Browse Trigonometry

$OAB$ is a sector of a circle with centre $O$ and radius $r$ cm
angle $AOB = 0.75$ radians
the arc $AB$ has length $10.5$ cm
$OBC$ is a triangle joined to the sector along the radius $OB$ , where $BC = 9$ cm and angle $OBC = 2.1$ radians

(a) Show that

r = 14

.

(2 marks)
(b) Find the area of the plan view of the screen, giving your answer to 3 significant figures.

(4 marks)

●●●●●Level 46 marksStart

A jeweller is designing a silver brooch.

The brooch is modelled as the region bounded by the chord

AB

and the arc

AB

of a circle with centre

O

and radius

16

cm, where angle

AOB = \dfrac{2\pi}{3}

radians.
(a) Use the cosine rule to show that

AB = 16\sqrt{3}

cm.

(2 marks)
(b) Find the exact perimeter of the brooch.

(2 marks)
(c) Find the exact area of the brooch, giving your answer in the form

(a\pi + b\sqrt{3})

^2

, where

a

and

b

are rational numbers.

(3 marks)

●●●●●Level 47 marksStart

The diagram shows a sector

OAB

of a circle with centre

O

and radius

5

cm. The angle

AOB

\theta

radians. The shaded segment is bounded by the chord

AB

and the arc

AB

, and the perimeter of the shaded segment is

21

cm.(a) Show that

\theta = 4.2 - 2\sin\left(\tfrac12\theta\right)

.

(4 marks)

The equation

\theta = 4.2 - 2\sin\left(\tfrac12\theta\right)

has only one root for

0 < \theta < \pi

.
(b) This root can be found by using the iterative formula

\theta_{n+1} = 4.2 - 2\sin\left(\tfrac12\theta_n\right)

with a starting value of

\theta_1 = 2

Write down the values of $\theta_2$ , $\theta_3$ and $\theta_4$ .

Hence find the value of this root correct to 3 significant figures.

(3 marks)

The diagram shows the graph of

y = 4.2 - 2\sin\left(\tfrac12\theta\right)

and the line

y = \theta

, for

0 \le \theta \le \pi

.(c) • Use this diagram to show how the iterative process used in (b) converges to this root.

State the type of convergence.

(3 marks)

●●●●●Level 410 marksStart

Figure 1 (not accurately drawn)

Two horizontal ropes are tied to a mooring post at the same point

O

. The ropes exert two horizontal forces,

\mathbf{P}

and

\mathbf{Q}

, on the post, and the forces are modelled as acting at

O

.

The angle between the direction of

\mathbf{P}

and the direction of

\mathbf{Q}

120°

, as shown in Figure 1.

Force $\mathbf{P}$ has magnitude $X$ newtons
Force $\mathbf{Q}$ has magnitude $60$ N
The resultant of $\mathbf{P}$ and $\mathbf{Q}$ has magnitude $20\sqrt{19}$ N

Find

(i) the value of

X

(ii) the angle between

\mathbf{Q}

and the resultant of

\mathbf{P}

and

\mathbf{Q}

, giving your answer to the nearest degree.

(8 marks)

●●●●●Level 48 marksStart

The point

A

is such that

|\overrightarrow{OA}| = 10

and

\overrightarrow{OA}

makes an angle of

135°

with the positive

x

-axis, measured anticlockwise.
(a) (i) On a sketch of the

x

- and

y

-axes, mark the position of the point

A

.

(ii) Find the position vector of

A

in terms of

\mathbf{i}

and

\mathbf{j}

.

(4 marks)

The point

B

has position vector

8\mathbf{i}

.
(b) Find the exact area of triangle

AOB

.

(2 marks)

OABC

is a parallelogram.
(c) Find the position vector of

C

in terms of

\mathbf{i}

and

\mathbf{j}

.

(2 marks)

●●●●●Level 48 marksStart

The diagram shows triangle

ABC

in which angle

BAC

120°

and the lengths of

AB

and

AC

are

(5 + h)

cm and

(3 - h)

cm respectively, where

0 < h < 3

.(a) Show that the length of

BC

\sqrt{p}

cm where

p = 49 + 2h + h^2

.

(2 marks)
(b) Hence show that, when

h

is small,

\sqrt{p} \approx 7 + mh + nh^2

, where

m

and

n

are rational numbers whose values are to be determined.

(4 marks)

●●●●●Level 46 marksStart

Figure 1

Diagram NOT accurately drawn

Figure 1 shows the plan view of a flower bed

ACD

.

The flower bed consists of triangle

BCD

joined to triangle

ABD

.

Given that

$BD = x$ m
$CD = (x + 3)$ m
$BC = 7$ m
angle $BCD = \theta°$

(a) show that

\cos \theta° = \frac{3x + 29}{7(x + 3)}

(2 marks)

Given also that

$x = 5$
angle $BAD = 35°$
$C$ , $B$ and $A$ lie on a straight line

(b) find the area of the flower bed

ACD

, giving your answer, in m², to one decimal place.

(5 marks)

●●●●●Level 47 marksStart

Figure 1

Diagram NOT accurately drawn

The shaded area in Figure 1 shows the plan view of a café terrace.

The terrace consists of a sector

AOB

of a circle centre

O

joined to a triangle

OBC

.

Given that

$OA = OB = 5$ m
$OC = 8$ m
$BC = 6$ m
$AOC$ is a straight line

(a) show that angle

BOC

0.847

radians to 3 decimal places.

(3 marks)
(b) Find the total area of the terrace. Give your answer in m

^2

to 3 significant figures.

(3 marks)
(c) Find the perimeter of the terrace. Give your answer in metres to 3 significant figures.

(3 marks)

●●●●●Level 49 marksStart

The points

A

B

and

C

have coordinates

(1, 0, 2)

(5, 8, -2)

and

(7, 6, k)

respectively, where

k

is a constant.
(a) Given that angle

ACB

is a right angle, find the possible values of

k

. (3 marks)
(b) Given also that

k > 0

, find the exact shortest distance from

C

to the line

AB

. (3 marks)
(c) The point

D

is the point on

AB

that is closest to

C

. Find the ratio

AD : DB

in its simplest form. (2 marks)

●●●●●Level 58 marksStart

Relative to a fixed origin

O

, the points

A

B

and

C

have position vectors

\mathbf{a} = -\mathbf{i} + \mathbf{j} + 4\mathbf{k}

\qquad \mathbf{b} = 5\mathbf{i} + 4\mathbf{j} - 2\mathbf{k}

\qquad \mathbf{c} = 5\mathbf{i} + 7\mathbf{j} + 4\mathbf{k}

.
(a) Show that angle

BAC = \dfrac{\pi}{4}

. (4 marks)

The point

P

lies on the line segment

AB

and is such that angle

APC = \dfrac{\pi}{2}

.
(b) Find the position vector of

P

. (3 marks)
(c) Find the area of triangle

ABC

. (2 marks)

●●●●●Level 59 marksStart

The diagram shows triangle

PQR

. The perpendicular from

R

PQ

meets

PQ

N

. Angle

PRN = x

, angle

NRQ = y

and the length

RN = h

.
(a) (i) Find the length

PR

in terms of

h

and

x

.

(ii) Write down the length

QR

in terms of

h

and

y

.

(iii) By finding the area of triangle

PQR

in two different ways, show that

\sin(x + y) \equiv \sin x\cos y + \cos x\sin y

.

(6 marks)
(b) Given that

\sin(\alpha + 60^\circ) = \cos(\alpha - 45^\circ)

, show that

\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

OAB

of a circle with centre

O

and radius

r

metres, together with the triangles

OAC

and

OBD

, which are joined to the sector along

OA

and along

OB

, where

angle $AOB = 1.5$ radians
the arc $AB$ has length $12$ m
triangle $OAC$ and triangle $OBD$ are congruent, with the sector $OAB$ lying between them
$C$ , $O$ and $D$ lie on a straight line, which forms the back edge of the platform

(a) Show that

r = 8

.

(2 marks)
(b) Show that the size of angle

AOC

0.821

radians, correct to 3 decimal places.

(2 marks)

The front edge of the platform runs in a straight line from

C

A

, then along the arc

AB

, then in a straight line from

B

D

.

Given that the total length of the front edge of the platform is

32

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

Figure 1

Diagram NOT accurately drawn

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

The tile is made from a triangle

OPQ

and a sector

OQR

of a circle with centre

O

, joined along

OQ

.

In the design

$OQ = 7.5$ cm
$PQ = 4.2$ cm
angle $POQ = 0.45$ radians
angle $OPQ > \dfrac{\pi}{2}$ radians

Making your method clear,
(a) find the size of angle

OPQ

, giving your answer in radians to 4 significant figures,

(3 marks)
(b) find the area of triangle

OPQ

, in cm

^2

, giving your answer to 3 significant figures.

(2 marks)

Given that the area of sector

OQR

and the area of triangle

OPQ

are in the ratio

5 : 3

(c) show that angle

QOR

0.397

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 exact value of

\tan\dfrac{2\pi}{3}

is

Circle your answer.

A $\sqrt3$
B $-\dfrac{1}{\sqrt3}$
C $-\sqrt3$
D $\dfrac{1}{\sqrt3}$

●●●●●Level 11 markStart

The exact value of

\cos\dfrac{5\pi}{6}

is

Circle your answer.

A $\dfrac{\sqrt{3}}{2}$
B $-\dfrac{\sqrt{3}}{2}$
C $-\dfrac{1}{2}$
D $\dfrac{1}{2}$

●●●●●Level 11 markStart

Write down the exact value of

\tan 45^\circ

●●●●●Level 21 markStart

Write down the exact values of

\sin 30^\circ

and

\cos 60^\circ

●●●●●Level 21 markStart

Solve the equation

\tan^2 x = 3

for

0° < x < 360°

(3 marks)

●●●●●Level 23 marksStart

Write down the exact value of

\sin 45^\circ

, rationalising the denominator.

●●●●●Level 31 markStart

Write down the exact value of

\cos 30^\circ

●●●●●Level 31 markStart

Write down the exact values of

\sin 60^\circ

and

\tan 60^\circ

●●●●●Level 32 marksStart

The diagram below shows a sector of a circle

OAB

.

The chord

AB

divides the sector into a triangle and a shaded segment.

Angle

AOB

\dfrac{\pi}{2}

radians.

The radius of the sector is 14 cm.Show that the area of the shaded segment is

k(\pi - 2) \text{ cm}^2

where

k

is an integer to be found.

(3 marks)

●●●●●Level 33 marksStart

Find the exact value of

\sin 30^\circ + \cos 60^\circ + \tan 45^\circ

●●●●●Level 32 marksStart

In this question you must show detailed reasoning.

Solve the equation

\sin\left(2x - \dfrac{\pi}{6}\right) = \cos 2x

for

0 \le x < 2\pi

, giving your answers in exact form. (5 marks)

●●●●●Level 45 marksStart

The curve

C

has parametric equations

x = 4\cos t, \qquad y = 3\sin t, \qquad 0 \leqslant t < 2\pi

(a) Find

\dfrac{\mathrm{d}y}{\mathrm{d}x}

in terms of

t

. (3 marks)
(b) The point

P

C

has parameter

t = \dfrac{\pi}{3}

.

Find an equation of the normal to

C

P

, giving your answer in the form

y = mx + c

, where

m

and

c

are exact constants. (6 marks)

●●●●●Level 49 marksStart

Solve the equation

2\cos\left(2x - \dfrac{\pi}{3}\right) = 1

for

0 \leqslant x \leqslant \pi

, giving your answers in terms of

\pi

. (5 marks)

●●●●●Level 45 marksStart

It is given that the angle

\theta

satisfies the equation

\sin(2\theta + 45^\circ) = 4\sin(2\theta - 45^\circ)

.
(a) Show that

\tan 2\theta = \dfrac53

.

(3 marks)
(b) Hence find, in surd form, the exact value of

\tan\theta

, given that

\theta

is an obtuse angle.

(5 marks)

●●●●●Level 48 marksStart

(a) Use the result

\cos(A + B) = \cos A\cos B - \sin A\sin B

to show that

\cos(A - B) = \cos A\cos B + \sin A\sin B

.

(2 marks)

The function

\mathrm{f}(\theta)

is defined as

\mathrm{f}(\theta) = 5 + 8\cos(\theta + 60)^\circ\cos(\theta - 60)^\circ

, where

\theta

is in degrees.
(b) Show that

\mathrm{f}(\theta) = 7 - 8\sin^2\theta

.

(3 marks)
(c) (i) Determine the following.

The maximum value of $\mathrm{f}(\theta)$

The smallest positive value of $\theta$ for which this maximum value occurs

(ii) Determine the following.

The minimum value of $\mathrm{f}(\theta)$

The smallest positive value of $\theta$ for which this minimum value occurs

(4 marks)

●●●●●Level 49 marksStart

It is given that

(5\sin\theta + 4\cos\theta)^2 + (2\sin\theta - 10\cos\theta)^2 = 58

and that

\theta

is obtuse.

Find the exact value of

\sin\theta

Fully justify your answer.

(6 marks)

●●●●●Level 46 marksStart

The curve with equation

y = \sin x°

for

-360 \le x \le 360

is shown below.(a) Point

A

on the curve has coordinates

\left(a, \dfrac{\sqrt{3}}{2}\right)

(a) (i) Find the value of

a

(ii) State the value of

\sin(180° - a°)

(3 marks)
(b) Point

B

on the curve has coordinates

\left(b, -\dfrac{\sqrt{7}}{4}\right)

(b) (i) Find the exact value of

\sin(b° - 180°)

(ii) Find the exact value of

\cos b°

(5 marks)

●●●●●Level 48 marksStart

A sequence

u_1, u_2, u_3, \ldots

is defined by

u_1 = 9

u_{n+1} = u_n + 3\cos\left(\frac{n\pi}{2}\right)

(a) Find the value of

u_2

, the value of

u_3

, the value of

u_4

and the value of

u_5

(2 marks)
(b) State the order of this periodic sequence.

(1 mark)
(c) Find the value of

\displaystyle\sum_{n=1}^{62} u_n

(3 marks)

●●●●●Level 46 marksStart

A jeweller is designing a silver brooch.

The brooch is modelled as the region bounded by the chord

AB

and the arc

AB

of a circle with centre

O

and radius

16

cm, where angle

AOB = \dfrac{2\pi}{3}

radians.
(a) Use the cosine rule to show that

AB = 16\sqrt{3}

cm.

(2 marks)
(b) Find the exact perimeter of the brooch.

(2 marks)
(c) Find the exact area of the brooch, giving your answer in the form

(a\pi + b\sqrt{3})

^2

, where

a

and

b

are rational numbers.

(3 marks)

●●●●●Level 47 marksStart

The first, second and fourth terms of an arithmetic progression are

u_1

u_2

and

u_4

respectively, where

u_1 = 3\sin\theta, \qquad u_2 = 5\cos\theta, \qquad u_4 = -21\sin\theta

and

\tfrac12\pi < \theta < \pi

.
(a) Determine the exact value of

\theta

.

(3 marks)
(b) Hence determine the value of

\displaystyle\sum_{r=1}^{20} u_r

.

(3 marks)

●●●●●Level 46 marksStart

The curve

C

has parametric equations

x = 6\sin t, \qquad y = 2\sin 2t, \qquad 0 \leqslant t \leqslant \dfrac{\pi}{2}

(a) The point

P

C

is the point at which the tangent to

C

is parallel to the

x

-axis.

Find the exact coordinates of

P

. (4 marks)
(b) The finite region

R

is bounded by

C

and the

x

-axis.

Use integration to find the exact area of

R

. (5 marks)

●●●●●Level 59 marksStart

The curve

C

has parametric equations

x = 3\sin\theta

\qquad y = \sin 2\theta

\qquad 0 \le \theta \le \dfrac{\pi}{2}

.
(a) Find the exact coordinates of the maximum point

M

C

. (4 marks)

The finite region

R

is bounded by

C

and the

x

-axis.
(b) Find the exact area of

R

. (3 marks)
(c) Show that the line through

M

parallel to the

y

-axis divides

R

into two regions whose areas are in the ratio

\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

22\cos\theta \qquad\qquad 7 + 8\sin\theta \qquad\qquad 11\tan\theta

(a) show that

64\sin^2\theta - 130\sin\theta + 49 = 0

(3 marks)

Given that

\theta

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

\theta

(2 marks)
(c) Find the sum to infinity of the series, giving your answer in the form

a + b\sqrt{3}

, where

a

and

b

are rational constants to be found.

(5 marks)

●●●●●Level 510 marksStart

The diagram shows triangle

PQR

. The perpendicular from

R

PQ

meets

PQ

N

. Angle

PRN = x

, angle

NRQ = y

and the length

RN = h

.
(a) (i) Find the length

PR

in terms of

h

and

x

.

(ii) Write down the length

QR

in terms of

h

and

y

.

(iii) By finding the area of triangle

PQR

in two different ways, show that

\sin(x + y) \equiv \sin x\cos y + \cos x\sin y

.

(6 marks)
(b) Given that

\sin(\alpha + 60^\circ) = \cos(\alpha - 45^\circ)

, show that

\tan\alpha = \sqrt6 + \sqrt3 - \sqrt2 - 2

.

(5 marks)

●●●●●Level 511 marksStart

Graphs of sine, cosine and tangent

31 questions

LessonNot started

The period of the curve

y = 4\sin(3x)

, where

x

is in radians, is

Circle your answer.

A $\dfrac{2\pi}{3}$
B $2\pi$
C $6\pi$
D $4$

●●●●●Level 11 markStart

State the amplitude and the period of the curve

y = 3\sin 2x^{\circ}

. (2 marks)

●●●●●Level 22 marksStart

The curve

y = a\cos x^{\circ} + b

has maximum value

7

and minimum value

-1

, where

a > 0

.

Find the values of

a

and

b

. (3 marks)

●●●●●Level 33 marksStart

On the same axes, sketch

y = 2\sin x^\circ

and

y = 1

for

0 \le x \le 360

. State the number of points of intersection, and find the values of

x

at which they occur.

●●●●●Level 32 marksStart

Sketch the graph of

y = \cot\left(x + \frac{\pi}{3}\right)

for

0 \le x \le 2\pi

(3 marks)

●●●●●Level 33 marksStart

Solve the equation

\sin(x - 30)^{\circ} = 0.5

for

0 \leqslant x \leqslant 360

. (4 marks)

●●●●●Level 34 marksStart

(a) State the number of solutions of

\sin x^{\circ} = 0.3

for

0 \leqslant x \leqslant 720

. (2 marks)
(b) Explain your answer with reference to the graph. (1 mark)

●●●●●Level 33 marksStart

(a) Sketch the graph of

y = \sin \tfrac{3}{2}x

for

0° \le x \le 360°

(2 marks)
(b) The equation

\sin \tfrac{3}{2}x = A

has exactly two solutions for

0° \le x \le 360°

State the possible values of

A

(1 mark)

●●●●●Level 33 marksStart

The curve

C

has equation

y = 4 - 3\cos(2x - 60)^\circ \qquad 0 \leqslant x \leqslant 360

(a) Write down the maximum value of

y

.

(1 mark)
(b) Find the coordinates of each of the minimum points of

C

.

(3 marks)

●●●●●Level 34 marksStart

Figure 1

Figure 1 shows a sketch of part of the curve with equation

y = 6\sin x

where

x

is measured in degrees.

The points

P

and

Q

lie on the curve and are shown in Figure 1.
(a) State the coordinates of

P

.

(1 mark)
(b) State the coordinates of

Q

.

(1 mark)
(c) State the number of solutions of the equation

(i)

6\sin x = 5

in the interval

0 < x \leqslant 9000°

(ii)

4 + 6\sin x = -2

in the interval

-1440° < x \leqslant 1440°

(iii)

6\sin x + 7 = 0

in the interval

-720° \leqslant x \leqslant 720°

(3 marks)

●●●●●Level 35 marksStart

Figure 1

Figure 1 shows a sketch of

the curve with equation $y = \tan x$
the straight line $l$ with equation $y = -2x$

in the interval

-\pi < x < \pi

, where

x

is measured in radians.
(a) Write down the equations of the asymptotes to the curve

y = \tan x

in the interval

-\pi < x < \pi

(2 marks)
(b) Write down the number of roots of the equation

(i)

\tan x + 2x = 0

in the interval

-\pi < x < \pi

(ii)

\tan x + 2x = 0

in the interval

-3\pi < x < 3\pi

(iii)

\tan x + 2x = 0

in the interval

-40\pi < x < 40\pi

(3 marks)

●●●●●Level 35 marksStart

State the number of solutions of the equation

5\cos x^\circ = 2

in the interval

0 \le x \le 900

●●●●●Level 32 marksStart

Describe fully the two transformations that map the graph of

y = \cos x^{\circ}

onto the graph of

y = 4\cos\left(\dfrac{x}{2}\right)^{\circ}

, and state the period of the new graph. (3 marks)

●●●●●Level 33 marksStart

Find the values of

x

in the interval

0 \leqslant x \leqslant 360

for which the graphs of

y = \sin x^{\circ}

and

y = \cos x^{\circ}

intersect. (3 marks)

●●●●●Level 33 marksStart

For the curve

y = \tan x^{\circ}

0 \leqslant x \leqslant 360

:
(a) state the period; (1 mark)
(b) write down the equations of the vertical asymptotes. (2 marks)

●●●●●Level 33 marksStart

(a) Use the result

\cos(A + B) = \cos A\cos B - \sin A\sin B

to show that

\cos(A - B) = \cos A\cos B + \sin A\sin B

.

(2 marks)

The function

\mathrm{f}(\theta)

is defined as

\mathrm{f}(\theta) = 5 + 8\cos(\theta + 60)^\circ\cos(\theta - 60)^\circ

, where

\theta

is in degrees.
(b) Show that

\mathrm{f}(\theta) = 7 - 8\sin^2\theta

.

(3 marks)
(c) (i) Determine the following.

The maximum value of $\mathrm{f}(\theta)$

The smallest positive value of $\theta$ for which this maximum value occurs

(ii) Determine the following.

The minimum value of $\mathrm{f}(\theta)$

The smallest positive value of $\theta$ for which this minimum value occurs

(4 marks)

●●●●●Level 49 marksStart

The depth of water,

d

metres, over a sandbar at the mouth of an estuary on one day is modelled by the equation

d = 4.5 + 2.3\cos\big(29(t - 3.2)\big)^\circ,

where

t

is the number of hours after midnight.
(a) (i) Find the least depth of water over the sandbar given by this model.

(ii) Find the value of

t

when the least depth first occurs.

(3 marks)
(b) A yacht can only cross the sandbar when the depth of water is at least

3.1

metres.

Determine the two periods of time during the day between which this yacht will not be able to cross the sandbar. Give your answers correct to the nearest minute.

(5 marks)

After heavy rain, more water flows down the estuary. An improved model uses the equation

d = 4.5 + 0.03r + 2.3\cos\big(29(t - 3.2)\big)^\circ,

where

r

is the rainfall, in millimetres, over the previous 24 hours.
(c) Explain how this new equation could give an improved model.

(1 mark)

●●●●●Level 49 marksStart

The function f is defined by

\text{f}(x) = \arccos\left(\frac{x}{2}\right) \quad \text{for } 0 \le x \le a

The curve with equation

y = \text{f}(x)

is shown below.(a) State the value of

a

(1 mark)
(b) (i) On the diagram above, sketch the curve with equation

y = 2\cos x \quad \text{for } 0 \le x \le \frac{\pi}{2}

and sketch the line with equation

y = x \quad \text{for } 0 \le x \le \frac{\pi}{2}

(ii) Explain why the solution to the equation

x - 2\cos x = 0

must also be a solution to the equation

2\cos x = \arccos\left(\frac{x}{2}\right)

(5 marks)
(c) Use the Newton-Raphson method with

x_0 = 0.5

to find an approximate solution,

x_3

, to the equation

x - 2\cos x = 0

Give your answer to four decimal places.

(3 marks)

●●●●●Level 49 marksStart

The curve with equation

y = \sin x°

for

-360 \le x \le 360

is shown below.(a) Point

A

on the curve has coordinates

\left(a, \dfrac{\sqrt{3}}{2}\right)

(a) (i) Find the value of

a

(ii) State the value of

\sin(180° - a°)

(3 marks)
(b) Point

B

on the curve has coordinates

\left(b, -\dfrac{\sqrt{7}}{4}\right)

(b) (i) Find the exact value of

\sin(b° - 180°)

(ii) Find the exact value of

\cos b°

(5 marks)

●●●●●Level 48 marksStart

(a) Given that

5\cos x + 3\sin x \equiv R\cos(x - \alpha)

, where

R > 0

and

0 < \alpha < \dfrac{\pi}{2}

, find

R

exactly and find

\alpha

, in radians, correct to 3 decimal places.

(3 marks)

An arithmetic sequence has first term

5\cos x

and common difference

\sin x

.

The sum of the first

7

terms of this sequence is

S_7

.
(b) (i) Using your answer to part (a), find the exact maximum value of

S_7

x

varies.

(ii) Write down the smallest positive value of

x

for which

S_7

takes this maximum value.

(3 marks)

●●●●●Level 46 marksStart

A wind turbine has three blades that rotate at a constant rate about a horizontal axis.

The height above the ground,

H

m, of the tip of one of the blades,

t

seconds after the turbine begins to rotate, is modelled by

H = \left|A\sin(bt + \alpha)^\circ\right|

where

A

b

and

\alpha

are constants.

In the model

the greatest height of the tip of the blade above the ground is $66$ m
the tip of the blade is $12$ m above the ground when the turbine begins to rotate
one complete rotation of the blades takes $6$ seconds, during which the tip of the blade is at its greatest height once

(a) Find a complete equation for the model, giving

A

and

b

exactly and giving

\alpha

to 3 significant figures.

(4 marks)

A student suggests that the model

H = \left|A\sin(bt + \alpha)^\circ\right| + d

where

d

is a positive constant, would be more appropriate.
(b) Explain why the student is correct.

(1 mark)

●●●●●Level 45 marksStart

The number of hours of daylight,

D

, in a town on the day that is

t

days after 21 March is modelled by the equation

D = A + B\sin(kt)^\circ

where

A

B

and

k

are positive constants.

Given that, according to the model,

$D$ is periodic with period $365$ days
the greatest number of hours of daylight in the town is $16.4$
the least number of hours of daylight in the town is $7.6$

(a) find a complete equation for the model.

(3 marks)
(b) Write down the value of

t

on which, according to the model, the town has the greatest number of hours of daylight.

(1 mark)
(c) Use the model to find the number of hours of daylight in the town when

t = 41

, giving your answer in hours and minutes to the nearest minute.

(2 marks)

On 21 December, when

t = 275

, the town actually had

7.9

hours of daylight.
(d) Use this information to comment on the model.

(2 marks)

●●●●●Level 48 marksStart

In a warehouse, storage crates are carried around a vertical loop of track on a carousel.

Each crate makes repeated circuits of the loop. In a model

the greatest height of a crate above the floor is $14$ m
each circuit begins with the crate $4$ m above the floor
the floor is horizontal

The height,

H

m, of a crate above the floor,

t

seconds after it begins its first circuit, is modelled by

H = a - b(t - 10)^2

where

a

and

b

are positive constants.
(a) Find a complete equation for the model.

(3 marks)
(b) Find, according to the model, the height of the crate above the floor when

t = 20

(1 mark)

An alternative model for the height of the crate,

t

seconds after it begins its first circuit, is

H = 5\cos(18t + \alpha)^\circ + \beta

where

\alpha

and

\beta

are constants and

0 \leqslant \alpha < 360

(c) Find a complete equation for the alternative model.

(2 marks)

The carousel runs continuously for

3

minutes.
(d) Explain why the alternative model is more suitable than the first model.

(1 mark)

●●●●●Level 47 marksStart

(a) Express

8\cos\theta + 15\sin\theta

in the form

R\cos(\theta - \alpha)

, where

R > 0

and

0 < \alpha < 90^\circ

State the value of

R

and give the value of

\alpha

, in degrees, to 2 decimal places.

(3 marks)

The air temperature,

T\ ^\circ

C, at a weather station in a desert,

t

hours after midnight on a particular day, is modelled by the equation

T = 15 - 8\cos(15t)^\circ - 15\sin(15t)^\circ \qquad 0 \leqslant t < 24

(b) Using your answer to part (a), find

(i) the maximum temperature at the weather station on that day according to the model,

(ii) the time of day at which this maximum temperature occurs according to the model, giving your answer to the nearest minute.

(3 marks)

On that day, the lowest temperature recorded at the weather station was

4\ ^\circ

C, at 06:30
(c) Use this information to comment on the model.

(2 marks)

●●●●●Level 48 marksStart

A biologist models the number,

N

, of water fleas in a pond

t

days after she starts observing them by

N = 5200 - 1800\cos(15t)^\circ.

(a) State the greatest number of water fleas in the pond according to the model.

(1 mark)
(b) Write down the difference between the greatest and least values of

N

.

(1 mark)
(c) Without giving any numerical values, describe briefly how the number of water fleas changes as

t

increases.

(1 mark)
(d) Find how long it takes for

N

to first return to its value at

t = 0

.

(2 marks)
(e) Determine the time on the second occasion when

N = 6000

.

(4 marks)

The biologist observes the pond for several months. She notices that, although the population rises and falls in a way similar to the model, each peak is larger than the peak before it.
(f) Suggest a change to the model that will take account of this observation.

(1 mark)

●●●●●Level 410 marksStart

Figure 1

Figure 1 shows a sketch of part of the curve

C_1

with equation

y = -3\sin x

where

x

is measured in radians.

The point

P

is the maximum point on

C_1

with the smallest positive

x

coordinate. The point

Q

is a minimum point on

C_1

, as shown in Figure 1.
(a) State

(i) the coordinates of

P

(ii) the coordinates of

Q

(2 marks)

The curve

C_2

has equation

y = -3\sin x + k

where

k

is a constant.

Given that the maximum value of

y

C_2

7

(b) (i) state the value of

k

(ii) state the coordinates of the minimum point on

C_2

with the smallest positive

x

coordinate.

(3 marks)

The curve

C_3

has equation

y = -3\sin(x + B)

where

B

is a positive constant.

The minimum point on

C_3

with the smallest positive

x

coordinate is

\left(\dfrac{\pi}{3},\ A\right)

, where

A

is a constant.
(c) Find

(i) the value of

A

(ii) the smallest possible value of

B

.

(2 marks)
(d) Using Figure 1, state the number of solutions of the equation

6\sin x + x = 0 \qquad -2\pi \leqslant x \leqslant 2\pi

giving a reason for your answer.

(2 marks)

●●●●●Level 49 marksStart

(a) Given that

7\cos\theta - 24\sin\theta \equiv K\cos(\theta + \alpha)

where

K > 0

and

0 < \alpha < 90^\circ

, find the value of

K

and the value of

\alpha

, giving

\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,

M

, is modelled by

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

where

t

is the number of months after the start of the year and

A

is a constant.

According to the model, the greatest number of mice on the farm during the year is

145

(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,

N

, on the farm during the same year is modelled by

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

The number of owls is at its lowest when

t = T

(d) Use the two models to find the number of mice on the farm when

t = T

(4 marks)

●●●●●Level 511 marksStart

(a) Given that

8\sin\theta + 6\cos\theta \equiv R\sin(\theta + \alpha)

where

R > 0

and

0 < \alpha < \dfrac{\pi}{2}

, find the exact value of

R

and find

\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\ ^\circ

C, is modelled by

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

where

t

is the number of months after the start of the year and

A

is a constant.

According to the model, the highest water temperature in the first pond during the year is

23\ ^\circ

C
(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\ ^\circ

C, during the same year is modelled by

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 = T

(d) Use the two models to find the water temperature in the first pond when

t = T

, giving your answer to 3 significant figures.

(4 marks)

●●●●●Level 511 marksStart

The depth of water in a harbour,

d

metres, is modelled by

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

where

t

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

6

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 < \pi

, the equation

3\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\%

, inside the gallery,

t

hours after midnight, is modelled by the equation

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

where

k

is a constant.

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

Given that

the relative humidity inside the gallery was $60\%$ at 5 am
$0 < k < 90$

(b) find all possible values for

k

, giving each answer to 2 decimal places.

(4 marks)

Given further that

0 < 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

6\cos 2x + 5\sin 2x

in the form

R\cos(2x - \alpha)

, where

R > 0

and

0 < \alpha < \tfrac12\pi

. Give the exact value of

R

and give the value of

\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 x

to the curve

y = 6\cos 2x + 5\sin 2x

.

(4 marks)
(c) Determine the greatest value of

6\cos 2x + 5\sin 2x

x

varies and give the smallest positive value of

x

for which it occurs.

(2 marks)
(d) Determine the least value of

6\cos 2x + 5\sin 2x

x

varies and give the smallest positive value of

x

for which it occurs.

(2 marks)

●●●●●Level 511 marksStart

Trigonometric identities

19 questions

LessonNot started

Solve the equation

4\sin^2\theta = 3 + \cos\theta

for

0^\circ \le \theta < 360^\circ

, giving your answers to 1 decimal place. (6 marks)

●●●●●Level 36 marksStart

Given that

\cos\theta \ne 0

, prove the identity

\frac{\sin\theta}{1 - \sin\theta} - \frac{\sin\theta}{1 + \sin\theta} \equiv 2\tan^2\theta

(4 marks)

●●●●●Level 34 marksStart

Show that

\tan^2 x - \sin^2 x \equiv \sin^2 x \tan^2 x

●●●●●Level 33 marksStart

Show that

\dfrac{\sin\theta}{1 + \cos\theta} + \dfrac{1 + \cos\theta}{\sin\theta} \equiv \dfrac{2}{\sin\theta}

●●●●●Level 33 marksStart

For

0 \le t < 2\pi

, a curve

C

has parametric equations

x = 5 - k + 3k\sin t, \qquad y = -2 + 6\cos t,

where

k

is a non-zero constant.
(a) Find a cartesian equation for

C

. You do not need to simplify your answer.

(2 marks)

You are given that

C

is a circle.
(b) (i) Determine the radius of

C

.

(ii) Find the possible coordinates of the centre of

C

.

(4 marks)

●●●●●Level 36 marksStart

Solve the equation

3\sin^{2}\theta + 4\cos\theta = 4

for

0^{\circ} \leqslant \theta \leqslant 360^{\circ}

. (5 marks)

●●●●●Level 35 marksStart

Solve, for

0 \leqslant x \leqslant \pi

, the equation

\sin 2x = \tan x

giving your answers in terms of

\pi

. (5 marks)

●●●●●Level 45 marksStart

(a) Show that the equation

6\cos^2\theta + \sin\theta - 5 = 0

can be written as

6\sin^2\theta - \sin\theta - 1 = 0

. (2 marks)
(b) Hence solve

6\cos^2\theta + \sin\theta - 5 = 0

for

0^\circ \le \theta < 360^\circ

, giving your answers to 1 decimal place where appropriate. (5 marks)

●●●●●Level 47 marksStart

It is given that

(5\sin\theta + 4\cos\theta)^2 + (2\sin\theta - 10\cos\theta)^2 = 58

and that

\theta

is obtuse.

Find the exact value of

\sin\theta

Fully justify your answer.

(6 marks)

●●●●●Level 46 marksStart

The curve with equation

y = \sin x°

for

-360 \le x \le 360

is shown below.(a) Point

A

on the curve has coordinates

\left(a, \dfrac{\sqrt{3}}{2}\right)

(a) (i) Find the value of

a

(ii) State the value of

\sin(180° - a°)

(3 marks)
(b) Point

B

on the curve has coordinates

\left(b, -\dfrac{\sqrt{7}}{4}\right)

(b) (i) Find the exact value of

\sin(b° - 180°)

(ii) Find the exact value of

\cos b°

(5 marks)

●●●●●Level 48 marksStart

The first three terms of a geometric series are

\cos\theta \qquad\qquad 1 + \sin\theta \qquad\qquad 4\cos\theta

(a) Show that

5\sin^2\theta + 2\sin\theta - 3 = 0

(3 marks)

Given that

0 < \theta < \dfrac{\pi}{2}

,
(b) find the value of

\theta

, giving your answer to 3 decimal places.

(2 marks)
(c) Find the sum of the first

10

terms of the series.

(3 marks)

●●●●●Level 48 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 < \pi

, the equation

5\sin(2x + 0.6) + 2 = 0

giving your answers in radians to 2 decimal places.

(4 marks)
(b) Solve, for

-180^\circ < \theta < 180^\circ

, the equation

3\tan\theta\sin\theta = 5 + 2\cos\theta

giving your answers, in degrees, to one decimal place.

(5 marks)

●●●●●Level 49 marksStart

In this question you must show all stages of your working.

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

4\cos\theta\,(2 - \tan\theta\sin\theta) = 2 + 3\cos\theta

may be written as

4\cos^2\theta + 5\cos\theta - 6 = 0

(3 marks)
(b) Hence solve, for

0 < x < 180^\circ

4\cos 3x\,(2 - \tan 3x\sin 3x) = 2 + 3\cos 3x

giving your answers to one decimal place.

(4 marks)

●●●●●Level 47 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 \leqslant x < \pi

, the equation

\sqrt{3}\tan\left(3x - \dfrac{\pi}{8}\right) = 1

giving the answers in radians in the form

k\pi

, where

k

is a rational constant to be found.

(3 marks)
(b) Solve, for

0 \leqslant \theta < 360^\circ

, the equation

4\sin\theta\tan\theta = 3 - \cos\theta

giving your answers, in degrees, to one decimal place.

(5 marks)

●●●●●Level 48 marksStart

In this question you must show detailed reasoning.

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

\tan\theta\,(4\sin\theta + 1) = 6\cos\theta

can be written as

10\sin^2\theta + \sin\theta - 6 = 0

(3 marks)
(b) Hence solve, for

-\pi < x < \pi

, the equation

\tan\left(x - \dfrac{\pi}{5}\right)\left(4\sin\left(x - \dfrac{\pi}{5}\right) + 1\right) = 6\cos\left(x - \dfrac{\pi}{5}\right)

giving your answers to 3 significant figures.

(4 marks)

●●●●●Level 47 marksStart

In this question you must show all stages of your working.

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

(\tan\theta + \sin\theta)(1 - \cos\theta) \equiv \sin^2\theta\tan\theta

(4 marks)
(b) Hence solve, for

0 < x < 2\pi

, the equation

(\tan x + \sin x)(1 - \cos x) = \tan x - 2\sin x\tan x

giving your answers to 3 significant figures.

(4 marks)

●●●●●Level 48 marksStart

(a) Given that

\sin\theta \ne \pm 1

, prove the identity

\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

A

for which the equation

\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

\theta

is obtuse and

\frac{3}{2 - 2\sin\theta} + \frac{3}{2 + 2\sin\theta} = 10

find the exact value of

\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

22\cos\theta \qquad\qquad 7 + 8\sin\theta \qquad\qquad 11\tan\theta

(a) show that

64\sin^2\theta - 130\sin\theta + 49 = 0

(3 marks)

Given that

\theta

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

\theta

(2 marks)
(c) Find the sum to infinity of the series, giving your answer in the form

a + b\sqrt{3}

, where

a

and

b

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 < \pi

, the equation

3\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\%

, inside the gallery,

t

hours after midnight, is modelled by the equation

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

where

k

is a constant.

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

Given that

the relative humidity inside the gallery was $60\%$ at 5 am
$0 < k < 90$

(b) find all possible values for

k

, giving each answer to 2 decimal places.

(4 marks)

Given further that

0 < k < 20

●●●●●Level 512 marksStart

Solving trigonometric equations

57 questions

LessonNot started

Solve the equation

\tan^2 x = 3

for

0° < x < 360°

(3 marks)

●●●●●Level 23 marksStart

Solve the equation

4\sin^2\theta = 3 + \cos\theta

for

0^\circ \le \theta < 360^\circ

, giving your answers to 1 decimal place. (6 marks)

●●●●●Level 36 marksStart

Solve

4\sin^2 x^\circ = 3

for

0 \le x \le 360

●●●●●Level 33 marksStart

(a) For a small angle

\theta

, where

\theta

is in radians, show that

3\cos 2\theta + 4\tan\theta \approx 3 + 4\theta - 6\theta^2

.

(3 marks)
(b) Hence determine an approximate solution to

3\cos 2\theta + 4\tan\theta = 7\sin 3\theta

.

(2 marks)

●●●●●Level 35 marksStart

(a) Sketch the graph of

y = \sin \tfrac{3}{2}x

for

0° \le x \le 360°

(2 marks)
(b) The equation

\sin \tfrac{3}{2}x = A

has exactly two solutions for

0° \le x \le 360°

State the possible values of

A

(1 mark)

●●●●●Level 33 marksStart

Solve

\sin 2x^\circ = 0.5

for

0 \le x \le 360

●●●●●Level 34 marksStart

In this question you must show detailed reasoning.

Solve the equation

2\tan^2 x + 5\sec x + 4 = 0

for

0^\circ \le x \le 360^\circ

. (6 marks)

●●●●●Level 36 marksStart

(a) For a small angle

\theta

, where

\theta

is in radians, show that

5\cos^2\theta - 3\tan 2\theta \approx 5 - 6\theta - 5\theta^2

.

(3 marks)
(b) Hence determine an approximate solution to

5\cos^2\theta - 3\tan 2\theta = 9\sin 2\theta

.

(2 marks)

●●●●●Level 35 marksStart

Solve, for

0 \leqslant \theta < 2\pi

, the equation

2\cos^2\theta + 5\sin\theta = 4

giving your answers in terms of

\pi

.

(4 marks)

●●●●●Level 34 marksStart

Solve the equation

3\sin^{2}\theta + 4\cos\theta = 4

for

0^{\circ} \leqslant \theta \leqslant 360^{\circ}

. (5 marks)

●●●●●Level 35 marksStart

Solve

\cos(x - 30)^\circ = 0.5

for

0 \le x \le 360

●●●●●Level 33 marksStart

In this question you must show all stages of your working.

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

5\sin(x - 30^\circ) = 2\cos(x - 60^\circ)

show that

\tan x = \frac{7\sqrt{3}}{9}

(4 marks)
(b) Hence or otherwise solve, for

0 \leqslant \theta < 180^\circ

, the equation

5\sin(2\theta - 30^\circ) = 2\cos(2\theta - 60^\circ)

giving your answers to one decimal place.

(4 marks)

●●●●●Level 48 marksStart

(a) Express

4\sin x - 3\cos x

in the form

R\sin(x - \alpha)

, where

R > 0

and

0 < \alpha < 90^\circ

. Give

\alpha

to 2 decimal places. (3 marks)
(b) Hence solve, for

0 \le x < 360^\circ

, the equation

4\sin x - 3\cos x = 2.5

,

giving your answers to 1 decimal place. (4 marks)
(c) Write down the maximum value of

\dfrac{20}{7 + 4\sin x - 3\cos x}

,

and justify your answer. (2 marks)

●●●●●Level 49 marksStart

Solve, for

0 < x < 360^\circ

, the equation

2\cot^2 x + 5\operatorname{cosec} x + 1 = 0

,

giving your answers to 1 decimal place. You must show that one possible value of

\operatorname{cosec} x

gives no solutions, with a reason. (7 marks)

●●●●●Level 47 marksStart

Solve, for

0 \leqslant x \leqslant \pi

, the equation

\sin 2x = \tan x

giving your answers in terms of

\pi

. (5 marks)

●●●●●Level 45 marksStart

(a) Prove that

\dfrac{1 - \cos 2x}{\sin 2x} \equiv \tan x

\quad \sin 2x \ne 0

. (4 marks)
(b) Hence solve, for

0 < \theta < 360^\circ

, the equation

\dfrac{1 - \cos 2\theta}{\sin 2\theta} = 3\sin\theta

,

giving your answers to 1 decimal place. (5 marks)

●●●●●Level 49 marksStart

In this question you must show detailed reasoning.

Solve the equation

\sin\left(2x - \dfrac{\pi}{6}\right) = \cos 2x

for

0 \le x < 2\pi

, giving your answers in exact form. (5 marks)

●●●●●Level 45 marksStart

Solve, for

-180^\circ \leqslant \theta < 180^\circ

, the equation

4\cos 2\theta + 14\sin\theta = 7

giving your answers to

1

decimal place. (7 marks)

●●●●●Level 47 marksStart

Solve the equation

2\cos\left(2x - \dfrac{\pi}{3}\right) = 1

for

0 \leqslant x \leqslant \pi

, giving your answers in terms of

\pi

. (5 marks)

●●●●●Level 45 marksStart

(a) Prove that

\dfrac{\sec\theta}{\sec\theta - \cos\theta} \equiv \operatorname{cosec}^2\theta, \qquad \theta \neq \dfrac{n\pi}{2},\; n \in \mathbb{Z}

(3 marks)
(b) Hence solve, for

0 < \theta < 2\pi

, the equation

\dfrac{\sec\theta}{\sec\theta - \cos\theta} = 4

giving your answers in terms of

\pi

. (3 marks)

●●●●●Level 46 marksStart

(a) Show that

(\sin\theta + \cos\theta)^2 \equiv 1 + \sin 2\theta

. (2 marks)
(b) Hence solve the equation

(\sin\theta + \cos\theta)^2 = \frac{3}{2}

for

0^\circ \le \theta < 360^\circ

. (4 marks)
(c) Determine the greatest possible value of

\sin\theta + \cos\theta

, justifying your answer. (2 marks)

●●●●●Level 48 marksStart

(a) Show that the equation

6\cos^2\theta + \sin\theta - 5 = 0

can be written as

6\sin^2\theta - \sin\theta - 1 = 0

. (2 marks)
(b) Hence solve

6\cos^2\theta + \sin\theta - 5 = 0

for

0^\circ \le \theta < 360^\circ

, giving your answers to 1 decimal place where appropriate. (5 marks)

●●●●●Level 47 marksStart

(a) Express

2\cos\theta + 3\sin\theta

in the form

R\cos(\theta - \alpha)

, where

R > 0

and

0^\circ < \alpha < 90^\circ

, giving the exact value of

R

and the value of

\alpha

correct to 2 decimal places. (3 marks)
(b) Hence solve the equation

2\cos\theta + 3\sin\theta = 1

for

0^\circ < \theta < 360^\circ

, giving your answers to 1 decimal place. (4 marks)
(c) Determine the minimum value of

2\cos\theta + 3\sin\theta

, and the value of

\theta

in the interval

0^\circ < \theta < 360^\circ

at which it occurs. (2 marks)

●●●●●Level 49 marksStart

In this question you must show detailed reasoning.
(a) Prove that

\tan\left(\theta + \dfrac{\pi}{4}\right) + \tan\left(\theta - \dfrac{\pi}{4}\right) \equiv 2\tan 2\theta

for all values of

\theta

for which both sides are defined. (4 marks)
(b) Hence solve the equation

\tan\left(\theta + \dfrac{\pi}{4}\right) + \tan\left(\theta - \dfrac{\pi}{4}\right) = -2\sqrt3

for

0 \leqslant \theta < \pi

, giving your answers in terms of

\pi

. (4 marks)

●●●●●Level 48 marksStart

The depth of water,

d

metres, over a sandbar at the mouth of an estuary on one day is modelled by the equation

d = 4.5 + 2.3\cos\big(29(t - 3.2)\big)^\circ,

where

t

is the number of hours after midnight.
(a) (i) Find the least depth of water over the sandbar given by this model.

(ii) Find the value of

t

when the least depth first occurs.

(3 marks)
(b) A yacht can only cross the sandbar when the depth of water is at least

3.1

d = 4.5 + 0.03r + 2.3\cos\big(29(t - 3.2)\big)^\circ,

where

r

is the rainfall, in millimetres, over the previous 24 hours.
(c) Explain how this new equation could give an improved model.

(1 mark)

●●●●●Level 49 marksStart

(a) Show that the expression

3\sin 2\theta \sec\theta - \cos 2\theta \, \operatorname{cosec}\,\theta

can be written as

8\sin\theta - \operatorname{cosec}\,\theta

where

\sin\theta \ne 0

and

\cos\theta \ne 0

(4 marks)
(b) A student is attempting to solve the equation

3\sin 2\theta \sec\theta - \cos 2\theta \, \operatorname{cosec}\,\theta = 7

for

0° \le \theta \le 360°

They use the result from part (a), and write the following incorrect solution:

Step 1

\quad 8\sin\theta - \operatorname{cosec}\,\theta = 7

Step 2

\quad 8\sin^2\theta - 1 = 7\sin\theta

Step 3

\quad 8\sin^2\theta - 7\sin\theta - 1 = 0

Step 4

\quad (8\sin\theta + 1)(\sin\theta - 1) = 0

, so

\sin\theta = -\dfrac{1}{8}

\sin\theta = 1

Step 5

\quad \theta = 90°,\ 187.2°,\ 352.8°

(b) (i) Explain why the student should reject one of their values for

\sin\theta

in Step 4.

(ii) State the correct solutions to the equation

3\sin 2\theta \sec\theta - \cos 2\theta \, \operatorname{cosec}\,\theta = 7

for

0° \le \theta \le 360°

(2 marks)

●●●●●Level 46 marksStart

(a) Prove that

\dfrac{1}{1 - \sin\theta} + \dfrac{1}{1 + \sin\theta} \equiv 2\sec^2\theta

(3 marks)
(b) Hence solve, for

0^\circ \leqslant \theta < 360^\circ

, the equation

\dfrac{1}{1 - \sin\theta} + \dfrac{1}{1 + \sin\theta} = 8

(4 marks)

●●●●●Level 47 marksStart

(a) Express

5\cos\theta - 12\sin\theta

in the form

R\cos(\theta + \alpha)

, where

R > 0

and

0^{\circ} < \alpha < 90^{\circ}

. Give

\alpha

correct to

2

decimal places. (3 marks)
(b) Hence solve the equation

5\cos\theta - 12\sin\theta = 4

for

0^{\circ} \leqslant \theta \leqslant 360^{\circ}

. (3 marks)

●●●●●Level 46 marksStart

The first three terms of a geometric series are

\cos\theta \qquad\qquad 1 + \sin\theta \qquad\qquad 4\cos\theta

(a) Show that

5\sin^2\theta + 2\sin\theta - 3 = 0

(3 marks)

Given that

0 < \theta < \dfrac{\pi}{2}

,
(b) find the value of

\theta

, giving your answer to 3 decimal places.

(2 marks)
(c) Find the sum of the first

10

terms of the series.

(3 marks)

●●●●●Level 48 marksStart

(a) Prove that

\dfrac{1 + \sec\theta}{\tan\theta + \sin\theta} \equiv \operatorname{cosec}\theta

.

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

Hence solve, for

0 < \theta < 2\pi

, the equation

\frac{1 + \sec\theta}{\tan\theta + \sin\theta} = 6 - 4\sin\theta.

Give your answers correct to 3 decimal places.

(5 marks)

●●●●●Level 49 marksStart

In this question you must show detailed reasoning.
(a) Show that the equation

4\sec\theta + m\sin\theta + 5\cos\theta = 0

, where

m

is a constant, can be expressed in the form

4\tan^2\theta + m\tan\theta + 9 = 0.

(3 marks)
(b) It is given that there is only one value of

\theta

, for

0^\circ < \theta < 180^\circ

, satisfying the equation

4\sec\theta + m\sin\theta + 5\cos\theta = 0

.

Given also that

m

is negative, find this value of

\theta

, correct to 3 significant figures.

(5 marks)

●●●●●Level 48 marksStart

A particle

P

moves in a straight line. At time

t

seconds, where

t \ge 0

, the velocity of

P

v

m s⁻¹, where

v = 4\cos t + 7\sin t.

(a) Express

4\cos t + 7\sin t

in the form

R\cos(t - \alpha)

, where

R > 0

and

0 < \alpha < \tfrac12\pi

. Give the value of

\alpha

correct to 4 significant figures.

(3 marks)
(b) Hence find the two smallest positive values of

t

for which

P

is moving, in either direction, with a speed of

6

m s⁻¹.

(3 marks)

●●●●●Level 46 marksStart

In this question you must show all stages of your working.

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

\frac{\cot\theta}{\operatorname{cosec}\theta - 1} - \frac{\cot\theta}{\operatorname{cosec}\theta + 1} \equiv 2\tan\theta \qquad \theta \neq \frac{n\pi}{2},\ n \in \mathbb{Z}

(3 marks)
(b) Hence solve, for

0 < x < 180^\circ

, the equation

\frac{4\cot x}{\operatorname{cosec} x - 1} - \frac{4\cot x}{\operatorname{cosec} x + 1} = 4\sec^2 x - 1

Give each answer, in degrees, to one decimal place.

(4 marks)

●●●●●Level 47 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

\frac{4\sin 2\theta}{1 + \tan^2\theta} + 8\cos^3\theta\tan\theta = 6\cos^2\theta\tan\theta

may be written as

\sin 2\theta\left(A\cos^2\theta + B\cos\theta + C\right) = 0

where

A

B

and

C

are constants to be found.

(3 marks)
(b) Hence solve, for

540^\circ \leqslant x \leqslant 720^\circ

, the equation

\frac{4\sin 2x}{1 + \tan^2 x} + 8\cos^3 x\tan x = 6\cos^2 x\tan x \qquad x \in \mathbb{R},\ x \neq 630^\circ

(4 marks)

●●●●●Level 47 marksStart

(a) Write

2x^2 - 8x + 11

in the form

p(x + q)^2 + r

, where

p

q

and

r

are constants.

(3 marks)
(b) State the coordinates of the minimum point on the graph of

y = 2x^2 - 8x + 11

.

(2 marks)
(c) Hence deduce

the minimum value of $2\tan^2\theta - 8\tan\theta + 11$ ,
the smallest positive value of $\theta$ , in degrees, for which the minimum value occurs.

(3 marks)

●●●●●Level 48 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 < \pi

, the equation

5\sin(2x + 0.6) + 2 = 0

giving your answers in radians to 2 decimal places.

(4 marks)
(b) Solve, for

-180^\circ < \theta < 180^\circ

, the equation

3\tan\theta\sin\theta = 5 + 2\cos\theta

giving your answers, in degrees, to one decimal place.

(5 marks)

●●●●●Level 49 marksStart

In this question you must show all stages of your working.

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

4\cos\theta\,(2 - \tan\theta\sin\theta) = 2 + 3\cos\theta

may be written as

4\cos^2\theta + 5\cos\theta - 6 = 0

(3 marks)
(b) Hence solve, for

0 < x < 180^\circ

4\cos 3x\,(2 - \tan 3x\sin 3x) = 2 + 3\cos 3x

giving your answers to one decimal place.

(4 marks)

●●●●●Level 47 marksStart

The first, second and fourth terms of an arithmetic progression are

u_1

u_2

and

u_4

respectively, where

u_1 = 3\sin\theta, \qquad u_2 = 5\cos\theta, \qquad u_4 = -21\sin\theta

and

\tfrac12\pi < \theta < \pi

.
(a) Determine the exact value of

\theta

.

(3 marks)
(b) Hence determine the value of

\displaystyle\sum_{r=1}^{20} u_r

.

(3 marks)

●●●●●Level 46 marksStart

A biologist models the number,

N

, of water fleas in a pond

t

days after she starts observing them by

N = 5200 - 1800\cos(15t)^\circ.

(a) State the greatest number of water fleas in the pond according to the model.

(1 mark)
(b) Write down the difference between the greatest and least values of

N

.

(1 mark)
(c) Without giving any numerical values, describe briefly how the number of water fleas changes as

t

increases.

(1 mark)
(d) Find how long it takes for

N

to first return to its value at

t = 0

.

(2 marks)
(e) Determine the time on the second occasion when

N = 6000

●●●●●Level 410 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 \leqslant x < \pi

, the equation

\sqrt{3}\tan\left(3x - \dfrac{\pi}{8}\right) = 1

giving the answers in radians in the form

k\pi

, where

k

is a rational constant to be found.

(3 marks)
(b) Solve, for

0 \leqslant \theta < 360^\circ

, the equation

4\sin\theta\tan\theta = 3 - \cos\theta

giving your answers, in degrees, to one decimal place.

(5 marks)

●●●●●Level 48 marksStart

In this question you must show detailed reasoning.

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

\tan\theta\,(4\sin\theta + 1) = 6\cos\theta

can be written as

10\sin^2\theta + \sin\theta - 6 = 0

(3 marks)
(b) Hence solve, for

-\pi < x < \pi

, the equation

\tan\left(x - \dfrac{\pi}{5}\right)\left(4\sin\left(x - \dfrac{\pi}{5}\right) + 1\right) = 6\cos\left(x - \dfrac{\pi}{5}\right)

giving your answers to 3 significant figures.

(4 marks)

●●●●●Level 47 marksStart

In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.

Figure 1Figure 1 shows a sketch of the curve with equation

y = \text{f}(x)

, where

\text{f}(x) = 3\cos x\,\text{e}^{2\sin x} \qquad 0 \leqslant x \leqslant \pi

The curve intersects the

x

-axis at the point

R

, as shown in Figure 1.
(a) State the coordinates of

R

(1 mark)

The curve has two turning points, at the point

P

and the point

Q

, also shown in Figure 1.
(b) Show that, at

P

and at

Q

a\sin^2 x + b\sin x + c = 0

where

a

b

and

c

are integers to be found.

(4 marks)
(c) Hence find the

x

coordinate of

Q

, giving your answer to 3 decimal places.

(2 marks)

●●●●●Level 47 marksStart

In this question you must show all stages of your working.

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

(\tan\theta + \sin\theta)(1 - \cos\theta) \equiv \sin^2\theta\tan\theta

(4 marks)
(b) Hence solve, for

0 < x < 2\pi

, the equation

(\tan x + \sin x)(1 - \cos x) = \tan x - 2\sin x\tan x

giving your answers to 3 significant figures.

(4 marks)

●●●●●Level 48 marksStart

(a) Prove that

\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

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

for

0 \le x < \pi

, giving your answers in exact form. (5 marks)

●●●●●Level 59 marksStart

(a) Express

2\cos x - 2\sqrt{3}\sin x

in the form

R\cos(x + \alpha)

, where

R > 0

and

0 < \alpha < \dfrac{\pi}{2}

. (3 marks)

The depth of water,

h

metres, in a harbour

t

hours after midnight is modelled by

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

7

metres.
(b) Find the total length of time, during these

24

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 \equiv 3\sin x - 4\sin^3 x

. (4 marks)
(b) Hence solve, for

0 \le x < 2\pi

, the equation

\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

\displaystyle\int_0^{\frac{\pi}{3}} \sin^3 x\,\mathrm{d}x

. (2 marks)

●●●●●Level 59 marksStart

(a) Prove that

\cos 3\theta \equiv 4\cos^3\theta - 3\cos\theta

(4 marks)
(b) Hence solve, for

0 \leqslant \theta < \pi

, the equation

8\cos^3\theta - 6\cos\theta + \sqrt{3} = 0

giving your answers in exact form in terms of

\pi

. (5 marks)

●●●●●Level 59 marksStart

The depth of water,

D

metres, at the entrance to a harbour is modelled by

D = 9 + 3\sin\left(\dfrac{\pi t}{6}\right) - 3\sqrt{3}\cos\left(\dfrac{\pi t}{6}\right)

\qquad 0 \le t < 24

,

where

t

is the time in hours after midnight.
(a) Express

3\sin\theta - 3\sqrt{3}\cos\theta

in the form

R\sin(\theta - \alpha)

, where

R > 0

and

0 < \alpha < \dfrac{\pi}{2}

, stating the exact values of

R

and

\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

12

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,

D

metres, in a harbour at a time

t

hours after midnight is modelled by

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

where the angle

0.5t

is measured in radians.
(a) Express

3\cos\theta - 4\sin\theta

in the form

R\cos(\theta + \alpha)

, where

R > 0

and

0 < \alpha < \dfrac{\pi}{2}

.

Give the exact value of

R

and the value of

\alpha

4

decimal places. (3 marks)
(b) Find the maximum depth of water in the harbour, and the value of

t

at which this maximum first occurs, giving

t

2

decimal places. (4 marks)
(c) A ship can enter the harbour only when the depth of water is at least

10

metres.

Find the two values of

t

between

t = 6

and

t = 18

for which the depth is exactly

10

metres, giving each value to

2

decimal places. (5 marks)

●●●●●Level 512 marksStart

(a) Prove that

\dfrac{1}{\sec\theta - \tan\theta} \equiv \sec\theta + \tan\theta

(3 marks)
(b) Hence solve the equation

\sec\theta - \tan\theta = 3

for

0 \leqslant \theta \leqslant 2\pi

, giving your answer to

3

decimal places. (4 marks)

●●●●●Level 57 marksStart

(a) Prove that

\cot\theta - \tan\theta \equiv 2\cot 2\theta

(4 marks)
(b) Hence solve, for

0 < \theta < \pi

, the equation

\cot\theta - \tan\theta = 2\sqrt3

giving your answers in terms of

\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

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

may be written as

3\cot^2\theta + 2\cot\theta - 8 = 0

(4 marks)
(b) Hence solve, for

180^\circ < x < 540^\circ

, the equation

\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 + 6\sin x)

, where

x

is in radians.
(a) State the values of

x

, for

0 \le x \le 2\pi

, for which

\ln(3 + 6\sin x)

is not defined.

(2 marks)
(b)

P

is the stationary point on the curve that has the smallest positive

x

-coordinate.

Determine the exact coordinates of

P

.

(4 marks)
(c) (i) Show that

\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 + 2\sin x

with domain

0 \le x \le 2\pi

.
(a) Solve the following equations.

(i)

f'(x) = 0

(ii)

f''(x) = 0

(7 marks)

The diagram shows the graph of the gradient function

y = f'(x)

for the domain

0 \le x \le 2\pi

. The points

A

C

and

E

are where the graph meets the

x

-axis, and

B

and

D

are minimum points.(b) Use your answers to parts (a)(i) and (a)(ii) to find the coordinates of points

A

B

C

D

and

E

.

(3 marks)
(c) (i) Explain how to use the graph of the gradient function to find the values of

x

for which

f(x)

is increasing.

(ii) Using set notation, write down the set of values of

x

for which

f(x)

is increasing in the domain

0 \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 < \pi

, the equation

3\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\%

, inside the gallery,

t

hours after midnight, is modelled by the equation

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

where

k

is a constant.

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

Given that

the relative humidity inside the gallery was $60\%$ at 5 am
$0 < k < 90$

(b) find all possible values for

k

, giving each answer to 2 decimal places.

(4 marks)

Given further that

0 < k < 20

●●●●●Level 512 marksStart

(a) Show that the equation

4\operatorname{cosec}x + \cot x\cos x - 40 = 0

can be expressed in the form

\sin^2 x + 40\sin x - 5 = 0.

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

Hence solve, for

0 < \theta < \tfrac12\pi

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 3\theta

is used to find an approximation for the smallest positive solution of the equation

4\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)

, and the double angle formulae, to show that

\cos 3\theta \equiv 4\cos^3\theta - 3\cos\theta

.

(ii) By writing

-\cos 2\theta

\cos(180^\circ - 2\theta)

, solve the equation

\cos 3\theta + \cos 2\theta = 0

for

0^\circ \le \theta \le 180^\circ

.

(5 marks)
(b) (i) Show that

4x^3 + 2x^2 - 3x - 1 = (x + 1)(4x^2 - 2x - 1)

.

(ii) Hence find the exact roots of the equation

4x^3 + 2x^2 - 3x - 1 = 0

.

(3 marks)
(c) Use the results from parts (a)(ii) and (b)(ii) to show that

\cos 36^\circ = \dfrac{1 + \sqrt5}{4}

.

(2 marks)

●●●●●Level 510 marksStart

Radian measure

31 questions

LessonNot started

The exact value of

\tan\dfrac{2\pi}{3}

is

Circle your answer.

A $\sqrt3$
B $-\dfrac{1}{\sqrt3}$
C $-\sqrt3$
D $\dfrac{1}{\sqrt3}$

●●●●●Level 11 markStart

The period of the curve

y = 4\sin(3x)

, where

x

is in radians, is

Circle your answer.

A $\dfrac{2\pi}{3}$
B $2\pi$
C $6\pi$
D $4$

●●●●●Level 11 markStart

The exact value of

\cos\dfrac{5\pi}{6}

is

Circle your answer.

A $\dfrac{\sqrt{3}}{2}$
B $-\dfrac{\sqrt{3}}{2}$
C $-\dfrac{1}{2}$
D $\dfrac{1}{2}$

●●●●●Level 11 markStart

A sector

OAB

of a circle with centre

O

has radius

r

cm and angle

\theta

radians. The arc

AB

has length

12

cm and the sector has area

36

^2

.
(a) Find

r

and

\theta

. (4 marks)
(b) Find the area of the segment bounded by the chord

AB

and the arc

AB

, giving your answer to 3 significant figures. (2 marks)

●●●●●Level 26 marksStart

Solve

2\cos x = \sqrt{3}

for

0 \le x \le 2\pi

, giving your answers as exact multiples of

\pi

●●●●●Level 23 marksStart

(a) Express

150^\circ

and

225^\circ

in radians, giving each answer as an exact multiple of

\pi

.
(b) Express

\frac{7\pi}{12}

radians in degrees.

●●●●●Level 23 marksStart

Without using a calculator, find the exact value of (a)

\cos\frac{5\pi}{6}

,
(b)

\sin\frac{3\pi}{4}

,
(c)

\tan\frac{7\pi}{6}

●●●●●Level 23 marksStart

Solve

\cos 2x = -\frac{1}{2}

for

0 \le x \le 2\pi

, giving your answers as exact multiples of

\pi

●●●●●Level 34 marksStart

Figure 1

In Figure 1,

OAB

is a sector of a circle with centre

O

and radius

6

cm. The angle

AOB

2.4

radians. The shaded region is the segment bounded by the chord

AB

and the arc

AB

.
(a) Find the perimeter of the shaded segment. (3 marks)
(b) Find the area of the shaded segment. (3 marks)

●●●●●Level 36 marksStart

Solve

2\sin^2 x = \sin x

for

0 \le x \le 2\pi

, giving your answers as exact multiples of

\pi

●●●●●Level 33 marksStart

Solve

4\cos^2 x - 3 = 0

for

0 \le x \le 2\pi

, giving your answers as exact multiples of

\pi

●●●●●Level 33 marksStart

A sector of a circle has radius

r

cm and angle

\theta

radians. The perimeter of the sector is

20

cm and its area is

21

cm².
(a) Show that

r^2 - 10r + 21 = 0

. (3 marks)
(b) Find the two possible values of

r

and, for each, the corresponding value of

\theta

. Give the values of

\theta

3

significant figures. (3 marks)

●●●●●Level 36 marksStart

Solve

\sqrt{2}\,\sin x + 1 = 0

for

0 \le x \le 2\pi

, giving your answers as exact multiples of

\pi

●●●●●Level 33 marksStart

Solve

\tan x = -1

for

0 \le x \le 2\pi

, giving your answers as exact multiples of

\pi

●●●●●Level 32 marksStart

Solve

\sin\!\left(x - \frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}

for

0 \le x \le 2\pi

, giving your answers as exact multiples of

\pi

●●●●●Level 33 marksStart

(a) Express

\dfrac{5\pi}{4}

radians in degrees.
(b) Without using a calculator, find the exact value of

\sin\dfrac{5\pi}{4}

●●●●●Level 33 marksStart

A landscape designer is planning a flower bed in the shape of a sector of a circle with radius

r

metres and angle

\theta

radians. The perimeter of the bed must be exactly

40

metres of edging.(a) Show that the area of the bed,

A

m², satisfies

A = 20r - r^2

. (3 marks)
(b) Hence find the maximum possible area of the bed, justifying that your value is a maximum, and find the value of

\theta

for which it occurs. (4 marks)

●●●●●Level 47 marksStart

Solve the equation

2\cos\left(2x - \dfrac{\pi}{3}\right) = 1

for

0 \leqslant x \leqslant \pi

, giving your answers in terms of

\pi

. (5 marks)

●●●●●Level 45 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 < \pi

, the equation

5\sin(2x + 0.6) + 2 = 0

giving your answers in radians to 2 decimal places.

(4 marks)
(b) Solve, for

-180^\circ < \theta < 180^\circ

, the equation

3\tan\theta\sin\theta = 5 + 2\cos\theta

giving your answers, in degrees, to one decimal place.

(5 marks)

●●●●●Level 49 marksStart

Figure 1

Diagram NOT accurately drawn

The shaded area in Figure 1 shows the plan view of a café terrace.

The terrace consists of a sector

AOB

of a circle centre

O

joined to a triangle

OBC

.

Given that

$OA = OB = 5$ m
$OC = 8$ m
$BC = 6$ m
$AOC$ is a straight line

(a) show that angle

BOC

0.847

radians to 3 decimal places.

(3 marks)
(b) Find the total area of the terrace. Give your answer in m

^2

to 3 significant figures.

(3 marks)
(c) Find the perimeter of the terrace. Give your answer in metres to 3 significant figures.

(3 marks)

●●●●●Level 49 marksStart

Figure 1

Diagram NOT accurately drawn

Figure 1 shows the plan view of a school playground.

The playground consists of a sector

COD

of a circle centre

O

joined to a sector

AOB

of a different circle, also with centre

O

. The point

B

lies on the line

OC

.

Given that

angle $COD = 0.65$ radians
angle $AOB = 1.9$ radians
arc length $CD = 9.1$ m
$OB : BC = 7 : 8$

(a) show that

OB = 6.53

m to 3 significant figures.

(3 marks)
(b) Find the perimeter of the playground, in m, to 3 significant figures.

(3 marks)
(c) Find the total area of the playground, giving your answer in m

^2

to the nearest whole number.

(3 marks)

●●●●●Level 49 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 \leqslant x < \pi

, the equation

\sqrt{3}\tan\left(3x - \dfrac{\pi}{8}\right) = 1

giving the answers in radians in the form

k\pi

, where

k

is a rational constant to be found.

(3 marks)
(b) Solve, for

0 \leqslant \theta < 360^\circ

, the equation

4\sin\theta\tan\theta = 3 - \cos\theta

giving your answers, in degrees, to one decimal place.

(5 marks)

●●●●●Level 48 marksStart

In this question you must show detailed reasoning.

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

\tan\theta\,(4\sin\theta + 1) = 6\cos\theta

can be written as

10\sin^2\theta + \sin\theta - 6 = 0

(3 marks)
(b) Hence solve, for

-\pi < x < \pi

, the equation

\tan\left(x - \dfrac{\pi}{5}\right)\left(4\sin\left(x - \dfrac{\pi}{5}\right) + 1\right) = 6\cos\left(x - \dfrac{\pi}{5}\right)

giving your answers to 3 significant figures.

(4 marks)

●●●●●Level 47 marksStart

In this question you must show all stages of your working.

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

(\tan\theta + \sin\theta)(1 - \cos\theta) \equiv \sin^2\theta\tan\theta

(4 marks)
(b) Hence solve, for

0 < x < 2\pi

, the equation

(\tan x + \sin x)(1 - \cos x) = \tan x - 2\sin x\tan x

giving your answers to 3 significant figures.

(4 marks)

●●●●●Level 48 marksStart

(a) Express

2\cos x - 2\sqrt{3}\sin x

in the form

R\cos(x + \alpha)

, where

R > 0

and

0 < \alpha < \dfrac{\pi}{2}

. (3 marks)

The depth of water,

h

metres, in a harbour

t

hours after midnight is modelled by

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

7

metres.
(b) Find the total length of time, during these

24

hours, for which the boat is able to enter the harbour. Fully justify your answer. (6 marks)

●●●●●Level 59 marksStart

The depth of water,

D

metres, at the entrance to a harbour is modelled by

D = 9 + 3\sin\left(\dfrac{\pi t}{6}\right) - 3\sqrt{3}\cos\left(\dfrac{\pi t}{6}\right)

\qquad 0 \le t < 24

,

where

t

is the time in hours after midnight.
(a) Express

3\sin\theta - 3\sqrt{3}\cos\theta

in the form

R\sin(\theta - \alpha)

, where

R > 0

and

0 < \alpha < \dfrac{\pi}{2}

, stating the exact values of

R

and

\alpha

12

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

OAB

of a circle with centre

O

and radius

r

. The angle

AOB

\theta

radians, where

0 < \theta < \tfrac12\pi

. The tangent to the circle at

A

meets the line

OB

extended at

T

. The region

R

is bounded by the arc

AB

and the lines

BT

and

AT

. The areas of region

R

and sector

OAB

are in the ratio

6 : 5

.(a) Show that

\theta = \tan^{-1}(2.2\theta)

.

(4 marks)

The equation

\theta = \tan^{-1}(2.2\theta)

has only one root for

0 < \theta < \tfrac12\pi

.
(b) This root can be found by using the iterative formula

\theta_{n+1} = \tan^{-1}(2.2\theta_n)

with a starting value of

\theta_1 = 1

Write down the values of $\theta_2$ , $\theta_3$ and $\theta_4$ .

Hence find the value of this root correct to 3 significant figures.

(3 marks)

The diagram shows the graph of

y = \tan^{-1}(2.2\theta)

and the line

y = \theta

, for

0 \le \theta \le \tfrac12\pi

.(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

\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

OAB

of a circle with centre

O

and radius

r

metres, together with the triangles

OAC

and

OBD

, which are joined to the sector along

OA

and along

OB

, where

angle $AOB = 1.5$ radians
the arc $AB$ has length $12$ m
triangle $OAC$ and triangle $OBD$ are congruent, with the sector $OAB$ lying between them
$C$ , $O$ and $D$ lie on a straight line, which forms the back edge of the platform

(a) Show that

r = 8

.

(2 marks)
(b) Show that the size of angle

AOC

0.821

radians, correct to 3 decimal places.

(2 marks)

The front edge of the platform runs in a straight line from

C

A

, then along the arc

AB

, then in a straight line from

B

D

.

Given that the total length of the front edge of the platform is

32

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

Figure 1

Diagram NOT accurately drawn

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

The tile is made from a triangle

OPQ

and a sector

OQR

of a circle with centre

O

, joined along

OQ

.

In the design

$OQ = 7.5$ cm
$PQ = 4.2$ cm
angle $POQ = 0.45$ radians
angle $OPQ > \dfrac{\pi}{2}$ radians

Making your method clear,
(a) find the size of angle

OPQ

, giving your answer in radians to 4 significant figures,

(3 marks)
(b) find the area of triangle

OPQ

, in cm

^2

, giving your answer to 3 significant figures.

(2 marks)

Given that the area of sector

OQR

and the area of triangle

OPQ

are in the ratio

5 : 3

(c) show that angle

QOR

0.397

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 < \pi

, the equation

3\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\%

, inside the gallery,

t

hours after midnight, is modelled by the equation

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

where

k

is a constant.

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

Given that

the relative humidity inside the gallery was $60\%$ at 5 am
$0 < k < 90$

(b) find all possible values for

k

, giving each answer to 2 decimal places.

(4 marks)

Given further that

0 < k < 20

●●●●●Level 512 marksStart

Figure 1

Diagram NOT accurately drawn

Figure 1 shows the plan view of a gravel path

PQRS

.

Given that

$OQR$ is a sector of a circle with centre $O$
$OPS$ is a sector of a smaller circle, also with centre $O$ , where $P$ lies on $OQ$ and $S$ lies on $OR$
angle $POS = 1.1$ radians
the area of sector $OPS$ is $4.95$ m $^2$

(a) show that

OP = 3

m

(2 marks)

Given also that

the area of the gravel path $PQRS$ is $40$ m $^2$
$PQ = x$ metres

(b) show that

11x^2 + 66x - 800 = 0

(3 marks)
(c) Solve the equation in part (b) to find the value of

x

, 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

OAB

of a circle of radius

12\text{ cm}

with angle

AOB = 75^\circ

. Find the exact length of the arc

AB

●●●●●Level 22 marksStart

A sector

OPQ

of a circle has centre

O

and radius

r

cm, where angle

POQ = 0.9

radians.

The area of the sector

OPQ

115.2

^2

.
(a) Show that

r = 16

.

(2 marks)
(b) Find the perimeter of the sector

OPQ

.

(2 marks)

●●●●●Level 24 marksStart

A sector

OAB

of a circle with centre

O

has radius

r

cm and angle

\theta

radians. The arc

AB

has length

12

cm and the sector has area

36

^2

.
(a) Find

r

and

\theta

. (4 marks)
(b) Find the area of the segment bounded by the chord

AB

and the arc

AB

, giving your answer to 3 significant figures. (2 marks)

●●●●●Level 26 marksStart

The diagram shows a sector

OAB

of a circle of radius

8\text{ cm}

, where the angle

AOB = 0.8

radians. Find
(a) the length of the arc

AB

,
(b) the area of the sector.

●●●●●Level 24 marksStart

The diagram shows a sector

OAB

of a circle of radius

6\text{ cm}

, in which the arc

AB

has length

15\text{ cm}

. Find
(a) the angle

AOB

in radians,
(b) the area of the sector.

●●●●●Level 24 marksStart

Two circles, each of radius

12

cm, have centres

A

and

B

, where

AB = 12

cm.

The two circles intersect at the points

P

and

Q

.

The region

R

is the region that lies inside both circles.

Find the exact area of

R

, giving your answer in the form

(a\pi + b\sqrt{3})

^2

, where

a

and

b

are rational numbers.

(4 marks)

●●●●●Level 34 marksStart

Figure 1

In Figure 1,

OAB

is a sector of a circle with centre

O

and radius

6

cm. The angle

AOB

2.4

radians. The shaded region is the segment bounded by the chord

AB

and the arc

AB

.
(a) Find the perimeter of the shaded segment. (3 marks)
(b) Find the area of the shaded segment. (3 marks)

●●●●●Level 36 marksStart

The diagram shows a sector

OAB

of a circle of radius

5\text{ cm}

with area

30\text{ cm}^2

. Find
(a) the angle

AOB

in radians,
(b) the perimeter of the sector.

●●●●●Level 34 marksStart

In triangle

ABC

AB = 8

cm,

AC = 11

cm and angle

BAC = 0.7

radians.
(a) Find the area of triangle

ABC

. (2 marks)
(b) Find the length of

BC

. (2 marks)

The circular arc

PQ

, with centre

A

and radius

6

cm, joins a point

P

AB

to a point

Q

AC

.(c) Find the area of the region enclosed by the arc

PQ

and the lines

PB

BC

and

CQ

, giving your answer to 3 significant figures. (2 marks)

●●●●●Level 36 marksStart

The diagram below shows a sector of a circle

OAB

.

The chord

AB

divides the sector into a triangle and a shaded segment.

Angle

AOB

\dfrac{\pi}{2}

radians.

The radius of the sector is 14 cm.Show that the area of the shaded segment is

k(\pi - 2) \text{ cm}^2

where

k

is an integer to be found.

(3 marks)

●●●●●Level 33 marksStart

A sector of a circle has radius

r

cm and angle

\theta

radians. The perimeter of the sector is

20

cm and its area is

21

cm².
(a) Show that

r^2 - 10r + 21 = 0

. (3 marks)
(b) Find the two possible values of

r

and, for each, the corresponding value of

\theta

. Give the values of

\theta

3

significant figures. (3 marks)

●●●●●Level 36 marksStart

The diagram shows a sector with radius

r\text{ cm}

, perimeter

28\text{ cm}

and area

40\text{ cm}^2

. Find the two possible values of

r

. (The diagram is not accurately drawn.)

●●●●●Level 34 marksStart

The diagram shows a circle of radius

6\text{ cm}

in which the chord

AB

subtends an angle of

\frac{\pi}{3}

at the centre

O

. Find the exact area of the shaded minor segment.

●●●●●Level 33 marksStart

The diagram shows a circle of radius

7\text{ cm}

in which the chord

AB

subtends an angle of

1.4

radians at the centre. Find the area of the shaded minor segment, giving your answer to 3 significant figures.

●●●●●Level 32 marksStart

The diagram shows a sector

OAB

of a circle of radius

8\text{ cm}

with angle

AOB = \frac{\pi}{2}

. The chord

AB

is drawn, cutting off the shaded segment. Find the exact perimeter of the shaded segment.

●●●●●Level 33 marksStart

A landscape designer is planning a flower bed in the shape of a sector of a circle with radius

r

metres and angle

\theta

radians. The perimeter of the bed must be exactly

40

metres of edging.(a) Show that the area of the bed,

A

m², satisfies

A = 20r - r^2

. (3 marks)
(b) Hence find the maximum possible area of the bed, justifying that your value is a maximum, and find the value of

\theta

for which it occurs. (4 marks)

●●●●●Level 47 marksStart

In this question you must show detailed reasoning.The diagram shows a sector

AOB

of a circle with centre

O

and radius

(x + 2)

cm. The angle

AOB

is 2 radians.

The area of sector

AOB

is less than

(5x + 34)

^2

.

Find the set of possible values of

x

. Give your answer in set notation.

(5 marks)

●●●●●Level 45 marksStart

The diagram below shows a sector of a circle

OAB

.

The chord

AB

divides the sector into a triangle and a shaded segment.

Angle

AOB

\theta

radians.(a) Given that the area of the shaded segment is one third of the area of the sector

OAB

, show that

2\theta = 3\sin\theta

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

2\theta = 3\sin\theta

lies in the interval given by

\theta \in [1.4, 1.6]

(2 marks)

●●●●●Level 46 marksStart

A circular metal disc has centre

O

and radius

9

cm.

The points

A

B

and

C

lie on the circumference of the disc so that triangle

ABC

is equilateral.

The triangle

ABC

is cut out of the disc, leaving three identical pieces of metal.
(a) Show that angle

AOB = \dfrac{2\pi}{3}

radians.

(2 marks)
(b) Find the exact total area of the three pieces of metal, giving your answer in the form

(a\pi + b\sqrt{3})

^2

, where

a

and

b

are rational numbers.

(4 marks)

●●●●●Level 46 marksStart

A metalworker is making a decorative garden screen.

The plan view of the screen is modelled as the region

OABC

, where

$OAB$ is a sector of a circle with centre $O$ and radius $r$ cm
angle $AOB = 0.75$ radians
the arc $AB$ has length $10.5$ cm
$OBC$ is a triangle joined to the sector along the radius $OB$ , where $BC = 9$ cm and angle $OBC = 2.1$ radians

(a) Show that

r = 14

.

(2 marks)
(b) Find the area of the plan view of the screen, giving your answer to 3 significant figures.

(4 marks)

●●●●●Level 46 marksStart

A jeweller is designing a silver brooch.

The brooch is modelled as the region bounded by the chord

AB

and the arc

AB

of a circle with centre

O

and radius

16

cm, where angle

AOB = \dfrac{2\pi}{3}

radians.
(a) Use the cosine rule to show that

AB = 16\sqrt{3}

cm.

(2 marks)
(b) Find the exact perimeter of the brooch.

(2 marks)
(c) Find the exact area of the brooch, giving your answer in the form

(a\pi + b\sqrt{3})

^2

, where

a

and

b

are rational numbers.

(3 marks)

●●●●●Level 47 marksStart

The diagram shows a sector

OAB

of a circle with centre

O

and radius

5

cm. The angle

AOB

\theta

radians. The shaded segment is bounded by the chord

AB

and the arc

AB

, and the perimeter of the shaded segment is

21

cm.(a) Show that

\theta = 4.2 - 2\sin\left(\tfrac12\theta\right)

.

(4 marks)

The equation

\theta = 4.2 - 2\sin\left(\tfrac12\theta\right)

has only one root for

0 < \theta < \pi

.
(b) This root can be found by using the iterative formula

\theta_{n+1} = 4.2 - 2\sin\left(\tfrac12\theta_n\right)

with a starting value of

\theta_1 = 2

Write down the values of $\theta_2$ , $\theta_3$ and $\theta_4$ .

Hence find the value of this root correct to 3 significant figures.

(3 marks)

The diagram shows the graph of

y = 4.2 - 2\sin\left(\tfrac12\theta\right)

and the line

y = \theta

, for

0 \le \theta \le \pi

.(c) • Use this diagram to show how the iterative process used in (b) converges to this root.

State the type of convergence.

(3 marks)

●●●●●Level 410 marksStart

Figure 1

Diagram NOT accurately drawn

The shaded area in Figure 1 shows the plan view of a café terrace.

The terrace consists of a sector

AOB

of a circle centre

O

joined to a triangle

OBC

.

Given that

$OA = OB = 5$ m
$OC = 8$ m
$BC = 6$ m
$AOC$ is a straight line

(a) show that angle

BOC

0.847

radians to 3 decimal places.

(3 marks)
(b) Find the total area of the terrace. Give your answer in m

^2

to 3 significant figures.

(3 marks)
(c) Find the perimeter of the terrace. Give your answer in metres to 3 significant figures.

(3 marks)

●●●●●Level 49 marksStart

Figure 1

Diagram NOT accurately drawn

Figure 1 shows the plan view of a school playground.

The playground consists of a sector

COD

of a circle centre

O

joined to a sector

AOB

of a different circle, also with centre

O

. The point

B

lies on the line

OC

.

Given that

angle $COD = 0.65$ radians
angle $AOB = 1.9$ radians
arc length $CD = 9.1$ m
$OB : BC = 7 : 8$

(a) show that

OB = 6.53

m to 3 significant figures.

(3 marks)
(b) Find the perimeter of the playground, in m, to 3 significant figures.

(3 marks)
(c) Find the total area of the playground, giving your answer in m

^2

to the nearest whole number.

(3 marks)

●●●●●Level 49 marksStart

The diagram shows a sector

OAB

of radius

r\text{ cm}

and angle

\theta

radians, with a fixed perimeter of

20\text{ cm}

.
(a) Show that the area of the sector is

A = 10r - r^2

.
(b) Hence find the value of

r

that maximises the area, and the maximum area. (The diagram is not accurately drawn.)

●●●●●Level 45 marksStart

A sector of a circle has angle

1.2

radians and perimeter

25.6

cm.
(a) Show that the radius of the sector is

8

cm.
(b) Find the area of the sector.

●●●●●Level 45 marksStart

The diagram shows a sector of a circle

OAB

.

The point

C

lies on

OB

such that

AC

is perpendicular to

OB

.

Angle

AOB

\theta

radians.(a) Given the area of the sector

OAB

\sqrt{2}

times the area of the triangle

OAC

, show that

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

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

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

lies in the interval given by

\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 2\theta = \sqrt{2}\,\theta

(c) (i) Using

\theta_1 = \dfrac{\pi}{4}

as a first approximation for

\theta

, apply the Newton-Raphson method twice to find the value of

\theta_3

Give your answer to three decimal places.

(ii) Explain how a more accurate approximation for

\theta

can be found using the Newton-Raphson method.

(iii) Explain why using

\theta_1 = \dfrac{\pi}{8}

as a first approximation in the Newton-Raphson method does not lead to a solution for

\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

P

O

and

B

lie on a straight line with

PO = 2OB

.

The shape consists of the sector

OBD

of a circle with centre

O

and radius

OB

, together with the triangle

OPD

.

The angle

POD

\theta

radians, where

0 < \theta < \pi

The weather vane is to be made with

PB = 1.5

metres.
(a) Show that the area of the weather vane,

A

square metres, is given by

A = \frac{1}{8}\left(\pi - \theta + 2\sin\theta\right)

(4 marks)
(b) (i) Show that the maximum value of

A

occurs when

\theta = \dfrac{\pi}{3}

Fully justify your answer.

(ii) Find the exact maximum value of

A

(8 marks)
(c) Without further calculation, state how your answers to parts (b)(i) and (b)(ii) would change if

PB

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

O

and radius

50

cm. The pipe contains some water. In the cross-section, the surface of the water is the chord

AB

and the angle

AOB

\theta

radians. The water occupies the shaded segment bounded by the chord

AB

and the arc

AB

, which has area

900

^2

.(a) Show that

\theta = \sin\theta + 0.72

.

(3 marks)
(b) Let

\text{F}(\theta) = \sin\theta + 0.72

.

By considering the value of

\text{F}'(\theta)

where

\theta = 1.7

, explain why using an iterative method based on the equation in part (a) will converge to the root, assuming that

1.7

is sufficiently close to the root.

(2 marks)
(c) Use the iterative formula

\theta_{n+1} = \sin\theta_n + 0.72

with a starting value of

1.7

to find the value of

\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

\theta

found in part (c) is correct to 4 significant figures.

(3 marks)

●●●●●Level 511 marksStart

The diagram shows a sector

OAB

of a circle with centre

O

and radius

r

. The angle

AOB

\theta

radians, where

0 < \theta < \tfrac12\pi

. The tangent to the circle at

A

meets the line

OB

extended at

T

. The region

R

is bounded by the arc

AB

and the lines

BT

and

AT

. The areas of region

R

and sector

OAB

are in the ratio

6 : 5

.(a) Show that

\theta = \tan^{-1}(2.2\theta)

.

(4 marks)

The equation

\theta = \tan^{-1}(2.2\theta)

has only one root for

0 < \theta < \tfrac12\pi

.
(b) This root can be found by using the iterative formula

\theta_{n+1} = \tan^{-1}(2.2\theta_n)

with a starting value of

\theta_1 = 1

Write down the values of $\theta_2$ , $\theta_3$ and $\theta_4$ .

Hence find the value of this root correct to 3 significant figures.

(3 marks)

The diagram shows the graph of

y = \tan^{-1}(2.2\theta)

and the line

y = \theta

, for

0 \le \theta \le \tfrac12\pi

.(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

\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

OAB

of a circle with centre

O

and radius

r

metres, together with the triangles

OAC

and

OBD

, which are joined to the sector along

OA

and along

OB

, where

angle $AOB = 1.5$ radians
the arc $AB$ has length $12$ m
triangle $OAC$ and triangle $OBD$ are congruent, with the sector $OAB$ lying between them
$C$ , $O$ and $D$ lie on a straight line, which forms the back edge of the platform

(a) Show that

r = 8

.

(2 marks)
(b) Show that the size of angle

AOC

0.821

radians, correct to 3 decimal places.

(2 marks)

The front edge of the platform runs in a straight line from

C

A

, then along the arc

AB

, then in a straight line from

B

D

.

Given that the total length of the front edge of the platform is

32

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

Figure 1

Diagram NOT accurately drawn

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

The tile is made from a triangle

OPQ

and a sector

OQR

of a circle with centre

O

, joined along

OQ

.

In the design

$OQ = 7.5$ cm
$PQ = 4.2$ cm
angle $POQ = 0.45$ radians
angle $OPQ > \dfrac{\pi}{2}$ radians

Making your method clear,
(a) find the size of angle

OPQ

, giving your answer in radians to 4 significant figures,

(3 marks)
(b) find the area of triangle

OPQ

, in cm

^2

, giving your answer to 3 significant figures.

(2 marks)

Given that the area of sector

OQR

and the area of triangle

OPQ

are in the ratio

5 : 3

(c) show that angle

QOR

0.397

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

Figure 1

Diagram NOT accurately drawn

Figure 1 shows the plan view of a gravel path

PQRS

.

Given that

$OQR$ is a sector of a circle with centre $O$
$OPS$ is a sector of a smaller circle, also with centre $O$ , where $P$ lies on $OQ$ and $S$ lies on $OR$
angle $POS = 1.1$ radians
the area of sector $OPS$ is $4.95$ m $^2$