(90° − θ) = 4 sin θ
M1
D is to the left of A and B to the right, so W = 6 cos θ + 4 sin θ, as required
A1
Part (b)
R cos α = 6 and R sin α = 4
M1
R = √52 = 2√13
A1
tan α = 4/6, α = 33.69°
A1
Part (c)
Wmax = 2√13 m (about 7.21 m)
B1
when θ − α = 0, so θ = 33.69° (ft their α)
B1
Part (d)
cos(θ − 33.69°) = 7/(2√13) = 0.9707…
M1
θ − 33.69° = ±13.90°
M1
θ = 47.6° or θ = 19.8°
A1
Exam tip: cos(θ − α) = k has two solutions symmetric about α: θ = α ± cos−1 k. Check both lie in the stated range (here 0° to 90°) before giving them.
Figure 1 shows the curves y = cos 2x and y = sin x for 0 ≤ x ≤ 2π, where x is in radians. The curves meet at A, B and C only.
(a)
Show that the x-coordinates of A, B and C satisfy 2 sin2x + sin x − 1 = 0.
(2)
(b)
Hence find the exact x-coordinates of A, B and C.
(3)
(c)
Explain why the curves touch at C rather than cross.
(1)
(d)
Use Figure 1 and your answers to part (b) to solve the inequality cos 2x > sin x for 0 ≤ x ≤ 2π.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
uses cos 2x = 1 − 2 sin2x
M1
1 − 2 sin2x = sin x ⇒ 2 sin2x + sin x − 1 = 0, as required
A1
Part (b)
(2 sin x − 1)(sin x + 1) = 0, so sin x = ½ or sin x = −1
M1
A: x = π/6 and B: x = 5π/6
A1
C: x = 3π/2
A1
Part (c)
cos 2x − sin x = (1 − 2 sin x)(1 + sin x); near x = 3π/2, 1 − 2 sin x > 0 and 1 + sin x ≥ 0, so cos 2x ≥ sin x on both sides of C: the curves meet at C but do not cross (accept: both curves have their least value −1 at C, and cos 2x > sin x just either side of C)
B1
Part (d)
0 ≤ x < π/6
B1
5π/6 < x < 3π/2 and 3π/2 < x ≤ 2π (accept 5π/6 < x ≤ 2π, x ≠ 3π/2)
B1
Common error: Including x = 3π/2 in the solution of the strict inequality. At C the curves are equal, so cos 2x > sin x is false there, even though cos 2x is above sin x on both sides of C.
Figure 1 shows the triangle ABC, in which AB = 5 cm and AC = 7 cm. Angle ABC = 2θ and angle ACB = θ.
(a)
Use the sine rule to show that cos θ = 7/10.
(3)
(b)
Find the exact value of cos 2θ, and hence state whether angle ABC is acute or obtuse.
(2)
(c)
Find the length BC.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
7/sin 2θ = 5/sin θ
M1
7 sin θ = 5(2 sin θ cos θ) = 10 sin θ cos θ
M1
sin θ ≠ 0, so cos θ = 7/10, as required
A1
Part (b)
cos 2θ = 2(7/10)2 − 1
M1
= −1/50, which is negative, so angle ABC is obtuse
A1
Part (c)
cosine rule at B: 72 = 52 + BC2 − 2(5)(BC) cos 2θ
M1
49 = 25 + BC2 + BC/5 ⇒ 5BC2 + BC − 120 = 0
M1
(5BC + 25)(BC − 4.8) = 0, BC > 0, so BC = 4.8 cm
A1
Exam tip: When one angle is double another, the sine rule produces sin 2θ next to sin θ. Expanding sin 2θ = 2 sin θ cos θ lets you cancel sin θ, which is never zero in a triangle.
Figure 1 shows the graph of y = f(x) for 0 ≤ x ≤ π, where f(x) = p sin2x + q, p and q are constants and x is in radians. The curve has minimum points A and C and a maximum point B.
(a)
Use Figure 1 to find the value of p and the value of q.
(2)
(b)
Show that f(x) can be written as 2 − 3 cos 2x.
(2)
(c)
Solve f(x) = 3.5 for 0 ≤ x ≤ π, giving exact answers.
(3)
(d)
Find the exact value of f(π/8).
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
A(0, −1) gives q = −1
B1
B(π/2, 5) gives p + q = 5, so p = 6
B1
Part (b)
uses sin2x = ½(1 − cos 2x)
M1
6 × ½(1 − cos 2x) − 1 = 3 − 3 cos 2x − 1 = 2 − 3 cos 2x, as required
A1
Part (c)
2 − 3 cos 2x = 3.5 ⇒ cos 2x = −½
M1
2x = 2π/3 or 4π/3, with 0 ≤ 2x ≤ 2π
M1
x = π/3 and x = 2π/3
A1
Part (d)
2 − 3 cos(π/4) = 2 − 3√2/2
B1
Exam tip: sin²x and cos²x cannot be put straight into an R-form or solved as a single cosine. Rewrite them with cos 2x first; the curve then has period π, not 2π, as Figure 1 shows.
Figure 1 shows a circle with centre O and radius 5 cm. The points A, B and C lie on the circle, with chord AB = 8 cm and chord BC = 6 cm. Angle AOB = θ and angle BOC = φ. Figure 1 is not drawn to scale. Give exact answers.
(a)
By considering the perpendicular from O to AB, show that sin(½θ) = 4/5, and write down cos(½θ).
(2)
(b)
Hence find the values of sin θ and cos θ.
(2)
(c)
Find the values of sin φ and cos φ.
(2)
(d)
Find cos(θ + φ), and hence show that AC is a diameter of the circle.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
the perpendicular bisects AB and angle AOB, so sin(½θ) = 4/5, as required
B1
cos(½θ) = 3/5
B1
Part (b)
sin θ = 2 sin(½θ) cos(½θ) = 2(4/5)(3/5) = 24/25
B1
cos θ = 1 − 2 sin2(½θ) = 1 − 32/25 = −7/25
B1
Part (c)
sin(½φ) = 3/5 and cos(½φ) = 4/5
M1
sin φ = 24/25 and cos φ = 1 − 2(9/25) = 7/25
A1
Part (d)
cos θ cos φ − sin θ sin φ = (−7/25)(7/25) − (24/25)(24/25)
M1
= −625/625 = −1, so θ + φ = 180° and A, O, C are collinear: AC is a diameter
A1
Exam tip: A chord of length c in a circle of radius r subtends an angle θ at the centre with sin(½θ) = c/(2r). The double-angle formulae then turn the half-angle ratios into sin θ and cos θ without a calculator.
Figure 1 shows the graph of y = √3 sin 2x − cos 2x for 0 ≤ x ≤ π, where x is in radians. The curve has a maximum point P and a minimum point Q.
(a)
Express √3 sin 2x − cos 2x in the form R sin(2x − α), where R > 0 and 0 < α < π/2. Give the exact values of R and α.
(3)
(b)
Find the exact coordinates of P and of Q.
(2)
(c)
Solve √3 sin 2x − cos 2x = 1 for 0 ≤ x ≤ π, giving exact answers.
(3)
(d)
Use Figure 1 to find the set of values of k for which the equation √3 sin 2x − cos 2x = k has exactly two solutions in 0 ≤ x ≤ π.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
R cos α = √3 and R sin α = 1
M1
R = 2
A1
tan α = 1/√3, α = π/6
A1
Part (b)
2x − π/6 = π/2 ⇒ P(π/3, 2)
B1
2x − π/6 = 3π/2 ⇒ Q(5π/6, −2)
B1
Part (c)
sin(2x − π/6) = ½
M1
2x − π/6 = π/6 or 5π/6, the values with −π/6 ≤ 2x − π/6 ≤ 11π/6
M1
x = π/6 and x = π/2
A1
Part (d)
−1 < k < 2
B1
−2 < k < −1
B1
Exam tip: For a 'how many solutions' question, draw horizontal lines y = k across the graph. The end-points of the interval matter: here both ends are at y = −1, so k = −1 gives three solutions, not two.
f(x) = 7 cos x − 24 sin x, where x is measured in degrees.
(a)
Express f(x) in the form R cos(x + α), where R > 0 and 0 < α < 90°. Give the value of α to 2 decimal places.
(3)
(b)
Solve f(x) = 10 for 0 ≤ x < 360°, giving your answers to 1 decimal place.
(4)
(c)
The function g is defined by g(x) = 30/(f(x) + 40). Find the greatest value of g(x) and the smallest positive value of x at which it occurs.
(3)
(d)
State the least value of g(x).
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
R cos α = 7 and R sin α = 24
M1
R = 25
A1
α = 73.74°
A1
Part (b)
cos(x + 73.74°) = 0.4
M1
x + α = 293.58° or 426.42°, the values in 73.74° ≤ x + α < 433.74°
M1
x = 219.8°
A1
x = 352.7°
A1
Part (c)
g is greatest when f(x) is least, f(x) = −25
M1
gmax = 30/15 = 2
A1
x + 73.74° = 180° ⇒ x = 106.26°
A1
Part (d)
f = 25 gives g = 30/65 = 6/13
B1
Model sketch
Figure: f(x) = 25 cos(x + 73.74°) and the line y = 10
Common error: Taking the greatest value of g where f is greatest. With a positive numerator, the fraction is largest when the denominator f(x) + 40 is smallest.
Exam tip: Use the forms sin²x = ½(1 − cos 2x) and 2 sin x cos x = sin 2x to turn squares and products into first powers of double angles; the R-form then applies.
Figure 1 shows a semicircle with centre O and radius 5 cm. The rectangle PQRS has its side PQ on the diameter, with OP = OQ, and its vertices R and S on the arc. Angle QOR = θ, where 0 < θ < 90°.
(a)
Show that the area of the rectangle is 25 sin 2θ cm2.
(2)
(b)
Write down the greatest possible area of the rectangle and the value of θ for which it occurs.
(2)
(c)
Show that the perimeter, L cm, of the rectangle is given by L = 20 cos θ + 10 sin θ, and express L in the form R cos(θ − α), where R > 0 and 0 < α < 90°. Give the exact value of R and the value of α to 2 decimal places.
(4)
(d)
Find the greatest possible perimeter of the rectangle and the value of θ for which it occurs.
(2)
(e)
Given that L = 20, show that tan θ = 4/3 and hence find the exact area of the rectangle.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
PQ = 2 × 5 cos θ = 10 cos θ and QR = 5 sin θ
M1
area = 50 sin θ cos θ = 25(2 sin θ cos θ) = 25 sin 2θ, as required
A1
Part (b)
25 cm2
B1
when 2θ = 90°, θ = 45°
B1
Part (c)
L = 2(10 cos θ) + 2(5 sin θ) = 20 cos θ + 10 sin θ
B1
R cos α = 20 and R sin α = 10
M1
R = √500 = 10√5
A1
tan α = ½, α = 26.57°
A1
Part (d)
Lmax = 10√5 cm (about 22.4 cm)
B1
θ = α = 26.57° (ft their α)
B1
Part (e)
cos(θ − α) = 20/(10√5) = cos α, so θ − α = ±α; θ = 0 is not allowed, so θ = 2α
M1
tan θ = tan 2α = (2 × ½)/(1 − ¼) = 4/3
A1
sin 2θ = 2(4/5)(3/5) = 24/25, so area = 25 × 24/25 = 24 cm2
A1
Exam tip: In part (e) the equation gives θ − α = ±α. One root is θ = 0, which is outside the stated range and gives a flat rectangle, so it is rejected; say so in your working.
Figure 1 shows the graph of y = cos 3x ÷ cos x for 0 < x < π, where x is in radians. The open circles show points that are not on the graph; one of them is labelled P.
(a)
By writing 3x as 2x + x, prove that, for cos x ≠ 0, cos 3x ÷ cos x ≡ 2 cos 2x − 1.
(5)
(b)
Explain why there is an open circle at P, and find the coordinates of P.
(2)
(c)
Solve cos 3x ÷ cos x = 0 for 0 < x < π, giving exact answers.
(2)
(d)
Find the set of values of k for which the equation cos 3x ÷ cos x = k has no solutions in 0 < x < π.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
cos 3x = cos 2x cos x − sin 2x sin x
M1
sin 2x sin x = 2 sin2x cos x
M1
dividing by cos x: cos 3x ÷ cos x = cos 2x − 2 sin2x
M1
uses 2 sin2x = 1 − cos 2x
M1
cos 2x − 1 + cos 2x = 2 cos 2x − 1, as required
A1
Part (b)
at x = π/2, cos x = 0, so cos 3x ÷ cos x is undefined there
B1
P is (π/2, −3), since 2 cos π − 1 = −3
B1
Part (c)
2 cos 2x − 1 = 0 ⇒ cos 2x = ½
M1
x = π/6 and x = 5π/6
A1
Part (d)
k ≥ 1, since y = 1 only at the excluded end-points x = 0 and x = π
B1
k ≤ −3, since y = −3 only at the excluded point P
B1
Common error: Stating the range as −3 ≤ y ≤ 1. The identity holds only where cos x ≠ 0, and the interval is open, so the values −3 and 1 are never reached. The open circles in Figure 1 show this.
Two signals are modelled by y1 = 3 sin t and y2 = 2 sin(t + c), where 0 < c < π and t is in radians. Figure 1 shows both graphs for 0 ≤ t ≤ 2π. The point M, on gridlines, is the first maximum point of y2. The combined signal is y = y1 + y2.
(a)
Use Figure 1 to show that c = π/3.
(2)
(b)
Show that y = 4 sin t + √3 cos t.
(2)
(c)
Express y in the form R sin(t + α), where R > 0 and 0 < α < π/2. Give the exact value of R and the value of α to 3 decimal places.
(3)
(d)
Write down the greatest value of y and find the smallest positive value of t at which it occurs, to 3 significant figures.
(2)
(e)
Solve y = 2 for 0 ≤ t ≤ 2π, giving your answers to 3 significant figures.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
M is (π/6, 2), and the maximum of y2 is where t + c = π/2
B1
c = π/2 − π/6 = π/3, as required
B1
Part (b)
2 sin(t + π/3) = 2 sin t cos(π/3) + 2 cos t sin(π/3) = sin t + √3 cos t
M1
y = 3 sin t + sin t + √3 cos t = 4 sin t + √3 cos t, as required
A1
Part (c)
R cos α = 4 and R sin α = √3
M1
R = √19
A1
tan α = √3/4, α = 0.409
A1
Part (d)
ymax = √19 (about 4.36)
B1
t + 0.4086 = π/2 ⇒ t = 1.16
B1
Part (e)
sin(t + α) = 2/√19, principal value 0.4767
M1
t + α = 0.4767 or π − 0.4767 = 2.6649, the values in 0.4086 ≤ t + α ≤ 2π + 0.4086
M1
t = 0.0680 and t = 2.26
A1
Exam tip: Keep α to at least 4 decimal places in part (e). Subtracting a rounded α from a small angle such as 0.4767 can change the third significant figure of the answer.
Figure 1 shows a screen fixed to a vertical wall. The bottom of the screen, B, is 1 m above the point F, and the top, T, is 3 m above B. A viewer's eye, E, is level with F and at a horizontal distance of x m from the wall. The angle BET = θ, where θ is acute.
(a)
Show that tan θ = 3x/(x2 + 4).
(3)
(b)
Find the values of x for which tan θ = 3/5.
(3)
(c)
By considering (x − 2)2, show that tan θ ≤ 3/4 for all x > 0. Hence find the value of x for which the angle θ is greatest, and the greatest value of θ to 1 decimal place.
(4)
Show mark scheme
Answer
Mark
Notes
Part (a)
angle FEB = α with tan α = 1/x, and angle FET = β with tan β = 4/x
B1
θ = β − α, tan θ = (4/x − 1/x)/(1 + 4/x2)
M1
multiplying through by x2: tan θ = 3x/(x2 + 4), as required
A1
Part (b)
3x/(x2 + 4) = 3/5 ⇒ 15x = 3x2 + 12
M1
x2 − 5x + 4 = 0 ⇒ (x − 1)(x − 4) = 0
M1
x = 1 or x = 4
A1
Part (c)
(x − 2)2 ≥ 0 ⇒ x2 + 4 ≥ 4x
M1
so 3x/(x2 + 4) ≤ 3x/(4x) = 3/4
A1
equality when x = 2; tan is increasing for acute angles, so θ is greatest there
B1
θmax = tan−1(3/4) = 36.9°
B1
Exam tip: Writing the angle as a difference of two angles of elevation turns a geometry problem into the tan(A − B) formula. Multiply the numerator and denominator by x² to clear the fractions inside the fraction.
We use essential cookies to make this site work. With your permission we'd also use analytics and advertising cookies to understand which resources help students most and how families find us. Nothing optional is set unless you accept. Privacy policy
EssentialNeeded for the site and your cookie choice to work.
Always on
AnalyticsAnonymous visit statistics that help improve the free resources.
MarketingMeasures whether ads reach the right families. Off unless you turn it on.