Browse Trigonometry

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The sine and cosine rules

29 questions

LessonNot started
The points AA, BB and CC have coordinates (1,0,2)(1, 0, 2), (5,8,−2)(5, 8, -2) and (7,6,k)(7, 6, k) respectively, where kk is a constant.
(a) Given that angle ACBACB is a right angle, find the possible values of kk. (3 marks)
(b) Given also that k>0k > 0, find the exact shortest distance from CC to the line ABAB. (3 marks)
(c) The point DD is the point on ABAB that is closest to CC. Find the ratio AD:DBAD : DB in its simplest form. (2 marks)
●●●●●Level 58 marksStart
Relative to a fixed origin OO, the points AA, BB and CC have position vectors

a=−i+j+4k\mathbf{a} = -\mathbf{i} + \mathbf{j} + 4\mathbf{k}, b=5i+4j−2k\qquad \mathbf{b} = 5\mathbf{i} + 4\mathbf{j} - 2\mathbf{k}, c=5i+7j+4k\qquad \mathbf{c} = 5\mathbf{i} + 7\mathbf{j} + 4\mathbf{k}.
(a) Show that angle BAC=π4BAC = \dfrac{\pi}{4}. (4 marks)

The point PP lies on the line segment ABAB and is such that angle APC=π2APC = \dfrac{\pi}{2}.
(b) Find the position vector of PP. (3 marks)
(c) Find the area of triangle ABCABC. (2 marks)
●●●●●Level 59 marksStart
hxyPQRNThe diagram shows triangle PQRPQR. The perpendicular from RR to PQPQ meets PQPQ at NN. Angle PRN=xPRN = x, angle NRQ=yNRQ = y and the length RN=hRN = h.
(a) (i) Find the length PRPR in terms of hh and xx.

(ii) Write down the length QRQR in terms of hh and yy.

(iii) By finding the area of triangle PQRPQR in two different ways, show that sin⁡(x+y)≡sin⁡xcos⁡y+cos⁡xsin⁡y\sin(x + y) \equiv \sin x\cos y + \cos x\sin y.

(6 marks)
(b) Given that sin⁡(α+60∘)=cos⁡(α−45∘)\sin(\alpha + 60^\circ) = \cos(\alpha - 45^\circ), show that tan⁡α=6+3−2−2\tan\alpha = \sqrt6 + \sqrt3 - \sqrt2 - 2.

(5 marks)
●●●●●Level 511 marksStart
A stage designer is planning a raised platform for an outdoor concert.

The plan view of the platform is modelled as the region made up of the sector OABOAB of a circle with centre OO and radius rr metres, together with the triangles OACOAC and OBDOBD, which are joined to the sector along OAOA and along OBOB, where
  • angle AOB=1.5AOB = 1.5 radians
  • the arc ABAB has length 1212 m
  • triangle OACOAC and triangle OBDOBD are congruent, with the sector OABOAB lying between them
  • CC, OO and DD lie on a straight line, which forms the back edge of the platform
(a) Show that r=8r = 8.

(2 marks)
(b) Show that the size of angle AOCAOC is 0.8210.821 radians, correct to 3 decimal places.

(2 marks)

The front edge of the platform runs in a straight line from CC to AA, then along the arc ABAB, then in a straight line from BB to DD.

Given that the total length of the front edge of the platform is 3232 m,
(c) find the total area of the plan view of the platform, giving your answer to the nearest square metre.

(6 marks)
●●●●●Level 510 marksStart
OR4.2 cm7.5 cm0.45 radPQFigure 1

Diagram NOT accurately drawn

Figure 1 shows a sketch of a design for a ceramic tile.

The tile is made from a triangle OPQOPQ and a sector OQROQR of a circle with centre OO, joined along OQOQ.

In the design
  • OQ=7.5OQ = 7.5 cm
  • PQ=4.2PQ = 4.2 cm
  • angle POQ=0.45POQ = 0.45 radians
  • angle OPQ>π2OPQ > \dfrac{\pi}{2} radians
Making your method clear,
(a) find the size of angle OPQOPQ, giving your answer in radians to 4 significant figures,

(3 marks)
(b) find the area of triangle OPQOPQ, in cm2^2, giving your answer to 3 significant figures.

(2 marks)

Given that the area of sector OQROQR and the area of triangle OPQOPQ are in the ratio 5:35 : 3
(c) show that angle QORQOR is 0.3970.397 radians to 3 significant figures.

(3 marks)
(d) Hence find the perimeter of the tile, in cm, to the nearest integer.

(5 marks)
●●●●●Level 513 marksStart

Exact values of trigonometric ratios

24 questions

LessonNot started
The curve CC has parametric equations

x=6sin⁡t,y=2sin⁡2t,0⩽t⩽π2x = 6\sin t, \qquad y = 2\sin 2t, \qquad 0 \leqslant t \leqslant \dfrac{\pi}{2}
(a) The point PP on CC is the point at which the tangent to CC is parallel to the xx-axis.

Find the exact coordinates of PP. (4 marks)
(b) The finite region RR is bounded by CC and the xx-axis.

Use integration to find the exact area of RR. (5 marks)
●●●●●Level 59 marksStart
The curve CC has parametric equations

x=3sin⁡θx = 3\sin\theta, y=sin⁡2θ\qquad y = \sin 2\theta, 0≤θ≤π2\qquad 0 \le \theta \le \dfrac{\pi}{2}.
(a) Find the exact coordinates of the maximum point MM of CC. (4 marks)

The finite region RR is bounded by CC and the xx-axis.
(b) Find the exact area of RR. (3 marks)
(c) Show that the line through MM parallel to the yy-axis divides RR into two regions whose areas are in the ratio (22−1):1\left(2\sqrt{2} - 1\right) : 1. (3 marks)
●●●●●Level 510 marksStart
In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.

Given that the first three terms of a geometric series are

22cos⁡θ7+8sin⁡θ11tan⁡θ22\cos\theta \qquad\qquad 7 + 8\sin\theta \qquad\qquad 11\tan\theta
(a) show that

64sin⁡2θ−130sin⁡θ+49=064\sin^2\theta - 130\sin\theta + 49 = 0

(3 marks)

Given that θ\theta is an obtuse angle measured in radians,
(b) solve the equation in part (a) to find the exact value of θ\theta

(2 marks)
(c) Find the sum to infinity of the series, giving your answer in the form a+b3a + b\sqrt{3}, where aa and bb are rational constants to be found.

(5 marks)
●●●●●Level 510 marksStart
hxyPQRNThe diagram shows triangle PQRPQR. The perpendicular from RR to PQPQ meets PQPQ at NN. Angle PRN=xPRN = x, angle NRQ=yNRQ = y and the length RN=hRN = h.
(a) (i) Find the length PRPR in terms of hh and xx.

(ii) Write down the length QRQR in terms of hh and yy.

(iii) By finding the area of triangle PQRPQR in two different ways, show that sin⁡(x+y)≡sin⁡xcos⁡y+cos⁡xsin⁡y\sin(x + y) \equiv \sin x\cos y + \cos x\sin y.

(6 marks)
(b) Given that sin⁡(α+60∘)=cos⁡(α−45∘)\sin(\alpha + 60^\circ) = \cos(\alpha - 45^\circ), show that tan⁡α=6+3−2−2\tan\alpha = \sqrt6 + \sqrt3 - \sqrt2 - 2.

(5 marks)
●●●●●Level 511 marksStart

Graphs of sine, cosine and tangent

31 questions

LessonNot started
(a) Given that

7cos⁡θ−24sin⁡θ≡Kcos⁡(θ+α)7\cos\theta - 24\sin\theta \equiv K\cos(\theta + \alpha)

where K>0K > 0 and 0<α<90∘0 < \alpha < 90^\circ, find the value of KK and the value of α\alpha, giving α\alpha to 2 decimal places.

(3 marks)

An ecologist records the number of mice and the number of owls living on a farm over one year.

The number of mice, MM, is modelled by

M=A+7cos⁡(30t)∘−24sin⁡(30t)∘M = A + 7\cos(30t)^\circ - 24\sin(30t)^\circ

where tt is the number of months after the start of the year and AA is a constant.

According to the model, the greatest number of mice on the farm during the year is 145145
(b) (i) Find a complete equation for the model.

(ii) Hence state the least number of mice on the farm during the year, according to the model.

(2 marks)

The ecologist finds that the number of mice on the farm is actually lowest in October.
(c) Use this information to comment on the model for the number of mice.

(2 marks)

The number of owls, NN, on the farm during the same year is modelled by

N=40+15sin⁡(30t+60)∘N = 40 + 15\sin(30t + 60)^\circ

The number of owls is at its lowest when t=Tt = T
(d) Use the two models to find the number of mice on the farm when t=Tt = T

(4 marks)
●●●●●Level 511 marksStart
(a) Given that

8sin⁡θ+6cos⁡θ≡Rsin⁡(θ+α)8\sin\theta + 6\cos\theta \equiv R\sin(\theta + \alpha)

where R>0R > 0 and 0<α<π20 < \alpha < \dfrac{\pi}{2}, find the exact value of RR and find α\alpha, in radians, correct to 3 decimal places.

(3 marks)

A scientist measures the temperature of the water in two garden ponds throughout one year.

For the first pond, the water temperature, S ∘S\ ^\circC, is modelled by

S=A−8sin⁡(πt6)−6cos⁡(πt6)S = A - 8\sin\left(\frac{\pi t}{6}\right) - 6\cos\left(\frac{\pi t}{6}\right)

where tt is the number of months after the start of the year and AA is a constant.

According to the model, the highest water temperature in the first pond during the year is 23 ∘23\ ^\circC
(b) (i) Find a complete equation for the model.

(ii) Hence state the lowest water temperature in the first pond during the year, according to the model.

(2 marks)

The scientist observes that the water temperature in the first pond is actually at its lowest towards the end of February.
(c) Use this information to comment on the model for the first pond.

(2 marks)

For the second pond, the water temperature, P ∘P\ ^\circC, during the same year is modelled by

P=12+9cos⁡(πt6−2.3)P = 12 + 9\cos\left(\frac{\pi t}{6} - 2.3\right)

The water temperature in the second pond is at its highest when t=Tt = T
(d) Use the two models to find the water temperature in the first pond when t=Tt = T, giving your answer to 3 significant figures.

(4 marks)
●●●●●Level 511 marksStart
The depth of water in a harbour, dd metres, is modelled by

d=5+2sin⁡(30t)∘d = 5 + 2\sin(30t)^{\circ}

where tt is the time in hours after midnight.
(a) Write down the maximum depth, and find the first time at which it occurs. (2 marks)
(b) Find the first time at which the depth is 66 metres. (2 marks)
(c) State the period of the model, and interpret it. (1 mark)
●●●●●Level 55 marksStart
In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.
(a) Solve, for 0<x<π0 < x < \pi, the equation

3cos⁡2x+5sin⁡x=43\cos^2 x + 5\sin x = 4

giving your answers in radians to 3 significant figures.

(5 marks)

The relative humidity inside a museum gallery is monitored on one particular day.

The relative humidity, H%H\%, inside the gallery, tt hours after midnight, is modelled by the equation

H=55+8sin⁡(kt−40)∘0⩽t<24H = 55 + 8\sin(kt - 40)^\circ \qquad 0 \leqslant t < 24

where kk is a constant.

Use the equation of the model to answer parts (b) to (d).

Given that
  • the relative humidity inside the gallery was 60%60\% at 5 am
  • 0<k<900 < k < 90
(b) find all possible values for kk, giving each answer to 2 decimal places.

(4 marks)

Given further that 0<k<200 < k < 20
(c) find the maximum relative humidity inside the gallery,

(1 mark)
(d) find the time of day at which this maximum relative humidity occurs. Give your answer to the nearest minute.

(2 marks)
●●●●●Level 512 marksStart
(a) Express 6cos⁡2x+5sin⁡2x6\cos 2x + 5\sin 2x in the form Rcos⁡(2x−α)R\cos(2x - \alpha), where R>0R > 0 and 0<α<12π0 < \alpha < \tfrac12\pi. Give the exact value of RR and give the value of α\alpha correct to 4 significant figures.

(3 marks)
(b) Give full details of a sequence of three transformations needed to transform the curve y=cos⁡xy = \cos x to the curve y=6cos⁡2x+5sin⁡2xy = 6\cos 2x + 5\sin 2x.

(4 marks)
(c) Determine the greatest value of 6cos⁡2x+5sin⁡2x6\cos 2x + 5\sin 2x as xx varies and give the smallest positive value of xx for which it occurs.

(2 marks)
(d) Determine the least value of 6cos⁡2x+5sin⁡2x6\cos 2x + 5\sin 2x as xx varies and give the smallest positive value of xx for which it occurs.

(2 marks)
●●●●●Level 511 marksStart

Trigonometric identities

19 questions

LessonNot started
(a) Given that sin⁡θ≠±1\sin\theta \ne \pm 1, prove the identity

32−2sin⁡θ+32+2sin⁡θ≡3sec⁡2θ\frac{3}{2 - 2\sin\theta} + \frac{3}{2 + 2\sin\theta} \equiv 3\sec^2\theta

(4 marks)
(b) Hence, find the set of values of AA for which the equation

32−2sin⁡θ+32+2sin⁡θ=A\frac{3}{2 - 2\sin\theta} + \frac{3}{2 + 2\sin\theta} = A

has real solutions.

Fully justify your answer.

(3 marks)
(c) Given that θ\theta is obtuse and

32−2sin⁡θ+32+2sin⁡θ=10\frac{3}{2 - 2\sin\theta} + \frac{3}{2 + 2\sin\theta} = 10

find the exact value of tan⁡θ\tan\theta

(3 marks)
●●●●●Level 510 marksStart
In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.

Given that the first three terms of a geometric series are

22cos⁡θ7+8sin⁡θ11tan⁡θ22\cos\theta \qquad\qquad 7 + 8\sin\theta \qquad\qquad 11\tan\theta
(a) show that

64sin⁡2θ−130sin⁡θ+49=064\sin^2\theta - 130\sin\theta + 49 = 0

(3 marks)

Given that θ\theta is an obtuse angle measured in radians,
(b) solve the equation in part (a) to find the exact value of θ\theta

(2 marks)
(c) Find the sum to infinity of the series, giving your answer in the form a+b3a + b\sqrt{3}, where aa and bb are rational constants to be found.

(5 marks)
●●●●●Level 510 marksStart
In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.
(a) Solve, for 0<x<π0 < x < \pi, the equation

3cos⁡2x+5sin⁡x=43\cos^2 x + 5\sin x = 4

giving your answers in radians to 3 significant figures.

(5 marks)

The relative humidity inside a museum gallery is monitored on one particular day.

The relative humidity, H%H\%, inside the gallery, tt hours after midnight, is modelled by the equation

H=55+8sin⁡(kt−40)∘0⩽t<24H = 55 + 8\sin(kt - 40)^\circ \qquad 0 \leqslant t < 24

where kk is a constant.

Use the equation of the model to answer parts (b) to (d).

Given that
  • the relative humidity inside the gallery was 60%60\% at 5 am
  • 0<k<900 < k < 90
(b) find all possible values for kk, giving each answer to 2 decimal places.

(4 marks)

Given further that 0<k<200 < k < 20
(c) find the maximum relative humidity inside the gallery,

(1 mark)
(d) find the time of day at which this maximum relative humidity occurs. Give your answer to the nearest minute.

(2 marks)
●●●●●Level 512 marksStart

Solving trigonometric equations

57 questions

LessonNot started
(a) Prove that

1−cos⁡2θ+sin⁡2θ1+cos⁡2θ+sin⁡2θ≡tan⁡θ\dfrac{1 - \cos 2\theta + \sin 2\theta}{1 + \cos 2\theta + \sin 2\theta} \equiv \tan\theta (4 marks)
(b) Hence solve the equation

1−cos⁡4x+sin⁡4x1+cos⁡4x+sin⁡4x=3tan⁡x\dfrac{1 - \cos 4x + \sin 4x}{1 + \cos 4x + \sin 4x} = 3\tan x

for 0≤x<π0 \le x < \pi, giving your answers in exact form. (5 marks)
●●●●●Level 59 marksStart
(a) Express 2cos⁡x−23sin⁡x2\cos x - 2\sqrt{3}\sin x in the form Rcos⁡(x+α)R\cos(x + \alpha), where R>0R > 0 and 0<α<π20 < \alpha < \dfrac{\pi}{2}. (3 marks)

The depth of water, hh metres, in a harbour tt hours after midnight is modelled by

h=5+2cos⁡(πt6)−23sin⁡(πt6),0≤t≤24.h = 5 + 2\cos\left(\dfrac{\pi t}{6}\right) - 2\sqrt{3}\sin\left(\dfrac{\pi t}{6}\right), \qquad 0 \le t \le 24.

A boat can enter the harbour only when the depth of water is at least 77 metres.
(b) Find the total length of time, during these 2424 hours, for which the boat is able to enter the harbour. Fully justify your answer. (6 marks)
●●●●●Level 59 marksStart
(a) Prove that

sin⁡3x≡3sin⁡x−4sin⁡3x\sin 3x \equiv 3\sin x - 4\sin^3 x. (4 marks)
(b) Hence solve, for 0≤x<2π0 \le x < 2\pi, the equation

sin⁡3x=2sin⁡x\sin 3x = 2\sin x,

giving your answers in exact form. (3 marks)
(c) Use the identity in part (a) to find the exact value of

∫0π3sin⁡3x dx\displaystyle\int_0^{\frac{\pi}{3}} \sin^3 x\,\mathrm{d}x. (2 marks)
●●●●●Level 59 marksStart
(a) Prove that

cos⁡3θ≡4cos⁡3θ−3cos⁡θ\cos 3\theta \equiv 4\cos^3\theta - 3\cos\theta

(4 marks)
(b) Hence solve, for 0⩽θ<π0 \leqslant \theta < \pi, the equation

8cos⁡3θ−6cos⁡θ+3=08\cos^3\theta - 6\cos\theta + \sqrt{3} = 0

giving your answers in exact form in terms of π\pi. (5 marks)
●●●●●Level 59 marksStart
The depth of water, DD metres, at the entrance to a harbour is modelled by

D=9+3sin⁡(πt6)−33cos⁡(πt6)D = 9 + 3\sin\left(\dfrac{\pi t}{6}\right) - 3\sqrt{3}\cos\left(\dfrac{\pi t}{6}\right), 0≤t<24\qquad 0 \le t < 24,

where tt is the time in hours after midnight.
(a) Express 3sin⁡θ−33cos⁡θ3\sin\theta - 3\sqrt{3}\cos\theta in the form Rsin⁡(θ−α)R\sin(\theta - \alpha), where R>0R > 0 and 0<α<π20 < \alpha < \dfrac{\pi}{2}, stating the exact values of RR and α\alpha. (3 marks)
(b) Find the maximum depth of water at the harbour entrance, and the first time after midnight at which this maximum depth occurs. (2 marks)

A ship can enter the harbour only when the depth of water is at least 1212 metres.
(c) Find the times during the day between which the ship can enter the harbour. (4 marks)
●●●●●Level 59 marksStart
The depth of water, DD metres, in a harbour at a time tt hours after midnight is modelled by

D=8+3cos⁡(0.5t)−4sin⁡(0.5t),0⩽t⩽24D = 8 + 3\cos(0.5t) - 4\sin(0.5t), \qquad 0 \leqslant t \leqslant 24

where the angle 0.5t0.5t is measured in radians.
(a) Express 3cos⁡θ−4sin⁡θ3\cos\theta - 4\sin\theta in the form Rcos⁡(θ+α)R\cos(\theta + \alpha), where R>0R > 0 and 0<α<π20 < \alpha < \dfrac{\pi}{2}.

Give the exact value of RR and the value of α\alpha to 44 decimal places. (3 marks)
(b) Find the maximum depth of water in the harbour, and the value of tt at which this maximum first occurs, giving tt to 22 decimal places. (4 marks)
(c) A ship can enter the harbour only when the depth of water is at least 1010 metres.

Find the two values of tt between t=6t = 6 and t=18t = 18 for which the depth is exactly 1010 metres, giving each value to 22 decimal places. (5 marks)
●●●●●Level 512 marksStart
(a) Prove that

1sec⁡θ−tan⁡θ≡sec⁡θ+tan⁡θ\dfrac{1}{\sec\theta - \tan\theta} \equiv \sec\theta + \tan\theta

(3 marks)
(b) Hence solve the equation

sec⁡θ−tan⁡θ=3\sec\theta - \tan\theta = 3

for 0⩽θ⩽2π0 \leqslant \theta \leqslant 2\pi, giving your answer to 33 decimal places. (4 marks)
●●●●●Level 57 marksStart
(a) Prove that

cot⁡θ−tan⁡θ≡2cot⁡2θ\cot\theta - \tan\theta \equiv 2\cot 2\theta

(4 marks)
(b) Hence solve, for 0<θ<π0 < \theta < \pi, the equation

cot⁡θ−tan⁡θ=23\cot\theta - \tan\theta = 2\sqrt3

giving your answers in terms of π\pi. (5 marks)
●●●●●Level 59 marksStart
In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.
(a) Show that the equation

2sin⁡2θ1−cos⁡2θ+3cosec⁡2θ=11\frac{2\sin 2\theta}{1 - \cos 2\theta} + 3\operatorname{cosec}^2\theta = 11

may be written as

3cot⁡2θ+2cot⁡θ−8=03\cot^2\theta + 2\cot\theta - 8 = 0

(4 marks)
(b) Hence solve, for 180∘<x<540∘180^\circ < x < 540^\circ, the equation

2sin⁡2x1−cos⁡2x+3cosec⁡2x=11x∈R, x≠360∘\frac{2\sin 2x}{1 - \cos 2x} + 3\operatorname{cosec}^2 x = 11 \qquad x \in \mathbb{R},\ x \neq 360^\circ

giving your answers to one decimal place.

(4 marks)
●●●●●Level 58 marksStart
A curve has equation y=ln⁡(3+6sin⁡x)y = \ln(3 + 6\sin x), where xx is in radians.
(a) State the values of xx, for 0≤x≤2π0 \le x \le 2\pi, for which ln⁡(3+6sin⁡x)\ln(3 + 6\sin x) is not defined.

(2 marks)
(b) PP is the stationary point on the curve that has the smallest positive xx-coordinate.

Determine the exact coordinates of PP.

(4 marks)
(c) (i) Show that d2ydx2+3e−y+27e−2y=0\dfrac{d^2y}{dx^2} + 3e^{-y} + 27e^{-2y} = 0.

(ii) State what can be deduced about all of the stationary points on this curve, giving a reason for your answer.

(6 marks)
●●●●●Level 512 marksStart
In this question you must show detailed reasoning.

The function f is defined by f(x)=sin⁡2x+2sin⁡xf(x) = \sin 2x + 2\sin x with domain 0≤x≤2π0 \le x \le 2\pi.
(a) Solve the following equations.

(i) f′(x)=0f'(x) = 0

(ii) f′′(x)=0f''(x) = 0

(7 marks)

The diagram shows the graph of the gradient function y=f′(x)y = f'(x) for the domain 0≤x≤2π0 \le x \le 2\pi. The points AA, CC and EE are where the graph meets the xx-axis, and BB and DD are minimum points.2πABCDEy = f′(x)x(b) Use your answers to parts (a)(i) and (a)(ii) to find the coordinates of points AA, BB, CC, DD and EE.

(3 marks)
(c) (i) Explain how to use the graph of the gradient function to find the values of xx for which f(x)f(x) is increasing.

(ii) Using set notation, write down the set of values of xx for which f(x)f(x) is increasing in the domain 0≤x≤2π0 \le x \le 2\pi.

(3 marks)
●●●●●Level 513 marksStart
In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.
(a) Solve, for 0<x<π0 < x < \pi, the equation

3cos⁡2x+5sin⁡x=43\cos^2 x + 5\sin x = 4

giving your answers in radians to 3 significant figures.

(5 marks)

The relative humidity inside a museum gallery is monitored on one particular day.

The relative humidity, H%H\%, inside the gallery, tt hours after midnight, is modelled by the equation

H=55+8sin⁡(kt−40)∘0⩽t<24H = 55 + 8\sin(kt - 40)^\circ \qquad 0 \leqslant t < 24

where kk is a constant.

Use the equation of the model to answer parts (b) to (d).

Given that
  • the relative humidity inside the gallery was 60%60\% at 5 am
  • 0<k<900 < k < 90
(b) find all possible values for kk, giving each answer to 2 decimal places.

(4 marks)

Given further that 0<k<200 < k < 20
(c) find the maximum relative humidity inside the gallery,

(1 mark)
(d) find the time of day at which this maximum relative humidity occurs. Give your answer to the nearest minute.

(2 marks)
●●●●●Level 512 marksStart
(a) Show that the equation 4cosec⁡x+cot⁡xcos⁡x−40=04\operatorname{cosec}x + \cot x\cos x - 40 = 0 can be expressed in the form sin⁡2x+40sin⁡x−5=0.\sin^2 x + 40\sin x - 5 = 0.

(3 marks)
(b) (i) In this question you must show detailed reasoning.

Hence solve, for 0<θ<12π0 < \theta < \tfrac12\pi, 4cosec⁡3θ+cot⁡3θcos⁡3θ−40=0.4\operatorname{cosec}3\theta + \cot 3\theta\cos 3\theta - 40 = 0. Give your answers correct to 3 decimal places.

The small angle approximation for sin⁡3θ\sin 3\theta is used to find an approximation for the smallest positive solution of the equation 4cosec⁡3θ+cot⁡3θcos⁡3θ−40=04\operatorname{cosec}3\theta + \cot 3\theta\cos 3\theta - 40 = 0.

(ii) Show that this approximate solution is accurate to 2 decimal places.

(6 marks)
●●●●●Level 59 marksStart
In this question you must show detailed reasoning.
(a) (i) Use the formula for cos⁡(A+B)\cos(A + B), and the double angle formulae, to show that cos⁡3θ≡4cos⁡3θ−3cos⁡θ\cos 3\theta \equiv 4\cos^3\theta - 3\cos\theta.

(ii) By writing −cos⁡2θ-\cos 2\theta as cos⁡(180∘−2θ)\cos(180^\circ - 2\theta), solve the equation cos⁡3θ+cos⁡2θ=0\cos 3\theta + \cos 2\theta = 0 for 0∘≤θ≤180∘0^\circ \le \theta \le 180^\circ.

(5 marks)
(b) (i) Show that 4x3+2x2−3x−1=(x+1)(4x2−2x−1)4x^3 + 2x^2 - 3x - 1 = (x + 1)(4x^2 - 2x - 1).

(ii) Hence find the exact roots of the equation 4x3+2x2−3x−1=04x^3 + 2x^2 - 3x - 1 = 0.

(3 marks)
(c) Use the results from parts (a)(ii) and (b)(ii) to show that cos⁡36∘=1+54\cos 36^\circ = \dfrac{1 + \sqrt5}{4}.

(2 marks)
●●●●●Level 510 marksStart

Radian measure

31 questions

LessonNot started
(a) Express 2cos⁡x−23sin⁡x2\cos x - 2\sqrt{3}\sin x in the form Rcos⁡(x+α)R\cos(x + \alpha), where R>0R > 0 and 0<α<π20 < \alpha < \dfrac{\pi}{2}. (3 marks)

The depth of water, hh metres, in a harbour tt hours after midnight is modelled by

h=5+2cos⁡(πt6)−23sin⁡(πt6),0≤t≤24.h = 5 + 2\cos\left(\dfrac{\pi t}{6}\right) - 2\sqrt{3}\sin\left(\dfrac{\pi t}{6}\right), \qquad 0 \le t \le 24.

A boat can enter the harbour only when the depth of water is at least 77 metres.
(b) Find the total length of time, during these 2424 hours, for which the boat is able to enter the harbour. Fully justify your answer. (6 marks)
●●●●●Level 59 marksStart
The depth of water, DD metres, at the entrance to a harbour is modelled by

D=9+3sin⁡(πt6)−33cos⁡(πt6)D = 9 + 3\sin\left(\dfrac{\pi t}{6}\right) - 3\sqrt{3}\cos\left(\dfrac{\pi t}{6}\right), 0≤t<24\qquad 0 \le t < 24,

where tt is the time in hours after midnight.
(a) Express 3sin⁡θ−33cos⁡θ3\sin\theta - 3\sqrt{3}\cos\theta in the form Rsin⁡(θ−α)R\sin(\theta - \alpha), where R>0R > 0 and 0<α<π20 < \alpha < \dfrac{\pi}{2}, stating the exact values of RR and α\alpha. (3 marks)
(b) Find the maximum depth of water at the harbour entrance, and the first time after midnight at which this maximum depth occurs. (2 marks)

A ship can enter the harbour only when the depth of water is at least 1212 metres.
(c) Find the times during the day between which the ship can enter the harbour. (4 marks)
●●●●●Level 59 marksStart
The diagram shows a sector OABOAB of a circle with centre OO and radius rr. The angle AOBAOB is θ\theta radians, where 0<θ<12π0 < \theta < \tfrac12\pi. The tangent to the circle at AA meets the line OBOB extended at TT. The region RR is bounded by the arc ABAB and the lines BTBT and ATAT. The areas of region RR and sector OABOAB are in the ratio 6:56 : 5.Oθ radrTRAB(a) Show that θ=tan⁡−1(2.2θ)\theta = \tan^{-1}(2.2\theta).

(4 marks)

The equation θ=tan⁡−1(2.2θ)\theta = \tan^{-1}(2.2\theta) has only one root for 0<θ<12π0 < \theta < \tfrac12\pi.
(b) This root can be found by using the iterative formula θn+1=tan⁡−1(2.2θn)\theta_{n+1} = \tan^{-1}(2.2\theta_n) with a starting value of θ1=1\theta_1 = 1.
  • Write down the values of θ2\theta_2, θ3\theta_3 and θ4\theta_4.
  • Hence find the value of this root correct to 3 significant figures.
(3 marks)

The diagram shows the graph of y=tan⁡−1(2.2θ)y = \tan^{-1}(2.2\theta) and the line y=θy = \theta, for 0≤θ≤12π0 \le \theta \le \tfrac12\pi.0.511.50.511.5θy(c) • Use this diagram to show how the iterative process used in (b) converges to this root.
  • State the type of convergence.
(3 marks)
(d) Draw a suitable diagram to show why using an iterative process with the formula θn+1=511tan⁡θn\theta_{n+1} = \tfrac{5}{11}\tan\theta_n does not converge to the root found in (b).

(2 marks)
●●●●●Level 512 marksStart
A stage designer is planning a raised platform for an outdoor concert.

The plan view of the platform is modelled as the region made up of the sector OABOAB of a circle with centre OO and radius rr metres, together with the triangles OACOAC and OBDOBD, which are joined to the sector along OAOA and along OBOB, where
  • angle AOB=1.5AOB = 1.5 radians
  • the arc ABAB has length 1212 m
  • triangle OACOAC and triangle OBDOBD are congruent, with the sector OABOAB lying between them
  • CC, OO and DD lie on a straight line, which forms the back edge of the platform
(a) Show that r=8r = 8.

(2 marks)
(b) Show that the size of angle AOCAOC is 0.8210.821 radians, correct to 3 decimal places.

(2 marks)

The front edge of the platform runs in a straight line from CC to AA, then along the arc ABAB, then in a straight line from BB to DD.

Given that the total length of the front edge of the platform is 3232 m,
(c) find the total area of the plan view of the platform, giving your answer to the nearest square metre.

(6 marks)
●●●●●Level 510 marksStart
OR4.2 cm7.5 cm0.45 radPQFigure 1

Diagram NOT accurately drawn

Figure 1 shows a sketch of a design for a ceramic tile.

The tile is made from a triangle OPQOPQ and a sector OQROQR of a circle with centre OO, joined along OQOQ.

In the design
  • OQ=7.5OQ = 7.5 cm
  • PQ=4.2PQ = 4.2 cm
  • angle POQ=0.45POQ = 0.45 radians
  • angle OPQ>π2OPQ > \dfrac{\pi}{2} radians
Making your method clear,
(a) find the size of angle OPQOPQ, giving your answer in radians to 4 significant figures,

(3 marks)
(b) find the area of triangle OPQOPQ, in cm2^2, giving your answer to 3 significant figures.

(2 marks)

Given that the area of sector OQROQR and the area of triangle OPQOPQ are in the ratio 5:35 : 3
(c) show that angle QORQOR is 0.3970.397 radians to 3 significant figures.

(3 marks)
(d) Hence find the perimeter of the tile, in cm, to the nearest integer.

(5 marks)
●●●●●Level 513 marksStart
In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.
(a) Solve, for 0<x<π0 < x < \pi, the equation

3cos⁡2x+5sin⁡x=43\cos^2 x + 5\sin x = 4

giving your answers in radians to 3 significant figures.

(5 marks)

The relative humidity inside a museum gallery is monitored on one particular day.

The relative humidity, H%H\%, inside the gallery, tt hours after midnight, is modelled by the equation

H=55+8sin⁡(kt−40)∘0⩽t<24H = 55 + 8\sin(kt - 40)^\circ \qquad 0 \leqslant t < 24

where kk is a constant.

Use the equation of the model to answer parts (b) to (d).

Given that
  • the relative humidity inside the gallery was 60%60\% at 5 am
  • 0<k<900 < k < 90
(b) find all possible values for kk, giving each answer to 2 decimal places.

(4 marks)

Given further that 0<k<200 < k < 20
(c) find the maximum relative humidity inside the gallery,

(1 mark)
(d) find the time of day at which this maximum relative humidity occurs. Give your answer to the nearest minute.

(2 marks)
●●●●●Level 512 marksStart
OQRPSx m1.1 radFigure 1

Diagram NOT accurately drawn

Figure 1 shows the plan view of a gravel path PQRSPQRS.

Given that
  • OQROQR is a sector of a circle with centre OO
  • OPSOPS is a sector of a smaller circle, also with centre OO, where PP lies on OQOQ and SS lies on OROR
  • angle POS=1.1POS = 1.1 radians
  • the area of sector OPSOPS is 4.954.95 m2^2
(a) show that OP=3OP = 3 m

(2 marks)

Given also that
  • the area of the gravel path PQRSPQRS is 4040 m2^2
  • PQ=xPQ = x metres
(b) show that

11x2+66x−800=011x^2 + 66x - 800 = 0

(3 marks)
(c) Solve the equation in part (b) to find the value of xx, giving your answer to 3 significant figures.

(Solutions relying entirely on calculator technology are not acceptable.)

(2 marks)
(d) Hence find the perimeter of the gravel path, giving your answer to the nearest metre.

(3 marks)
●●●●●Level 510 marksStart

Arc length and sector area

33 questions

LessonNot started
The diagram shows a sector of a circle OABOAB.

The point CC lies on OBOB such that ACAC is perpendicular to OBOB.

Angle AOBAOB is θ\theta radians.OBAθC(a) Given the area of the sector OABOAB is 2\sqrt{2} times the area of the triangle OACOAC, show that

sin⁡2θ=2 θ\sin 2\theta = \sqrt{2}\,\theta

(4 marks)
(b) Use a suitable change of sign to show that a solution to the equation

sin⁡2θ=2 θ\sin 2\theta = \sqrt{2}\,\theta

lies in the interval given by θ∈[π5,π4]\theta \in \left[\dfrac{\pi}{5}, \dfrac{\pi}{4}\right]

(2 marks)
(c) The Newton-Raphson method is used to find an approximate solution to the equation sin⁡2θ=2 θ\sin 2\theta = \sqrt{2}\,\theta
(c) (i) Using θ1=π4\theta_1 = \dfrac{\pi}{4} as a first approximation for θ\theta, apply the Newton-Raphson method twice to find the value of θ3\theta_3

Give your answer to three decimal places.

(ii) Explain how a more accurate approximation for θ\theta can be found using the Newton-Raphson method.

(iii) Explain why using θ1=π8\theta_1 = \dfrac{\pi}{8} as a first approximation in the Newton-Raphson method does not lead to a solution for θ\theta.

(6 marks)
●●●●●Level 512 marksStart
A flat metal weather vane is to be made in the shape shown in the diagram below.

The points PP, OO and BB lie on a straight line with PO=2OBPO = 2OB.

The shape consists of the sector OBDOBD of a circle with centre OO and radius OBOB, together with the triangle OPDOPD.

The angle PODPOD is θ\theta radians, where 0<θ<π0 < \theta < \piOBDθPThe weather vane is to be made with PB=1.5PB = 1.5 metres.
(a) Show that the area of the weather vane, AA square metres, is given by

A=18(π−θ+2sin⁡θ)A = \frac{1}{8}\left(\pi - \theta + 2\sin\theta\right)

(4 marks)
(b) (i) Show that the maximum value of AA occurs when θ=π3\theta = \dfrac{\pi}{3}

Fully justify your answer.

(ii) Find the exact maximum value of AA

(8 marks)
(c) Without further calculation, state how your answers to parts (b)(i) and (b)(ii) would change if PBPB were increased to 3 metres.

(2 marks)
●●●●●Level 514 marksStart
The diagram shows the cross-section of a horizontal pipe, which is a circle with centre OO and radius 5050 cm. The pipe contains some water. In the cross-section, the surface of the water is the chord ABAB and the angle AOBAOB is θ\theta radians. The water occupies the shaded segment bounded by the chord ABAB and the arc ABAB, which has area 900900 cm2^2.OABθ rad50 cm(a) Show that θ=sin⁡θ+0.72\theta = \sin\theta + 0.72.

(3 marks)
(b) Let F(θ)=sin⁡θ+0.72\text{F}(\theta) = \sin\theta + 0.72.

By considering the value of F′(θ)\text{F}'(\theta) where θ=1.7\theta = 1.7, explain why using an iterative method based on the equation in part (a) will converge to the root, assuming that 1.71.7 is sufficiently close to the root.

(2 marks)
(c) Use the iterative formula θn+1=sin⁡θn+0.72\theta_{n+1} = \sin\theta_n + 0.72 with a starting value of 1.71.7 to find the value of θ\theta correct to 4 significant figures.

You should show the result of each iteration.

(3 marks)
(d) Use a change of sign method to show that the value of θ\theta found in part (c) is correct to 4 significant figures.

(3 marks)
●●●●●Level 511 marksStart
The diagram shows a sector OABOAB of a circle with centre OO and radius rr. The angle AOBAOB is θ\theta radians, where 0<θ<12π0 < \theta < \tfrac12\pi. The tangent to the circle at AA meets the line OBOB extended at TT. The region RR is bounded by the arc ABAB and the lines BTBT and ATAT. The areas of region RR and sector OABOAB are in the ratio 6:56 : 5.Oθ radrTRAB(a) Show that θ=tan⁡−1(2.2θ)\theta = \tan^{-1}(2.2\theta).

(4 marks)

The equation θ=tan⁡−1(2.2θ)\theta = \tan^{-1}(2.2\theta) has only one root for 0<θ<12π0 < \theta < \tfrac12\pi.
(b) This root can be found by using the iterative formula θn+1=tan⁡−1(2.2θn)\theta_{n+1} = \tan^{-1}(2.2\theta_n) with a starting value of θ1=1\theta_1 = 1.
  • Write down the values of θ2\theta_2, θ3\theta_3 and θ4\theta_4.
  • Hence find the value of this root correct to 3 significant figures.
(3 marks)

The diagram shows the graph of y=tan⁡−1(2.2θ)y = \tan^{-1}(2.2\theta) and the line y=θy = \theta, for 0≤θ≤12π0 \le \theta \le \tfrac12\pi.0.511.50.511.5θy(c) • Use this diagram to show how the iterative process used in (b) converges to this root.
  • State the type of convergence.
(3 marks)
(d) Draw a suitable diagram to show why using an iterative process with the formula θn+1=511tan⁡θn\theta_{n+1} = \tfrac{5}{11}\tan\theta_n does not converge to the root found in (b).

(2 marks)
●●●●●Level 512 marksStart
A stage designer is planning a raised platform for an outdoor concert.

The plan view of the platform is modelled as the region made up of the sector OABOAB of a circle with centre OO and radius rr metres, together with the triangles OACOAC and OBDOBD, which are joined to the sector along OAOA and along OBOB, where
  • angle AOB=1.5AOB = 1.5 radians
  • the arc ABAB has length 1212 m
  • triangle OACOAC and triangle OBDOBD are congruent, with the sector OABOAB lying between them
  • CC, OO and DD lie on a straight line, which forms the back edge of the platform
(a) Show that r=8r = 8.

(2 marks)
(b) Show that the size of angle AOCAOC is 0.8210.821 radians, correct to 3 decimal places.

(2 marks)

The front edge of the platform runs in a straight line from CC to AA, then along the arc ABAB, then in a straight line from BB to DD.

Given that the total length of the front edge of the platform is 3232 m,
(c) find the total area of the plan view of the platform, giving your answer to the nearest square metre.

(6 marks)
●●●●●Level 510 marksStart
OR4.2 cm7.5 cm0.45 radPQFigure 1

Diagram NOT accurately drawn

Figure 1 shows a sketch of a design for a ceramic tile.

The tile is made from a triangle OPQOPQ and a sector OQROQR of a circle with centre OO, joined along OQOQ.

In the design
  • OQ=7.5OQ = 7.5 cm
  • PQ=4.2PQ = 4.2 cm
  • angle POQ=0.45POQ = 0.45 radians
  • angle OPQ>π2OPQ > \dfrac{\pi}{2} radians
Making your method clear,
(a) find the size of angle OPQOPQ, giving your answer in radians to 4 significant figures,

(3 marks)
(b) find the area of triangle OPQOPQ, in cm2^2, giving your answer to 3 significant figures.

(2 marks)

Given that the area of sector OQROQR and the area of triangle OPQOPQ are in the ratio 5:35 : 3
(c) show that angle QORQOR is 0.3970.397 radians to 3 significant figures.

(3 marks)
(d) Hence find the perimeter of the tile, in cm, to the nearest integer.

(5 marks)
●●●●●Level 513 marksStart
OQRPSx m1.1 radFigure 1

Diagram NOT accurately drawn

Figure 1 shows the plan view of a gravel path PQRSPQRS.

Given that
  • OQROQR is a sector of a circle with centre OO
  • OPSOPS is a sector of a smaller circle, also with centre OO, where PP lies on OQOQ and SS lies on OROR
  • angle POS=1.1POS = 1.1 radians
  • the area of sector OPSOPS is 4.954.95 m2^2
(a) show that OP=3OP = 3 m

(2 marks)

Given also that
  • the area of the gravel path PQRSPQRS is 4040 m2^2
  • PQ=xPQ = x metres
(b) show that

11x2+66x−800=011x^2 + 66x - 800 = 0

(3 marks)
(c) Solve the equation in part (b) to find the value of xx, giving your answer to 3 significant figures.

(Solutions relying entirely on calculator technology are not acceptable.)

(2 marks)
(d) Hence find the perimeter of the gravel path, giving your answer to the nearest metre.

(3 marks)
●●●●●Level 510 marksStart

Small angle approximations

23 questions

LessonNot started
(a) Show that the equation 4cosec⁡x+cot⁡xcos⁡x−40=04\operatorname{cosec}x + \cot x\cos x - 40 = 0 can be expressed in the form sin⁡2x+40sin⁡x−5=0.\sin^2 x + 40\sin x - 5 = 0.

(3 marks)
(b) (i) In this question you must show detailed reasoning.

Hence solve, for 0<θ<12π0 < \theta < \tfrac12\pi, 4cosec⁡3θ+cot⁡3θcos⁡3θ−40=0.4\operatorname{cosec}3\theta + \cot 3\theta\cos 3\theta - 40 = 0. Give your answers correct to 3 decimal places.

The small angle approximation for sin⁡3θ\sin 3\theta is used to find an approximation for the smallest positive solution of the equation 4cosec⁡3θ+cot⁡3θcos⁡3θ−40=04\operatorname{cosec}3\theta + \cot 3\theta\cos 3\theta - 40 = 0.

(ii) Show that this approximate solution is accurate to 2 decimal places.

(6 marks)
●●●●●Level 59 marksStart

Reciprocal trigonometric functions

12 questions

LessonNot started
(a) Given that y=sec⁡θy = \sec\theta
(a) (i) Express yy in terms of cos⁡θ\cos\theta.

(ii) Hence, prove that

dydθ=sec⁡θtan⁡θ\frac{\mathrm{d}y}{\mathrm{d}\theta} = \sec\theta\tan\theta

(iii) Show that

y2−1y=sin⁡θfor 0<θ<π2\frac{\sqrt{y^2 - 1}}{y} = \sin\theta \quad \text{for } 0 < \theta < \frac{\pi}{2}

(7 marks)
(b) (i) Use the substitution x=4sec⁡ux = 4\sec u to show that

∫1(x2−16)32 dxfor x>4\int \frac{1}{\left(x^2 - 16\right)^{\frac{3}{2}}} \, \mathrm{d}x \quad \text{for } x > 4

can be written as

k∫cosec⁡ucot⁡u duk\int \operatorname{cosec} u \cot u \, \mathrm{d}u

where kk is a constant to be found.

(ii) Hence, show that

∫1(x2−16)32 dx=−x16x2−16+cfor x>4\int \frac{1}{\left(x^2 - 16\right)^{\frac{3}{2}}} \, \mathrm{d}x = -\frac{x}{16\sqrt{x^2 - 16}} + c \quad \text{for } x > 4

where cc is a constant.

(9 marks)
●●●●●Level 516 marksStart
The functions f and g are defined for 0<x<12π0 < x < \tfrac12\pi by

f(x)=2cot⁡2xf(x) = 2\cot^2 x \quad and g(x)=cosec⁡x\quad g(x) = \operatorname{cosec} x.
(a) State the range of

(i) f,

(ii) g.

(2 marks)
(b) (i) Show that gf(1.1)=2.02gf(1.1) = 2.02, correct to 3 significant figures.

(ii) Explain why gf−1(−3)gf^{-1}(-3) is not defined.

(3 marks)
(c) In this question you must show detailed reasoning.

Solve the equation f(x)=3g(x)+3f(x) = 3g(x) + 3.

(5 marks)
●●●●●Level 510 marksStart

The sec/cosec/cot identities

26 questions

LessonNot started
A ball is kicked from a point OO on horizontal ground with speed 1414 m s⁻¹ at an angle θ\theta above the horizontal, where 0<θ<90∘0 < \theta < 90^\circ. The ball is modelled as a particle moving freely under gravity. Take g=9.8g = 9.8 m s⁻².

After tt seconds the horizontal and vertical displacements of the ball from OO are xx metres and yy metres respectively.
(a) Show that y=xtan⁡θ−x240(1+tan⁡2θ)y = x\tan\theta - \dfrac{x^2}{40}\left(1 + \tan^2\theta\right). (3 marks)

The ball passes through the point PP, which is 1010 m horizontally from OO and 55 m above the ground.
(b) Show that tan⁡2θ−4tan⁡θ+3=0\tan^2\theta - 4\tan\theta + 3 = 0, and hence find the two possible values of θ\theta. (4 marks)
(c) Show that, for either value of θ\theta, the speed of the ball as it passes through PP is 727\sqrt{2} m s⁻¹. (3 marks)
(d) For the smaller value of θ\theta, determine whether the ball is rising, falling or at its highest point as it passes through PP. Fully justify your answer. (2 marks)
●●●●●Level 512 marksStart
A small ball is projected with speed 14 m s−114\ \text{m s}^{-1} from a point OO on horizontal ground in a park.

After moving for TT seconds, the ball passes through the point AA, which is the top of a tree.

The point AA is 88 m horizontally and 88 m vertically from the point OO.

The motion of the ball from OO to AA is modelled as that of a particle moving freely under gravity.

Given that the ball is projected at an angle α\alpha to the horizontal, use the model to
(a) show that T=47cos⁡αT = \dfrac{4}{7\cos\alpha}

(2 marks)
(b) show that tan⁡2α−5tan⁡α+6=0\tan^2\alpha - 5\tan\alpha + 6 = 0

(5 marks)
(c) find the greatest possible height of the ball above the ground.

(3 marks)

The model does not include air resistance.
(d) State one other limitation of the model.

(1 mark)
●●●●●Level 511 marksStart
(a) Given that sin⁡θ≠±1\sin\theta \ne \pm 1, prove the identity

32−2sin⁡θ+32+2sin⁡θ≡3sec⁡2θ\frac{3}{2 - 2\sin\theta} + \frac{3}{2 + 2\sin\theta} \equiv 3\sec^2\theta

(4 marks)
(b) Hence, find the set of values of AA for which the equation

32−2sin⁡θ+32+2sin⁡θ=A\frac{3}{2 - 2\sin\theta} + \frac{3}{2 + 2\sin\theta} = A

has real solutions.

Fully justify your answer.

(3 marks)
(c) Given that θ\theta is obtuse and

32−2sin⁡θ+32+2sin⁡θ=10\frac{3}{2 - 2\sin\theta} + \frac{3}{2 + 2\sin\theta} = 10

find the exact value of tan⁡θ\tan\theta

(3 marks)
●●●●●Level 510 marksStart
(a) Prove that

1sec⁡θ−tan⁡θ≡sec⁡θ+tan⁡θ\dfrac{1}{\sec\theta - \tan\theta} \equiv \sec\theta + \tan\theta

(3 marks)
(b) Hence solve the equation

sec⁡θ−tan⁡θ=3\sec\theta - \tan\theta = 3

for 0⩽θ⩽2π0 \leqslant \theta \leqslant 2\pi, giving your answer to 33 decimal places. (4 marks)
●●●●●Level 57 marksStart
In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.
(a) Show that the equation

2sin⁡2θ1−cos⁡2θ+3cosec⁡2θ=11\frac{2\sin 2\theta}{1 - \cos 2\theta} + 3\operatorname{cosec}^2\theta = 11

may be written as

3cot⁡2θ+2cot⁡θ−8=03\cot^2\theta + 2\cot\theta - 8 = 0

(4 marks)
(b) Hence solve, for 180∘<x<540∘180^\circ < x < 540^\circ, the equation

2sin⁡2x1−cos⁡2x+3cosec⁡2x=11x∈R, x≠360∘\frac{2\sin 2x}{1 - \cos 2x} + 3\operatorname{cosec}^2 x = 11 \qquad x \in \mathbb{R},\ x \neq 360^\circ

giving your answers to one decimal place.

(4 marks)
●●●●●Level 58 marksStart
The functions f and g are defined for 0<x<12π0 < x < \tfrac12\pi by

f(x)=2cot⁡2xf(x) = 2\cot^2 x \quad and g(x)=cosec⁡x\quad g(x) = \operatorname{cosec} x.
(a) State the range of

(i) f,

(ii) g.

(2 marks)
(b) (i) Show that gf(1.1)=2.02gf(1.1) = 2.02, correct to 3 significant figures.

(ii) Explain why gf−1(−3)gf^{-1}(-3) is not defined.

(3 marks)
(c) In this question you must show detailed reasoning.

Solve the equation f(x)=3g(x)+3f(x) = 3g(x) + 3.

(5 marks)
●●●●●Level 510 marksStart
(a) Show that the equation 4cosec⁡x+cot⁡xcos⁡x−40=04\operatorname{cosec}x + \cot x\cos x - 40 = 0 can be expressed in the form sin⁡2x+40sin⁡x−5=0.\sin^2 x + 40\sin x - 5 = 0.

(3 marks)
(b) (i) In this question you must show detailed reasoning.

Hence solve, for 0<θ<12π0 < \theta < \tfrac12\pi, 4cosec⁡3θ+cot⁡3θcos⁡3θ−40=0.4\operatorname{cosec}3\theta + \cot 3\theta\cos 3\theta - 40 = 0. Give your answers correct to 3 decimal places.

The small angle approximation for sin⁡3θ\sin 3\theta is used to find an approximation for the smallest positive solution of the equation 4cosec⁡3θ+cot⁡3θcos⁡3θ−40=04\operatorname{cosec}3\theta + \cot 3\theta\cos 3\theta - 40 = 0.

(ii) Show that this approximate solution is accurate to 2 decimal places.

(6 marks)
●●●●●Level 59 marksStart

Inverse trigonometric functions

12 questions

LessonNot started
The diagram shows a sector OABOAB of a circle with centre OO and radius rr. The angle AOBAOB is θ\theta radians, where 0<θ<12π0 < \theta < \tfrac12\pi. The tangent to the circle at AA meets the line OBOB extended at TT. The region RR is bounded by the arc ABAB and the lines BTBT and ATAT. The areas of region RR and sector OABOAB are in the ratio 6:56 : 5.Oθ radrTRAB(a) Show that θ=tan⁡−1(2.2θ)\theta = \tan^{-1}(2.2\theta).

(4 marks)

The equation θ=tan⁡−1(2.2θ)\theta = \tan^{-1}(2.2\theta) has only one root for 0<θ<12π0 < \theta < \tfrac12\pi.
(b) This root can be found by using the iterative formula θn+1=tan⁡−1(2.2θn)\theta_{n+1} = \tan^{-1}(2.2\theta_n) with a starting value of θ1=1\theta_1 = 1.
  • Write down the values of θ2\theta_2, θ3\theta_3 and θ4\theta_4.
  • Hence find the value of this root correct to 3 significant figures.
(3 marks)

The diagram shows the graph of y=tan⁡−1(2.2θ)y = \tan^{-1}(2.2\theta) and the line y=θy = \theta, for 0≤θ≤12π0 \le \theta \le \tfrac12\pi.0.511.50.511.5θy(c) • Use this diagram to show how the iterative process used in (b) converges to this root.
  • State the type of convergence.
(3 marks)
(d) Draw a suitable diagram to show why using an iterative process with the formula θn+1=511tan⁡θn\theta_{n+1} = \tfrac{5}{11}\tan\theta_n does not converge to the root found in (b).

(2 marks)
●●●●●Level 512 marksStart

Addition formulae

31 questions

LessonNot started
(a) Prove that

sin⁡3x≡3sin⁡x−4sin⁡3x\sin 3x \equiv 3\sin x - 4\sin^3 x. (4 marks)
(b) Hence solve, for 0≤x<2π0 \le x < 2\pi, the equation

sin⁡3x=2sin⁡x\sin 3x = 2\sin x,

giving your answers in exact form. (3 marks)
(c) Use the identity in part (a) to find the exact value of

∫0π3sin⁡3x dx\displaystyle\int_0^{\frac{\pi}{3}} \sin^3 x\,\mathrm{d}x. (2 marks)
●●●●●Level 59 marksStart
(a) Prove that

cos⁡3θ≡4cos⁡3θ−3cos⁡θ\cos 3\theta \equiv 4\cos^3\theta - 3\cos\theta

(4 marks)
(b) Hence solve, for 0⩽θ<π0 \leqslant \theta < \pi, the equation

8cos⁡3θ−6cos⁡θ+3=08\cos^3\theta - 6\cos\theta + \sqrt{3} = 0

giving your answers in exact form in terms of π\pi. (5 marks)
●●●●●Level 59 marksStart
(a) Show that sin⁡(A+B)−sin⁡(A−B)≡2cos⁡Asin⁡B\sin(A+B) - \sin(A-B) \equiv 2\cos A\sin B.
(b) By writing 3x=2x+x3x = 2x + x and x=2x−xx = 2x - x, solve sin⁡3x=sin⁡x\sin 3x = \sin x for 0≤x≤π0 \le x \le \pi, giving your answers as exact multiples of π\pi.
●●●●●Level 56 marksStart
hxyPQRNThe diagram shows triangle PQRPQR. The perpendicular from RR to PQPQ meets PQPQ at NN. Angle PRN=xPRN = x, angle NRQ=yNRQ = y and the length RN=hRN = h.
(a) (i) Find the length PRPR in terms of hh and xx.

(ii) Write down the length QRQR in terms of hh and yy.

(iii) By finding the area of triangle PQRPQR in two different ways, show that sin⁡(x+y)≡sin⁡xcos⁡y+cos⁡xsin⁡y\sin(x + y) \equiv \sin x\cos y + \cos x\sin y.

(6 marks)
(b) Given that sin⁡(α+60∘)=cos⁡(α−45∘)\sin(\alpha + 60^\circ) = \cos(\alpha - 45^\circ), show that tan⁡α=6+3−2−2\tan\alpha = \sqrt6 + \sqrt3 - \sqrt2 - 2.

(5 marks)
●●●●●Level 511 marksStart
In this question you must show detailed reasoning.

The diagram shows the circle with equation x2+y2−12x−8y+42=0x^2 + y^2 - 12x - 8y + 42 = 0 and the point P(0,−4)P(0, -4). The tangent from PP touching the circle at AA has equation y=mx−4y = mx - 4, where m>1m > 1.PABOxy(a) (i) Show that the xx-coordinate of AA satisfies (m2+1)x2−4(4m+3)x+90=0(m^2 + 1)x^2 - 4(4m + 3)x + 90 = 0.

(ii) Hence determine the equation of this tangent.

(6 marks)

The other tangent from PP touches the circle at BB and has equation y=nx−4y = nx - 4, where n<1n < 1.
(b) Determine the exact value of tan⁡APB\tan APB.

(4 marks)
●●●●●Level 510 marksStart
In this question you must show detailed reasoning.
(a) (i) Use the formula for cos⁡(A+B)\cos(A + B), and the double angle formulae, to show that cos⁡3θ≡4cos⁡3θ−3cos⁡θ\cos 3\theta \equiv 4\cos^3\theta - 3\cos\theta.

(ii) By writing −cos⁡2θ-\cos 2\theta as cos⁡(180∘−2θ)\cos(180^\circ - 2\theta), solve the equation cos⁡3θ+cos⁡2θ=0\cos 3\theta + \cos 2\theta = 0 for 0∘≤θ≤180∘0^\circ \le \theta \le 180^\circ.

(5 marks)
(b) (i) Show that 4x3+2x2−3x−1=(x+1)(4x2−2x−1)4x^3 + 2x^2 - 3x - 1 = (x + 1)(4x^2 - 2x - 1).

(ii) Hence find the exact roots of the equation 4x3+2x2−3x−1=04x^3 + 2x^2 - 3x - 1 = 0.

(3 marks)
(c) Use the results from parts (a)(ii) and (b)(ii) to show that cos⁡36∘=1+54\cos 36^\circ = \dfrac{1 + \sqrt5}{4}.

(2 marks)
●●●●●Level 510 marksStart

Double-angle formulae

42 questions

LessonNot started
(a) Prove that

1−cos⁡2θ+sin⁡2θ1+cos⁡2θ+sin⁡2θ≡tan⁡θ\dfrac{1 - \cos 2\theta + \sin 2\theta}{1 + \cos 2\theta + \sin 2\theta} \equiv \tan\theta (4 marks)
(b) Hence solve the equation

1−cos⁡4x+sin⁡4x1+cos⁡4x+sin⁡4x=3tan⁡x\dfrac{1 - \cos 4x + \sin 4x}{1 + \cos 4x + \sin 4x} = 3\tan x

for 0≤x<π0 \le x < \pi, giving your answers in exact form. (5 marks)
●●●●●Level 59 marksStart
(a) Prove that

cos⁡2θcos⁡θ+sin⁡θ≡cos⁡θ−sin⁡θ,cos⁡θ+sin⁡θ≠0\dfrac{\cos 2\theta}{\cos\theta + \sin\theta} \equiv \cos\theta - \sin\theta, \qquad \cos\theta + \sin\theta \neq 0

(3 marks)
(b) Hence solve, for 0⩽θ⩽2π0 \leqslant \theta \leqslant 2\pi, the equation

cos⁡2θcos⁡θ+sin⁡θ=12\dfrac{\cos 2\theta}{\cos\theta + \sin\theta} = \dfrac12

giving your answers to 33 decimal places. (5 marks)
●●●●●Level 58 marksStart
(a) Using the double-angle formula for cosine, show that

cos⁡2x≡1+cos⁡2x2\cos^2 x \equiv \dfrac{1 + \cos 2x}{2}. (2 marks)

The curve SS has equation y=6cos⁡2xy = 6\cos^2 x,  0≤x≤π3\ 0 \le x \le \dfrac{\pi}{3}.0.20.40.60.81123456xy(b) Find the exact area of the region bounded by SS, the coordinate axes and the line x=π3x = \dfrac{\pi}{3}. (4 marks)
(c) The horizontal line y=3y = 3 cuts SS at a single point in this interval. Find the exact area of the region enclosed by SS, the yy-axis and the line y=3y = 3. (4 marks)
●●●●●Level 510 marksStart
(a) Prove that

sin⁡3x≡3sin⁡x−4sin⁡3x\sin 3x \equiv 3\sin x - 4\sin^3 x. (4 marks)
(b) Hence solve, for 0≤x<2π0 \le x < 2\pi, the equation

sin⁡3x=2sin⁡x\sin 3x = 2\sin x,

giving your answers in exact form. (3 marks)
(c) Use the identity in part (a) to find the exact value of

∫0π3sin⁡3x dx\displaystyle\int_0^{\frac{\pi}{3}} \sin^3 x\,\mathrm{d}x. (2 marks)
●●●●●Level 59 marksStart
The diagram shows a sector of a circle OABOAB.

The point CC lies on OBOB such that ACAC is perpendicular to OBOB.

Angle AOBAOB is θ\theta radians.OBAθC(a) Given the area of the sector OABOAB is 2\sqrt{2} times the area of the triangle OACOAC, show that

sin⁡2θ=2 θ\sin 2\theta = \sqrt{2}\,\theta

(4 marks)
(b) Use a suitable change of sign to show that a solution to the equation

sin⁡2θ=2 θ\sin 2\theta = \sqrt{2}\,\theta

lies in the interval given by θ∈[π5,π4]\theta \in \left[\dfrac{\pi}{5}, \dfrac{\pi}{4}\right]

(2 marks)
(c) The Newton-Raphson method is used to find an approximate solution to the equation sin⁡2θ=2 θ\sin 2\theta = \sqrt{2}\,\theta
(c) (i) Using θ1=π4\theta_1 = \dfrac{\pi}{4} as a first approximation for θ\theta, apply the Newton-Raphson method twice to find the value of θ3\theta_3

Give your answer to three decimal places.

(ii) Explain how a more accurate approximation for θ\theta can be found using the Newton-Raphson method.

(iii) Explain why using θ1=π8\theta_1 = \dfrac{\pi}{8} as a first approximation in the Newton-Raphson method does not lead to a solution for θ\theta.

(6 marks)
●●●●●Level 512 marksStart
The curve CC has parametric equations

x=4cos⁡t,y=2sin⁡2t,0⩽t⩽π2x = 4\cos t, \qquad y = 2\sin 2t, \qquad 0 \leqslant t \leqslant \dfrac{\pi}{2}

The point PP lies on CC where t=π6t = \dfrac{\pi}{6}.
(a) Show that the tangent to CC at PP has equation x+y=33x + y = 3\sqrt{3}. (4 marks)

The tangent to CC at PP meets the xx-axis at the point QQ, and CC meets the positive xx-axis at the point AA.

The finite region RR is bounded by the arc of CC from AA to PP, the line segment PQPQ and the line segment QAQA.
(b) Find the exact area of RR. (6 marks)
●●●●●Level 510 marksStart
(a) Prove that

cos⁡3θ≡4cos⁡3θ−3cos⁡θ\cos 3\theta \equiv 4\cos^3\theta - 3\cos\theta

(4 marks)
(b) Hence solve, for 0⩽θ<π0 \leqslant \theta < \pi, the equation

8cos⁡3θ−6cos⁡θ+3=08\cos^3\theta - 6\cos\theta + \sqrt{3} = 0

giving your answers in exact form in terms of π\pi. (5 marks)
●●●●●Level 59 marksStart
The curve CC has parametric equations

x=12sin⁡2ty=4sin⁡2t+sin⁡t0⩽t⩽π2x = 12\sin^2 t \qquad\qquad y = 4\sin 2t + \sin t \qquad\qquad 0 \leqslant t \leqslant \frac{\pi}{2}

The region RR is bounded by CC, the xx-axis and the line with equation x=9x = 9.
(a) Show that the area of RR is given by

∫0a(48sin⁡22t+24sin⁡2tcos⁡t)dt\int_{0}^{a} \left(48\sin^2 2t + 24\sin^2 t\cos t\right) dt

where aa is a constant to be found.

(5 marks)
(b) Hence, using algebraic integration, find the exact area of RR.

(4 marks)
●●●●●Level 59 marksStart
(a) Prove that

cot⁡θ−tan⁡θ≡2cot⁡2θ\cot\theta - \tan\theta \equiv 2\cot 2\theta

(4 marks)
(b) Hence solve, for 0<θ<π0 < \theta < \pi, the equation

cot⁡θ−tan⁡θ=23\cot\theta - \tan\theta = 2\sqrt3

giving your answers in terms of π\pi. (5 marks)
●●●●●Level 59 marksStart
In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.
(a) Show that the equation

2sin⁡2θ1−cos⁡2θ+3cosec⁡2θ=11\frac{2\sin 2\theta}{1 - \cos 2\theta} + 3\operatorname{cosec}^2\theta = 11

may be written as

3cot⁡2θ+2cot⁡θ−8=03\cot^2\theta + 2\cot\theta - 8 = 0

(4 marks)
(b) Hence solve, for 180∘<x<540∘180^\circ < x < 540^\circ, the equation

2sin⁡2x1−cos⁡2x+3cosec⁡2x=11x∈R, x≠360∘\frac{2\sin 2x}{1 - \cos 2x} + 3\operatorname{cosec}^2 x = 11 \qquad x \in \mathbb{R},\ x \neq 360^\circ

giving your answers to one decimal place.

(4 marks)
●●●●●Level 58 marksStart
In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.

Figure 1PlCOxyFigure 1 shows a sketch of the curve CC with parametric equations

x=cos⁡2ty=sin⁡tcos⁡2t−π2<t<π2x = \cos 2t \qquad y = \sin t\cos 2t \qquad -\dfrac{\pi}{2} < t < \dfrac{\pi}{2}

where tt is a parameter.

The point PP lies on CC where t=π6t = \dfrac{\pi}{6}

The line ll, shown in Figure 1, is the tangent to CC at PP.
(a) Use parametric differentiation to show that

(i) dydx=psin⁡t+qcosec⁡t\dfrac{dy}{dx} = p\sin t + q\operatorname{cosec} t where pp and qq are constants to be found

(ii) an equation for ll is 2x−8y+1=02x - 8y + 1 = 0

(6 marks)

The line ll intersects the curve CC again at the point QQ.
(b) Using algebra and showing detailed reasoning, find the exact coordinates of QQ.

(6 marks)
●●●●●Level 512 marksStart
In this question you must show detailed reasoning.

The function f is defined by f(x)=sin⁡2x+2sin⁡xf(x) = \sin 2x + 2\sin x with domain 0≤x≤2π0 \le x \le 2\pi.
(a) Solve the following equations.

(i) f′(x)=0f'(x) = 0

(ii) f′′(x)=0f''(x) = 0

(7 marks)

The diagram shows the graph of the gradient function y=f′(x)y = f'(x) for the domain 0≤x≤2π0 \le x \le 2\pi. The points AA, CC and EE are where the graph meets the xx-axis, and BB and DD are minimum points.2πABCDEy = f′(x)x(b) Use your answers to parts (a)(i) and (a)(ii) to find the coordinates of points AA, BB, CC, DD and EE.

(3 marks)
(c) (i) Explain how to use the graph of the gradient function to find the values of xx for which f(x)f(x) is increasing.

(ii) Using set notation, write down the set of values of xx for which f(x)f(x) is increasing in the domain 0≤x≤2π0 \le x \le 2\pi.

(3 marks)
●●●●●Level 513 marksStart
In this question you must show detailed reasoning.
(a) (i) Use the formula for cos⁡(A+B)\cos(A + B), and the double angle formulae, to show that cos⁡3θ≡4cos⁡3θ−3cos⁡θ\cos 3\theta \equiv 4\cos^3\theta - 3\cos\theta.

(ii) By writing −cos⁡2θ-\cos 2\theta as cos⁡(180∘−2θ)\cos(180^\circ - 2\theta), solve the equation cos⁡3θ+cos⁡2θ=0\cos 3\theta + \cos 2\theta = 0 for 0∘≤θ≤180∘0^\circ \le \theta \le 180^\circ.

(5 marks)
(b) (i) Show that 4x3+2x2−3x−1=(x+1)(4x2−2x−1)4x^3 + 2x^2 - 3x - 1 = (x + 1)(4x^2 - 2x - 1).

(ii) Hence find the exact roots of the equation 4x3+2x2−3x−1=04x^3 + 2x^2 - 3x - 1 = 0.

(3 marks)
(c) Use the results from parts (a)(ii) and (b)(ii) to show that cos⁡36∘=1+54\cos 36^\circ = \dfrac{1 + \sqrt5}{4}.

(2 marks)
●●●●●Level 510 marksStart

The R-form (a sinθ + b cosθ)

28 questions

LessonNot started
(a) Express 2cos⁡x−23sin⁡x2\cos x - 2\sqrt{3}\sin x in the form Rcos⁡(x+α)R\cos(x + \alpha), where R>0R > 0 and 0<α<π20 < \alpha < \dfrac{\pi}{2}. (3 marks)

The depth of water, hh metres, in a harbour tt hours after midnight is modelled by

h=5+2cos⁡(πt6)−23sin⁡(πt6),0≤t≤24.h = 5 + 2\cos\left(\dfrac{\pi t}{6}\right) - 2\sqrt{3}\sin\left(\dfrac{\pi t}{6}\right), \qquad 0 \le t \le 24.

A boat can enter the harbour only when the depth of water is at least 77 metres.
(b) Find the total length of time, during these 2424 hours, for which the boat is able to enter the harbour. Fully justify your answer. (6 marks)
●●●●●Level 59 marksStart
(a) Prove that

cos⁡2θcos⁡θ+sin⁡θ≡cos⁡θ−sin⁡θ,cos⁡θ+sin⁡θ≠0\dfrac{\cos 2\theta}{\cos\theta + \sin\theta} \equiv \cos\theta - \sin\theta, \qquad \cos\theta + \sin\theta \neq 0

(3 marks)
(b) Hence solve, for 0⩽θ⩽2π0 \leqslant \theta \leqslant 2\pi, the equation

cos⁡2θcos⁡θ+sin⁡θ=12\dfrac{\cos 2\theta}{\cos\theta + \sin\theta} = \dfrac12

giving your answers to 33 decimal places. (5 marks)
●●●●●Level 58 marksStart
(a) Express 7cos⁡θ+24sin⁡θ7\cos\theta + 24\sin\theta in the form Rcos⁡(θ−α)R\cos(\theta - \alpha), with R>0R > 0 and 0<α<90∘0 < \alpha < 90^\circ.
(b) State the maximum value and the value of θ\theta (to 1 d.p.) at which it occurs.
(c) Solve 7cos⁡θ+24sin⁡θ=157\cos\theta + 24\sin\theta = 15 for 0≤θ≤360∘0 \le \theta \le 360^\circ, to 1 d.p.
●●●●●Level 58 marksStart
The depth of water, DD metres, at the entrance to a harbour is modelled by

D=9+3sin⁡(πt6)−33cos⁡(πt6)D = 9 + 3\sin\left(\dfrac{\pi t}{6}\right) - 3\sqrt{3}\cos\left(\dfrac{\pi t}{6}\right), 0≤t<24\qquad 0 \le t < 24,

where tt is the time in hours after midnight.
(a) Express 3sin⁡θ−33cos⁡θ3\sin\theta - 3\sqrt{3}\cos\theta in the form Rsin⁡(θ−α)R\sin(\theta - \alpha), where R>0R > 0 and 0<α<π20 < \alpha < \dfrac{\pi}{2}, stating the exact values of RR and α\alpha. (3 marks)
(b) Find the maximum depth of water at the harbour entrance, and the first time after midnight at which this maximum depth occurs. (2 marks)

A ship can enter the harbour only when the depth of water is at least 1212 metres.
(c) Find the times during the day between which the ship can enter the harbour. (4 marks)
●●●●●Level 59 marksStart
The depth of water, DD metres, in a harbour at a time tt hours after midnight is modelled by

D=8+3cos⁡(0.5t)−4sin⁡(0.5t),0⩽t⩽24D = 8 + 3\cos(0.5t) - 4\sin(0.5t), \qquad 0 \leqslant t \leqslant 24

where the angle 0.5t0.5t is measured in radians.
(a) Express 3cos⁡θ−4sin⁡θ3\cos\theta - 4\sin\theta in the form Rcos⁡(θ+α)R\cos(\theta + \alpha), where R>0R > 0 and 0<α<π20 < \alpha < \dfrac{\pi}{2}.

Give the exact value of RR and the value of α\alpha to 44 decimal places. (3 marks)
(b) Find the maximum depth of water in the harbour, and the value of tt at which this maximum first occurs, giving tt to 22 decimal places. (4 marks)
(c) A ship can enter the harbour only when the depth of water is at least 1010 metres.

Find the two values of tt between t=6t = 6 and t=18t = 18 for which the depth is exactly 1010 metres, giving each value to 22 decimal places. (5 marks)
●●●●●Level 512 marksStart
(a) Given that

7cos⁡θ−24sin⁡θ≡Kcos⁡(θ+α)7\cos\theta - 24\sin\theta \equiv K\cos(\theta + \alpha)

where K>0K > 0 and 0<α<90∘0 < \alpha < 90^\circ, find the value of KK and the value of α\alpha, giving α\alpha to 2 decimal places.

(3 marks)

An ecologist records the number of mice and the number of owls living on a farm over one year.

The number of mice, MM, is modelled by

M=A+7cos⁡(30t)∘−24sin⁡(30t)∘M = A + 7\cos(30t)^\circ - 24\sin(30t)^\circ

where tt is the number of months after the start of the year and AA is a constant.

According to the model, the greatest number of mice on the farm during the year is 145145
(b) (i) Find a complete equation for the model.

(ii) Hence state the least number of mice on the farm during the year, according to the model.

(2 marks)

The ecologist finds that the number of mice on the farm is actually lowest in October.
(c) Use this information to comment on the model for the number of mice.

(2 marks)

The number of owls, NN, on the farm during the same year is modelled by

N=40+15sin⁡(30t+60)∘N = 40 + 15\sin(30t + 60)^\circ

The number of owls is at its lowest when t=Tt = T
(d) Use the two models to find the number of mice on the farm when t=Tt = T

(4 marks)
●●●●●Level 511 marksStart
(a) Given that

8sin⁡θ+6cos⁡θ≡Rsin⁡(θ+α)8\sin\theta + 6\cos\theta \equiv R\sin(\theta + \alpha)

where R>0R > 0 and 0<α<π20 < \alpha < \dfrac{\pi}{2}, find the exact value of RR and find α\alpha, in radians, correct to 3 decimal places.

(3 marks)

A scientist measures the temperature of the water in two garden ponds throughout one year.

For the first pond, the water temperature, S ∘S\ ^\circC, is modelled by

S=A−8sin⁡(πt6)−6cos⁡(πt6)S = A - 8\sin\left(\frac{\pi t}{6}\right) - 6\cos\left(\frac{\pi t}{6}\right)

where tt is the number of months after the start of the year and AA is a constant.

According to the model, the highest water temperature in the first pond during the year is 23 ∘23\ ^\circC
(b) (i) Find a complete equation for the model.

(ii) Hence state the lowest water temperature in the first pond during the year, according to the model.

(2 marks)

The scientist observes that the water temperature in the first pond is actually at its lowest towards the end of February.
(c) Use this information to comment on the model for the first pond.

(2 marks)

For the second pond, the water temperature, P ∘P\ ^\circC, during the same year is modelled by

P=12+9cos⁡(πt6−2.3)P = 12 + 9\cos\left(\frac{\pi t}{6} - 2.3\right)

The water temperature in the second pond is at its highest when t=Tt = T
(d) Use the two models to find the water temperature in the first pond when t=Tt = T, giving your answer to 3 significant figures.

(4 marks)
●●●●●Level 511 marksStart
(a) Express 5sin⁡θ−12cos⁡θ5\sin\theta - 12\cos\theta in the form Rsin⁡(θ−α)R\sin(\theta - \alpha), where R>0R > 0 and 0<α<π20 < \alpha < \dfrac{\pi}{2}. Give α\alpha correct to 4 significant figures.
(b) Hence solve 5sin⁡θ−12cos⁡θ=6.55\sin\theta - 12\cos\theta = 6.5 for 0≤θ≤2π0 \le \theta \le 2\pi, giving your answers correct to 3 significant figures.
●●●●●Level 57 marksStart
(a) Express 6cos⁡2x+5sin⁡2x6\cos 2x + 5\sin 2x in the form Rcos⁡(2x−α)R\cos(2x - \alpha), where R>0R > 0 and 0<α<12π0 < \alpha < \tfrac12\pi. Give the exact value of RR and give the value of α\alpha correct to 4 significant figures.

(3 marks)
(b) Give full details of a sequence of three transformations needed to transform the curve y=cos⁡xy = \cos x to the curve y=6cos⁡2x+5sin⁡2xy = 6\cos 2x + 5\sin 2x.

(4 marks)
(c) Determine the greatest value of 6cos⁡2x+5sin⁡2x6\cos 2x + 5\sin 2x as xx varies and give the smallest positive value of xx for which it occurs.

(2 marks)
(d) Determine the least value of 6cos⁡2x+5sin⁡2x6\cos 2x + 5\sin 2x as xx varies and give the smallest positive value of xx for which it occurs.

(2 marks)
●●●●●Level 511 marksStart

Proving trigonometric identities

26 questions

LessonNot started
(a) Prove that

1−cos⁡2θ+sin⁡2θ1+cos⁡2θ+sin⁡2θ≡tan⁡θ\dfrac{1 - \cos 2\theta + \sin 2\theta}{1 + \cos 2\theta + \sin 2\theta} \equiv \tan\theta (4 marks)
(b) Hence solve the equation

1−cos⁡4x+sin⁡4x1+cos⁡4x+sin⁡4x=3tan⁡x\dfrac{1 - \cos 4x + \sin 4x}{1 + \cos 4x + \sin 4x} = 3\tan x

for 0≤x<π0 \le x < \pi, giving your answers in exact form. (5 marks)
●●●●●Level 59 marksStart
(a) Prove that

cos⁡2θcos⁡θ+sin⁡θ≡cos⁡θ−sin⁡θ,cos⁡θ+sin⁡θ≠0\dfrac{\cos 2\theta}{\cos\theta + \sin\theta} \equiv \cos\theta - \sin\theta, \qquad \cos\theta + \sin\theta \neq 0

(3 marks)
(b) Hence solve, for 0⩽θ⩽2π0 \leqslant \theta \leqslant 2\pi, the equation

cos⁡2θcos⁡θ+sin⁡θ=12\dfrac{\cos 2\theta}{\cos\theta + \sin\theta} = \dfrac12

giving your answers to 33 decimal places. (5 marks)
●●●●●Level 58 marksStart
(a) Prove that

sin⁡3x≡3sin⁡x−4sin⁡3x\sin 3x \equiv 3\sin x - 4\sin^3 x. (4 marks)
(b) Hence solve, for 0≤x<2π0 \le x < 2\pi, the equation

sin⁡3x=2sin⁡x\sin 3x = 2\sin x,

giving your answers in exact form. (3 marks)
(c) Use the identity in part (a) to find the exact value of

∫0π3sin⁡3x dx\displaystyle\int_0^{\frac{\pi}{3}} \sin^3 x\,\mathrm{d}x. (2 marks)
●●●●●Level 59 marksStart
(a) Prove that

cos⁡3θ≡4cos⁡3θ−3cos⁡θ\cos 3\theta \equiv 4\cos^3\theta - 3\cos\theta

(4 marks)
(b) Hence solve, for 0⩽θ<π0 \leqslant \theta < \pi, the equation

8cos⁡3θ−6cos⁡θ+3=08\cos^3\theta - 6\cos\theta + \sqrt{3} = 0

giving your answers in exact form in terms of π\pi. (5 marks)
●●●●●Level 59 marksStart
(a) Given that sin⁡θ≠±1\sin\theta \ne \pm 1, prove the identity

32−2sin⁡θ+32+2sin⁡θ≡3sec⁡2θ\frac{3}{2 - 2\sin\theta} + \frac{3}{2 + 2\sin\theta} \equiv 3\sec^2\theta

(4 marks)
(b) Hence, find the set of values of AA for which the equation

32−2sin⁡θ+32+2sin⁡θ=A\frac{3}{2 - 2\sin\theta} + \frac{3}{2 + 2\sin\theta} = A

has real solutions.

Fully justify your answer.

(3 marks)
(c) Given that θ\theta is obtuse and

32−2sin⁡θ+32+2sin⁡θ=10\frac{3}{2 - 2\sin\theta} + \frac{3}{2 + 2\sin\theta} = 10

find the exact value of tan⁡θ\tan\theta

(3 marks)
●●●●●Level 510 marksStart
(a) Prove that

1sec⁡θ−tan⁡θ≡sec⁡θ+tan⁡θ\dfrac{1}{\sec\theta - \tan\theta} \equiv \sec\theta + \tan\theta

(3 marks)
(b) Hence solve the equation

sec⁡θ−tan⁡θ=3\sec\theta - \tan\theta = 3

for 0⩽θ⩽2π0 \leqslant \theta \leqslant 2\pi, giving your answer to 33 decimal places. (4 marks)
●●●●●Level 57 marksStart
(a) Prove that

cot⁡θ−tan⁡θ≡2cot⁡2θ\cot\theta - \tan\theta \equiv 2\cot 2\theta

(4 marks)
(b) Hence solve, for 0<θ<π0 < \theta < \pi, the equation

cot⁡θ−tan⁡θ=23\cot\theta - \tan\theta = 2\sqrt3

giving your answers in terms of π\pi. (5 marks)
●●●●●Level 59 marksStart
hxyPQRNThe diagram shows triangle PQRPQR. The perpendicular from RR to PQPQ meets PQPQ at NN. Angle PRN=xPRN = x, angle NRQ=yNRQ = y and the length RN=hRN = h.
(a) (i) Find the length PRPR in terms of hh and xx.

(ii) Write down the length QRQR in terms of hh and yy.

(iii) By finding the area of triangle PQRPQR in two different ways, show that sin⁡(x+y)≡sin⁡xcos⁡y+cos⁡xsin⁡y\sin(x + y) \equiv \sin x\cos y + \cos x\sin y.

(6 marks)
(b) Given that sin⁡(α+60∘)=cos⁡(α−45∘)\sin(\alpha + 60^\circ) = \cos(\alpha - 45^\circ), show that tan⁡α=6+3−2−2\tan\alpha = \sqrt6 + \sqrt3 - \sqrt2 - 2.

(5 marks)
●●●●●Level 511 marksStart