Free worked solutions · 2026 Edition

Chapter 10: Trigonometry worked solutions

Worked solutions to the practice questions in Chapter 10 of The Ultimate Guide for the IB: Mathematics: Analysis and Approaches HL (2026 Edition). Each solution shows every step with the mark it earns, the way an IB mark scheme does.

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Basics

One idea at a time, in the order the chapter teaches them.

Question 1

[6 marks]

The points P and Q have coordinates P(1,\ -2,\ 4) and Q(7,\ 1,\ -2).

(a) Find the distance PQ.

[2]

(b) Find the coordinates of the midpoint of [PQ].

[2]

(c) The point S is such that Q is the midpoint of [PS]. Find the coordinates of S.

[2]

Show solutionHide solution

(a) PQ=\sqrt{(7-1)^2+(1+2)^2+(-2-4)^2}=\sqrt{36+9+36}

(b) \left(\dfrac{1+7}{2},\ \dfrac{-2+1}{2},\ \dfrac{4-2}{2}\right)

=\left(4,\ -\dfrac{1}{2},\ 1\right)

(c) \dfrac{1+x}{2}=7,\quad \dfrac{-2+y}{2}=1,\quad \dfrac{4+z}{2}=-2

S(13,\ 4,\ -8)

Note: in (c) accept the step method: from P to Q is +6,\ +3,\ -6, so add the same again to Q.

Final answer

(a) PQ=9
(b) \left(4,-\frac12,1\right)
(c) S(13,4,-8)

Question 2

[10 marks]

A right pyramid has a square base of side 10 cm, and each of its four slant edges has length 13 cm.

(a) Find the exact distance from a corner of the base to the centre of the base.

[2]

(b) Find the exact vertical height of the pyramid.

[2]

(c) Find the exact volume of the pyramid.

[2]

(d) Find the height of one triangular face.

[2]

(e) Hence find the total surface area of the pyramid.

[2]

Show solutionHide solution

(a) half of the diagonal: \dfrac{1}{2}\sqrt{10^2+10^2}

=5\sqrt{2} cm

(b) h=\sqrt{13^2-(5\sqrt{2})^2}=\sqrt{169-50}

=\sqrt{119} cm

(c) \dfrac{1}{3}(100)\sqrt{119}

=\dfrac{100\sqrt{119}}{3}\text{ cm}^3

(d) each face is isosceles; the height bisects the base side: \sqrt{13^2-5^2}

=12 cm

(e) 100+4\times\dfrac{1}{2}(10)(12)

=340\text{ cm}^2

Note: in (b) the slant edge is the hypotenuse, so SUBTRACT. Accept 363.6\ldots for (c) only if the exact form is also seen.

Final answer

(a) 5\sqrt2 cm
(b) \sqrt{119} cm
(c) \frac{100\sqrt{119}}{3} cm^3
(d) 12 cm
(e) 340 cm^2

Question 3

[5 marks]

A vertical floodlight pole FT of height 12 m stands on level ground, with foot F and top T.

(a) From a point A on the ground, the angle of elevation of T is 30^\circ. Find the exact distance AF.

[2]

(b) A second point C on the ground is 12 m from F. Find the angle of elevation of T from C.

[2]

(c) Find the exact distance CT.

[1]

Show solutionHide solution

(a) \tan 30^\circ=\dfrac{12}{AF}\Rightarrow AF=\dfrac{12}{1/\sqrt{3}}

AF=12\sqrt{3} m

(b) \tan\theta=\dfrac{12}{12}=1

\theta=45^\circ

(c) CT=\sqrt{12^2+12^2}=12\sqrt{2} m

Final answer

(a) 12\sqrt3 m
(b) 45^\circ
(c) 12\sqrt2 m

Question 4

[5 marks]

A sector of a circle has an angle of 120^\circ at the centre and an area of 12\pi\ \text{cm}^2.

(a) Find the radius of the sector.

[3]

(b) Find the exact perimeter of the sector.

[2]

Show solutionHide solution

(a) 120^\circ=\dfrac{2\pi}{3}

\dfrac{1}{2}r^2\left(\dfrac{2\pi}{3}\right)=12\pi \Rightarrow \dfrac{\pi r^2}{3}=12\pi

r^2=36, r=6 cm

(b) arc =6\times\dfrac{2\pi}{3}=4\pi (M1)

perimeter =12+4\pi cm

Note: reject r=-6 (a length is positive); it need not be written.

Final answer

(a) r=6 cm
(b) 12+4\pi cm

Question 5

[8 marks]

The points A and B lie on a circle with centre O and radius 7 cm, and A\hat{O}B=2.

(a) Write down the length of the minor arc AB.

[1]

(b) Find the length of the chord [AB].

[2]

(c) Find the area of triangle OAB.

[2]

(d) Find the area of the minor segment cut off by [AB].

[2]

(e) Find the perimeter of this minor segment.

[1]

Show solutionHide solution

(a) 7(2)=14 cm

(b) AB^2=7^2+7^2-2(7)(7)\cos2

AB=11.780\ldots=11.8 cm

(c) \dfrac{1}{2}(7)(7)\sin2

=22.277\ldots=22.3\ \text{cm}^2

(d) sector area =\dfrac{1}{2}(49)(2)=49; segment =49-22.277\ldots

=26.722\ldots=26.7\ \text{cm}^2

(e) 14+11.780\ldots=25.8 cm

Note: the GDC must be in radian mode for \cos2 and \sin2; degree mode gives AB=0.244, which earns M1 A0.

Final answer

(a) 14 cm
(b) 11.8 cm
(c) 22.3 cm^2
(d) 26.7 cm^2
(e) 25.8 cm

Question 6

[5 marks]

The angle \theta satisfies 0<\theta<\dfrac{\pi}{2} and \tan\theta=\dfrac{3}{2}.

(a) Find the exact values of \sin\theta and \cos\theta.

[3]

(b) By dividing the numerator and the denominator by \cos\theta, or otherwise, find the value of \dfrac{2\sin\theta+\cos\theta}{4\cos\theta-\sin\theta}.

[2]

Show solutionHide solution

(a) right-angled triangle with opposite 3, adjacent 2: hypotenuse \sqrt{13}

\sin\theta=\dfrac{3}{\sqrt{13}}=\dfrac{3\sqrt{13}}{13}

\cos\theta=\dfrac{2}{\sqrt{13}}=\dfrac{2\sqrt{13}}{13}

(b) \dfrac{2\tan\theta+1}{4-\tan\theta}=\dfrac{2\left(\frac32\right)+1}{4-\frac32}

=\dfrac{4}{5/2}=\dfrac{8}{5}

Note: substituting the values from (a) also earns M1: \dfrac{(6+2)/\sqrt{13}}{(8-3)/\sqrt{13}}=\dfrac85. Accept unrationalised answers in (a).

Final answer

(a) \sin\theta=\frac{3\sqrt{13}}{13}, \cos\theta=\frac{2\sqrt{13}}{13}
(b) \frac85

Question 7

[9 marks]

The temperature, T\,{}^\circ\mathrm{C}, inside a greenhouse t hours after midnight is modelled by

T(t)=21+6\sin\left(\dfrac{\pi}{12}(t-8)\right),\quad 0 \le t \le 24.

(a) Find the temperature at 06:30.

[2]

(b) Find the maximum temperature and the time of day at which it occurs.

[3]

The roof vents are open whenever the temperature is above 25\,{}^\circ\mathrm{C}.

(c) Find the length of time for which the vents are open during the day.

[3]

(d) Write down the period of the model and explain what it means in this context.

[1]

Show solutionHide solution

(a) T(6.5)=21+6\sin\left(\dfrac{\pi}{12}(-1.5)\right) (M1)

=18.703\ldots=18.7\,{}^\circ\mathrm{C}

(b) maximum 27\,{}^\circ\mathrm{C}

when \dfrac{\pi}{12}(t-8)=\dfrac{\pi}{2} (M1)

t=14, i.e. at 14:00

(c) GDC: T(t)=25 at t=10.787\ldots

and t=17.212\ldots

17.212\ldots-10.787\ldots=6.43 hours

Note: accept 6 hours 26 minutes. Use of rounded intermediate values may lose the final A1.

(d) \dfrac{2\pi}{\pi/12}=24 hours: the temperature pattern repeats every day

Final answer

(a) 18.7\,{}^\circC
(b) 27\,{}^\circC at 14:00
(c) 6.43 hours
(d) 24 hours: the pattern repeats every day

Question 8

[8 marks]

Solve each equation, giving your answers as exact multiples of \pi.

(a) 2\cos x=\sqrt2, for 0\le x\le 2\pi.

[2]

(b) \sqrt3\tan x+3=0, for -\pi\le x\le \pi.

[3]

(c) 4\sin x+2\sqrt3=0, for 0\le x\le 2\pi.

[3]

Show solutionHide solution

(a) \cos x=\dfrac{\sqrt2}{2}, reference angle \dfrac{\pi}{4}; \cos positive in the first and fourth quadrants

x=\dfrac{\pi}{4}

x=2\pi-\dfrac{\pi}{4}=\dfrac{7\pi}{4}

(b) \tan x=-\dfrac{3}{\sqrt3}=-\sqrt3

reference angle \dfrac{\pi}{3}; \tan negative in the second and fourth quadrants (M1)

x=-\dfrac{\pi}{3},\ \dfrac{2\pi}{3}

(c) \sin x=-\dfrac{2\sqrt3}{4}=-\dfrac{\sqrt3}{2}

reference angle \dfrac{\pi}{3}; \sin negative in the third and fourth quadrants (M1)

x=\pi+\dfrac{\pi}{3}=\dfrac{4\pi}{3},\quad x=2\pi-\dfrac{\pi}{3}=\dfrac{5\pi}{3}

Note: in (b), -\dfrac{\pi}{3} is the fourth-quadrant angle written inside -\pi\le x\le\pi; do not accept \dfrac{5\pi}{3}, which is outside the interval.

Final answer

(a) \frac\pi4,\ \frac{7\pi}{4}
(b) -\frac\pi3,\ \frac{2\pi}{3}
(c) \frac{4\pi}{3},\ \frac{5\pi}{3}

Question 9

[7 marks]

Consider the equation 3\tan^2x-7\sec x+5=0, for -\pi\le x\le\pi.

(a) Show that the equation can be written as 3\sec^2x-7\sec x+2=0.

[2]

(b) Hence solve the equation, giving exact answers.

[5]

Show solutionHide solution

(a) using \tan^2x=\sec^2x-1: 3(\sec^2x-1)-7\sec x+5=0

3\sec^2x-7\sec x+2=0

A1AG

(b) (3\sec x-1)(\sec x-2)=0

\sec x=\dfrac13 or \sec x=2

\sec x=\dfrac13 is impossible because \lvert\sec x\rvert\ge1 (it would need \cos x=3)

\cos x=\dfrac12: x=-\dfrac{\pi}{3}

x=\dfrac{\pi}{3}

Note: award the final A1 only if no extra values are given in the interval. Award the R1 for rejecting \sec x=\dfrac13 (or \cos x=3) with a reason.

Final answer

(a) shown
(b) x=-\frac\pi3,\ \frac\pi3 (\sec x=\frac13 is impossible)

Tutorial

Exam-style questions on each question type, with the marks the IB would give.

Question 10

[7 marks]

In triangle ABC, AB = 12 cm, BC = 9 cm and \angle BAC = 40^\circ.

(a) Find the two possible sizes of \angle ACB.

[3]

(b) For each of the two triangles, find the length of AC.

[2]

(c) Find the area of the larger of the two possible triangles.

[2]

Show solutionHide solution

Degree mode.

(a) Sine rule: \dfrac{\sin C}{12} = \dfrac{\sin 40^\circ}{9}, so \sin C = \dfrac{12\sin 40^\circ}{9} = 0.8570\ldots

[1]

C = \sin^{-1}(0.8570\ldots) = 58.99^\circ\ldots \approx 59.0^\circ.

[1]

The calculator returns only the acute value; the other solution of \sin C = 0.857 is 180^\circ - 58.99^\circ = 121.0^\circ, and it is valid because 40^\circ + 121.0^\circ < 180^\circ.

[1]

(b) Acute case: B = 180^\circ - 40^\circ - 58.99^\circ = 81.01^\circ, so AC = \dfrac{9\sin 81.01^\circ}{\sin 40^\circ} = 13.8 cm.

[1]

Obtuse case: B = 180^\circ - 40^\circ - 121.01^\circ = 18.99^\circ, so AC = \dfrac{9\sin 18.99^\circ}{\sin 40^\circ} = 4.56 cm.

[1]

(c) The larger triangle has AC = 13.83\ldots

[1]

Area = \tfrac12(12)(13.83)\sin 40^\circ = 53.3\ldots \approx 53.3 cm^2.

[1]

Final answer

(a) 59.0^\circ, 121.0^\circ
(b) 13.8 cm, 4.56 cm
(c) 53.3 cm^2

Question 11

[6 marks]

A sector of a circle has area 40 cm^2 and perimeter 26 cm. The radius of the circle is r cm and the angle of the sector is \theta radians.

Find the two possible pairs of values of r and \theta.

Show solutionHide solution

Area: \tfrac12 r^2\theta = 40. Perimeter: 2r + r\theta = 26.

[1]

From the perimeter, r\theta = 26 - 2r. Then \tfrac12 r(r\theta) = \tfrac12 r(26 - 2r) = 40.

[1]

13r - r^2 = 40, so r^2 - 13r + 40 = 0.

[1]

(r - 5)(r - 8) = 0, so r = 5 or r = 8.

[2]

r = 5: \theta = \dfrac{26 - 10}{5} = \dfrac{16}{5}. r = 8: \theta = \dfrac{26 - 16}{8} = \dfrac54 (both angles are less than 2\pi, so both pairs are valid).

[1]

Final answer

r=5,\ \theta=\frac{16}{5}
or
r=8,\ \theta=\frac54

Question 12

[7 marks]

A sector OAB of a circle has radius 9.5 cm and area 60 cm^2.

(a) Find \angle AOB, in radians.

[2]

(b) Find the perimeter of the sector.

[2]

(c) Find the area of triangle OAB, and hence the area of the segment cut off by the chord AB.

[3]

Show solutionHide solution

Radian mode.

(a) \tfrac12 r^2\theta = 60, so \tfrac12(9.5)^2\theta = 60.

[1]

\theta = \dfrac{120}{90.25} = 1.3296\ldots \approx 1.33 radians.

[1]

(b) Arc length = r\theta = 9.5 \times 1.3296\ldots = 12.63\ldots

[1]

Perimeter = 2(9.5) + 12.63 = 31.63\ldots \approx 31.6 cm.

[1]

(c) Triangle area = \tfrac12 r^2\sin\theta = \tfrac12(90.25)\sin 1.3296\ldots = 43.82\ldots \approx 43.8 cm^2.

[2]

Segment area = 60 - 43.82\ldots = 16.18\ldots \approx 16.2 cm^2.

[1]

Final answer

(a) 1.33
(b) 31.6 cm
(c) 43.8 cm^2 and 16.2 cm^2

Question 13

[8 marks]

The angles A and B satisfy \sin A = \dfrac{5}{13}, with A obtuse, and \cos B = \dfrac45, with B acute.

(a) Find the exact values of \cos A and \sin B.

[2]

(b) Find the exact value of \sin(A - B).

[2]

(c) Find the exact value of \cos(A + B).

[2]

(d) Hence find the exact value of \tan(A + B).

[2]

Show solutionHide solution

; (a) A obtuse so \cos A < 0: \cos A = -\sqrt{1 - \dfrac{25}{169}} = -\dfrac{12}{13}.

[1]

B acute so \sin B > 0: \sin B = \sqrt{1 - \dfrac{16}{25}} = \dfrac35.

[1]

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

[1]

= \dfrac{5}{13}\cdot\dfrac45 - \left(-\dfrac{12}{13}\right)\dfrac35 = \dfrac{20}{65} + \dfrac{36}{65} = \dfrac{56}{65}.

[1]

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

[1]

= \left(-\dfrac{12}{13}\right)\dfrac45 - \dfrac{5}{13}\cdot\dfrac35 = -\dfrac{48}{65} - \dfrac{15}{65} = -\dfrac{63}{65}.

[1]

(d) \sin(A + B) = \sin A\cos B + \cos A\sin B = \dfrac{20}{65} - \dfrac{36}{65} = -\dfrac{16}{65}.

[1]

\tan(A + B) = \dfrac{-16/65}{-63/65} = \dfrac{16}{63}.

[1]

Final answer

(a) \cos A=-\frac{12}{13}, \sin B=\frac35
(b) \frac{56}{65}
(c) -\frac{63}{65}
(d) \frac{16}{63}

Question 14

[8 marks]

(a) Show that \dfrac{\tan\theta}{1+\sec\theta}+\dfrac{1+\sec\theta}{\tan\theta}\equiv 2\operatorname{cosec}\theta.

[3]

(b) Hence find the value of \dfrac{\tan\theta}{1+\sec\theta}+\dfrac{1+\sec\theta}{\tan\theta} when \theta is acute and \tan\theta=\dfrac34.

[2]

(c) Show that \sec\theta-\cos\theta\equiv\sin\theta\tan\theta.

[3]

Show solutionHide solution

(a) Common denominator: \dfrac{\tan^2\theta+(1+\sec\theta)^2}{\tan\theta(1+\sec\theta)}

[1]

Numerator =(\sec^2\theta-1)+1+2\sec\theta+\sec^2\theta=2\sec\theta(\sec\theta+1) (using \tan^2\theta=\sec^2\theta-1).

[1]

So the expression is \dfrac{2\sec\theta}{\tan\theta}=\dfrac{2}{\cos\theta}\cdot\dfrac{\cos\theta}{\sin\theta}=2\operatorname{cosec}\theta.

AG[1]

(b) The expression equals 2\operatorname{cosec}\theta; with \tan\theta=\dfrac34 and \theta acute, the right-angled triangle has sides 3,4,5, so \sin\theta=\dfrac35.

[1]

2\operatorname{cosec}\theta=\dfrac{10}{3}.

[1]

(c) \sec\theta-\cos\theta=\dfrac{1}{\cos\theta}-\cos\theta=\dfrac{1-\cos^2\theta}{\cos\theta}

[1]

=\dfrac{\sin^2\theta}{\cos\theta}

[1]

=\sin\theta\cdot\dfrac{\sin\theta}{\cos\theta}=\sin\theta\tan\theta.

AG[1]

Final answer

(a) shown
(b) \frac{10}{3}
(c) shown

Question 15

[6 marks]

(a) Solve \sin 2x = \sin x for 0 \le x \le 2\pi.

[4]

(b) A student solves the equation in part (a) by dividing both sides by \sin x and then solving 2\cos x = 1. State which solutions the student misses, and explain why they are lost.

[2]

Show solutionHide solution

(a) 2\sin x\cos x = \sin x, so 2\sin x\cos x - \sin x = 0.

[1]

\sin x(2\cos x - 1) = 0.

[1]

\sin x = 0: x = 0,\ \pi,\ 2\pi.

[1]

\cos x = \dfrac12: x = \dfrac{\pi}{3},\ \dfrac{5\pi}{3}.

[1]

(b) The student misses x = 0, \pi and 2\pi, the solutions of \sin x = 0.

[1]

Dividing by \sin x is only valid when \sin x \ne 0; at these values both sides of the equation are 0, so the division throws those solutions away.

[1]

Final answer

(a) 0,\ \frac\pi3,\ \pi,\ \frac{5\pi}{3},\ 2\pi
(b) x=0,\ \pi,\ 2\pi, where \sin x=0: dividing by \sin x assumes it is not zero

Question 16

[7 marks]

A buoy bobs up and down on the waves. Its height h metres above the seabed at time t seconds is modelled by

h(t) = a\sin(bt) + d,\qquad a > 0,\ b > 0.

At t = 0 the buoy is at its mean height of 2.0 m and rising. Its greatest height is 2.6 m, and consecutive greatest heights are 5.2 seconds apart.

(a) Find the values of a, b and d.

[3]

(b) Find the height of the buoy when t = 1.5.

[1]

(c) Find the total time, in the first 5.2 seconds, for which the buoy is higher than 2.4 m.

[3]

Show solutionHide solution

Radian mode.

(a) d = 2.0 (the mean height), and the amplitude is a = 2.6 - 2.0 = 0.6.

[1]

The period is 5.2, so b = \dfrac{2\pi}{5.2} = 1.2083\ldots \approx 1.21.

[1]

Starting at the mean height and rising is the shape of \sin, with no phase shift, consistent with h(t) = 0.6\sin(bt) + 2.0.

[1]

(b) h(1.5) = 0.6\sin(1.2083\times 1.5) + 2.0 = 2.5825\ldots \approx 2.58 m.

[1]

(c) Intersect y = h(t) with y = 2.4 on the GDC (window 0 \le t \le 5.2): t = 0.6039\ldots and t = 1.9960\ldots.

[2]

The graph is above 2.4 between these times (and nowhere else in the first period), so the total time is 1.9961 - 0.6039 = 1.392\ldots \approx 1.39 seconds.

[1]

Final answer

(a) a=0.6, b=\frac{2\pi}{5.2}\approx1.21, d=2.0
(b) 2.58 m
(c) 1.39 s

Question 17

[7 marks]

Let f(x) = 3\tan\left(x - \dfrac{\pi}{4}\right) for -\dfrac{\pi}{4} < x < \dfrac{3\pi}{4}.

(a) Write down the equations of the two vertical asymptotes of the graph of y = f(x).

[2]

(b) Find the x-intercept of the graph.

[1]

(c) Solve f(x) = 3.

[2]

(d) Using your answers to parts (a) to (c), sketch the graph of y = f(x).

[2]

Show solutionHide solution

(a) The argument x - \dfrac{\pi}{4} must stay between -\dfrac{\pi}{2} and \dfrac{\pi}{2}, and the asymptotes are where it reaches these values.

[1]

x = -\dfrac{\pi}{4} and x = \dfrac{3\pi}{4}.

[1]

(b) \tan\left(x - \dfrac{\pi}{4}\right) = 0 \Rightarrow x - \dfrac{\pi}{4} = 0, so x = \dfrac{\pi}{4}.

[1]

(c) \tan\left(x - \dfrac{\pi}{4}\right) = 1 \Rightarrow x - \dfrac{\pi}{4} = \dfrac{\pi}{4}.

[1]

x = \dfrac{\pi}{2} (the only solution in the interval).

[1]

(d)

Asymptotes drawn and labelled, with the curve increasing between them

[1]

Curve passing through \left(\dfrac{\pi}{4}, 0\right) and \left(\dfrac{\pi}{2}, 3\right)

[1]

Final answer

(a) x=-\frac\pi4, x=\frac{3\pi}{4}
(b) \frac\pi4
(c) \frac\pi2
(d) sketch through \left(\frac\pi4,0\right) and \left(\frac\pi2,3\right) between the asymptotes

Question 18

[7 marks]

Find the exact value of

(a) \cos\left(\arcsin\dfrac{5}{13}\right),

[2]

(b) \tan\left(\arccos\left(-\dfrac35\right)\right),

[2]

(c) \sin\left(2\arctan\dfrac34\right).

[3]

Show solutionHide solution

; (a) Let \alpha = \arcsin\dfrac{5}{13}, so \sin\alpha = \dfrac{5}{13} and \alpha \in \left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right], where \cos\alpha \ge 0.

[1]

\cos\alpha = \sqrt{1 - \dfrac{25}{169}} = \dfrac{12}{13}.

[1]

(b) Let \beta = \arccos\left(-\dfrac35\right), so \cos\beta = -\dfrac35 and \beta \in [0, \pi], where \sin\beta \ge 0: \sin\beta = \dfrac45.

[1]

\tan\beta = \dfrac{4/5}{-3/5} = -\dfrac43.

[1]

(c) Let \gamma = \arctan\dfrac34, so \tan\gamma = \dfrac34 with \gamma acute: a 3–4–5 triangle gives \sin\gamma = \dfrac35 and \cos\gamma = \dfrac45.

[2]

\sin 2\gamma = 2\sin\gamma\cos\gamma = 2\cdot\dfrac35\cdot\dfrac45 = \dfrac{24}{25}.

[1]

Final answer

(a) \frac{12}{13}
(b) -\frac43
(c) \frac{24}{25}

Mastery

Full exam questions that mix question types, at Paper 1 and Paper 2 standard.

Question 19

[6 marks]

A chord AB of a circle with centre O and radius 8 cm subtends an angle \theta radians at O, where 0<\theta<\pi. The area of the minor segment cut off by AB is half the area of triangle OAB.

(a) Show that 2\theta=3\sin\theta.

[2]

(b) Find the value of \theta.

[2]

(c) Find the length of the chord AB.

[2]

Show solutionHide solution

Radian mode.

(a) Segment area =\tfrac12 r^2(\theta-\sin\theta) and triangle area =\tfrac12 r^2\sin\theta.

[1]

\tfrac12 r^2(\theta-\sin\theta)=\tfrac12\cdot\tfrac12 r^2\sin\theta, so \theta-\sin\theta=\tfrac12\sin\theta and 2\theta=3\sin\theta.

[1]

(b) Solve 2\theta=3\sin\theta for 0<\theta<\pi on the GDC (intersect y=2x and y=3\sin x, or solve 2x-3\sin x=0).

[1]

\theta=1.4957\ldots\approx 1.50.

[1]

(c) AB=2\times 8\sin\dfrac{\theta}{2} (or the cosine rule)

[1]

=16\sin(0.7478\ldots)=10.9 cm (3 s.f.).

[1]

Final answer

(a) shown
(b) \theta=1.50
(c) 10.9 cm

Question 20

[10 marks]

In triangle ABC, B\hat{A}C=\dfrac{\pi}{3} and AB=10 cm. The length BC=a cm can vary.

(a) Find the least possible value of a.

[2]

(b) Write down the set of values of a for which two different triangles ABC are possible.

[2]

(c) Let a=9 and AC=b cm. Show that b^2-10b+19=0, and hence find the two possible exact values of b.

[4]

(d) Show that, for every value of a in part (b), the two possible values of AC add up to 10 cm.

[2]

Show solutionHide solution

(a) least a is the perpendicular distance from B to the line AC: 10\sin\dfrac{\pi}{3}

=5\sqrt{3}

(b) 5\sqrt{3}<a

a<10

(c) 81=100+b^2-2(10)b\cos\dfrac{\pi}{3}

\cos\dfrac{\pi}{3}=\dfrac{1}{2}: 81=100+b^2-10b, so b^2-10b+19=0

A1AG

b=\dfrac{10\pm\sqrt{100-76}}{2}

b=5\pm\sqrt{6}

(d) in general b^2-10b+100-a^2=0, so b=\dfrac{10\pm\sqrt{4a^2-300}}{2}=5\pm\dfrac{\sqrt{4a^2-300}}{2}

the sum of the two values is 5+5=10

A1AG

Note: in (b) award A1A0 for 5\sqrt{3}\le a<10 or 5\sqrt{3}<a\le10. Geometric argument in (d) is also fine: the two positions of C are symmetric about the foot of the perpendicular from B, which is 10\cos\dfrac{\pi}{3}=5 from A.

Final answer

(a) 5\sqrt3 cm
(b) 5\sqrt3<a<10
(c) b=5\pm\sqrt6
(d) shown

Question 21

[10 marks]

A ship leaves a harbour H and sails in a straight line on a bearing of 110^\circ at 15 km per hour. A reef R is 9 km from H on a bearing of 145^\circ.

(a) Sketch a diagram showing H, R, the north line at H and the path of the ship.

[2]

(b) Find the shortest distance between the reef and the path of the ship.

[3]

A warning sounds whenever the ship is within 6 km of the reef. Let d km be the distance the ship has sailed from H.

(c) Show that, when the ship is exactly 6 km from the reef, d^2-(18\cos 35^\circ)d+45=0.

[2]

(d) Find, to the nearest minute, how long the warning sounds.

[3]

Show solutionHide solution

(a)path from H on 110^\circ and HR on 145^\circ from the north line, with R clockwise of the path

9 km and the angle 35^\circ between the path and HR marked

(b) angle between the path and HR =145^\circ-110^\circ=35^\circ

shortest distance RN=9\sin 35^\circ, where RN is perpendicular to the path

=5.16 km (5.162\ldots)

(c) cosine rule in the triangle with sides d, 9 and 6: 6^2=d^2+9^2-2(9)d\cos 35^\circ

36=d^2+81-18d\cos 35^\circ

d^2-(18\cos 35^\circ)d+45=0

(d) d=4.3142\ldots or d=10.430\ldots

distance inside the warning zone =10.430\ldots-4.3142\ldots=6.1161\ldots km

time =\dfrac{6.1161\ldots}{15}\times 60=24 minutes (24.46\ldots)

Note: in (b), 5.16<6 confirms the ship does enter the warning zone. In (d) accept the chord route 2\sqrt{6^2-5.162\ldots^2}=6.116\ldots.

Final answer

(a) sketch
(b) 5.16 km
(c) shown
(d) 24 minutes

Question 22

[8 marks]

(a) Prove that

\tan\left(x + \dfrac{\pi}{4}\right) - \tan\left(x - \dfrac{\pi}{4}\right) = 2\sec 2x

for all values of x for which \tan x, \tan\left(x + \dfrac{\pi}{4}\right) and \tan\left(x - \dfrac{\pi}{4}\right) are defined.

[5]

(b) Hence solve the equation \tan\left(x + \dfrac{\pi}{4}\right) - \tan\left(x - \dfrac{\pi}{4}\right) = 4 for 0 < x < \pi, x \ne \dfrac{\pi}{4}, x \ne \dfrac{\pi}{2}, x \ne \dfrac{3\pi}{4}.

[3]

Show solutionHide solution

(a) let t = \tan x: \tan\left(x + \dfrac{\pi}{4}\right) = \dfrac{t + 1}{1 - t} and \tan\left(x - \dfrac{\pi}{4}\right) = \dfrac{t - 1}{1 + t}

M1A1

LHS = \dfrac{(1 + t)^2 - (t - 1)(1 - t)}{(1 - t)(1 + t)} = \dfrac{(1 + t)^2 + (1 - t)^2}{1 - t^2} = \dfrac{2(1 + t^2)}{1 - t^2}

using 1 + \tan^2 x = \sec^2 x: \dfrac{2\sec^2 x}{1 - \tan^2 x}

multiplying the numerator and the denominator by \cos^2 x: \dfrac{2}{\cos^2 x - \sin^2 x} = \dfrac{2}{\cos 2x} = 2\sec 2x

A1AG

(b) 2\sec 2x = 4 \Rightarrow \cos 2x = \dfrac{1}{2}

0 < 2x < 2\pi: 2x = \dfrac{\pi}{3} or 2x = \dfrac{5\pi}{3}, so x = \dfrac{\pi}{6}

or x = \dfrac{5\pi}{6}

Note: in (a) accept \dfrac{2(1 + t^2)}{1 - t^2} = \dfrac{2}{\cos 2x} via \cos 2x = \dfrac{1 - \tan^2 x}{1 + \tan^2 x} shown from \cos 2x = \cos^2 x - \sin^2 x. Do not award the final A1 if working backwards from the result. In (b) award the final A1 only if no extra values are given in the interval.

Final answer

(a) shown
(b) x=\frac\pi6,\ \frac{5\pi}{6}

Question 23

[10 marks]

(a) Show that \arctan\dfrac{1}{4} + \arctan\dfrac{3}{5} = \dfrac{\pi}{4}.

[4]

(b) Solve the equation \arctan x + \arctan 3x = \dfrac{3\pi}{4}.

[6]

Show solutionHide solution

(a) \tan\left(\arctan\dfrac{1}{4} + \arctan\dfrac{3}{5}\right) = \dfrac{\frac{1}{4} + \frac{3}{5}}{1 - \frac{1}{4} \times \frac{3}{5}}

= \dfrac{\frac{17}{20}}{\frac{17}{20}} = 1

0 < \dfrac{1}{4} < 1 and 0 < \dfrac{3}{5} < 1, so both angles lie in \left(0, \dfrac{\pi}{4}\right) and their sum lies in \left(0, \dfrac{\pi}{2}\right)

the only angle in \left(0, \dfrac{\pi}{2}\right) with tangent 1 is \dfrac{\pi}{4}

A1AG

(b) taking the tangent of both sides: \dfrac{x + 3x}{1 - 3x^2} = \tan\dfrac{3\pi}{4} = -1

4x = 3x^2 - 1 \Rightarrow 3x^2 - 4x - 1 = 0

x = \dfrac{4 \pm \sqrt{16 + 12}}{6} = \dfrac{2 \pm \sqrt{7}}{3} (M1)

x = \dfrac{2 - \sqrt{7}}{3} < 0 makes both arctans negative, so the sum is negative and cannot be \dfrac{3\pi}{4}: rejected

x = \dfrac{2 + \sqrt{7}}{3}

Note: in (b), for x = \dfrac{2 + \sqrt{7}}{3} > 0 both arctans lie in \left(0, \dfrac{\pi}{2}\right) and the tangent of the sum is -1, so the sum is \dfrac{3\pi}{4} (not -\dfrac{\pi}{4}). Award the final A1 only if the negative root is rejected. (1 - 3x^2 \ne 0 for both roots, so the tangent formula applies.)

Final answer

(a) shown
(b) x=\frac{2+\sqrt7}{3} (the root \frac{2-\sqrt7}{3} is rejected)

Question 24

[15 marks]

A function is given by f(x)=2\cos 2x+2\sin x+1, with domain 0\le x\le 2\pi.

(a) Show that f(x)=3+2\sin x-4\sin^2x.

[2]

(b) Solve the equation f(x)=3.

[5]

It is given that f(x)=\dfrac{13}{4}-4\left(\sin x-\dfrac{1}{4}\right)^2.

(c) Write down the maximum value of f(x) and the value of \sin x at which it occurs.

[2]

(d) Find the minimum value of f(x) and the value of x at which it occurs.

[2]

(e) Find all values of k such that the horizontal line y=k meets the graph of y=f(x) at exactly two points.

[4]

Show solutionHide solution

(a) 2\left(1-2\sin^2x\right)+2\sin x+1

=2-4\sin^2x+2\sin x+1=3+2\sin x-4\sin^2x

A1AG

(b) 2\sin x-4\sin^2x=0

2\sin x\left(1-2\sin x\right)=0

\sin x=0: x=0,\ \pi,\ 2\pi

\sin x=\dfrac{1}{2}: x=\dfrac{\pi}{6}

x=\dfrac{5\pi}{6}

(c) maximum value \dfrac{13}{4} (the squared term is 0)

when \sin x=\dfrac{1}{4} (possible, since -1\le\dfrac{1}{4}\le1)

(d) \left(\sin x-\dfrac{1}{4}\right)^2 is largest when \sin x=-1: f=\dfrac{13}{4}-4\left(\dfrac{25}{16}\right)=-3

at x=\dfrac{3\pi}{2}

(e) let s=\sin x. Each value of s with -1<s<1, s\ne0, gives two values of x; s=0 gives three (0,\pi,2\pi); s=\pm1 gives one

g(s)=3+2s-4s^2 has g(-1)=-3, g\left(-\dfrac{1}{2}\right)=1, g(1)=1, g\left(\dfrac{1}{4}\right)=\dfrac{13}{4}; for -3<k<1 the only root in [-1,1] is one value of s in \left(-1,-\dfrac{1}{2}\right)

-3<k<1

or k=\dfrac{13}{4} (the single value s=\dfrac{1}{4})

Note: for 1<k<\dfrac{13}{4} there are two values of s in (-1,1), giving four solutions (five when k=3); k=1 gives three. Award A1 for -3<k<1 with correct strict signs.

Final answer

(a) shown
(b) 0,\ \frac\pi6,\ \frac{5\pi}{6},\ \pi,\ 2\pi
(c) \frac{13}{4} when \sin x=\frac14
(d) -3 at x=\frac{3\pi}{2}
(e) -3<k<1 or k=\frac{13}{4}

Question 25

[6 marks]

A coaster is the shape bounded by three circular arcs. The points A, B and C are the vertices of an equilateral triangle with sides of 6 cm; the arc BC has centre A, the arc CA has centre B and the arc AB has centre C.

(a) Find the exact perimeter of the coaster.

[2]

(b) Show that the area of the coaster is 18(\pi-\sqrt3) cm^2.

[4]

Show solutionHide solution

(a) Each arc has radius 6 and angle \frac{\pi}{3}, so length 6\cdot\frac{\pi}{3}=2\pi.

Perimeter =3\times2\pi=6\pi cm.

(b) Each of the three sectors has area \frac12(6^2)\frac{\pi}{3}=6\pi.

The triangle has area \frac12(6)(6)\sin\frac{\pi}{3}=9\sqrt3.

The three sectors cover the triangle three times and the rest of the coaster once, so adding them counts the triangle two times too many.

Area =3(6\pi)-2(9\sqrt3)=18\pi-18\sqrt3=18(\pi-\sqrt3) cm^2.

Final answer

(a) 6\pi cm
(b) shown

Question 26

[14 marks]

The number of hours of daylight, D, at a town on day t of the year, where t is the number of days after 1 January, is modelled by D(t)=a\cos\bigl(b(t-c)\bigr)+d, with a>0, b>0 and 0\le c<365. The longest day has 15.8 hours of daylight and occurs when t=172. The shortest day has 8.6 hours of daylight and the model has a period of 365 days.

(a) Find the values of a, b, c and d.

[4]

(b) Find the number of hours of daylight when t=100.

[1]

A solar water heater runs only on days with at least 13 hours of daylight.

(c) Find the values of t, for 0\le t\le 365, on which the heater can run, giving your answers to the nearest 0.1.

[3]

(d) Without differentiating, find the value of t, for 0\le t\le 172, at which the number of hours of daylight is increasing most rapidly. Give a reason for your answer.

[2]

A second town has daylight E(t)=11.4+2.3\cos\bigl(b(t-160)\bigr) hours, with the same value of b.

(e) Find the values of t, for 0\le t\le 365, for which the first town has more daylight than the second.

[2]

(f) Find the greatest value of D(t)-E(t) for 0\le t\le 365, and the value of t at which it occurs.

[2]

Show solutionHide solution

Radian mode.

(a) a=\dfrac{15.8-8.6}{2}=3.6 and d=\dfrac{15.8+8.6}{2}=12.2.

[2]

b=\dfrac{2\pi}{365}=0.017214\ldots\approx 0.0172.

[1]

A cosine takes its greatest value when its argument is zero, so c=172.

[1]

(b) D(100)=3.6\cos\left(\dfrac{2\pi}{365}(100-172)\right)+12.2=13.37\ldots\approx 13.4 hours.

[1]

(c) Solve 3.6\cos\bigl(b(t-172)\bigr)+12.2=13, using the GDC intersection tool (or \cos\bigl(b(t-172)\bigr)=\tfrac29).

[1]

t=93.8 and t=250.2 (to the nearest 0.1).

[1]

The heater can run for 93.8\le t\le 250.2.

[1]

(d) t=172-\dfrac{365}{4}=80.75.

[1]

On a sinusoid the gradient is steepest where the graph crosses its principal axis; the rising crossing comes a quarter of a period before the maximum.

[1]

(e) Graph y=D(t)-E(t) and find its zeros (equivalently intersect D and E): t=65.8\ldots and t=317.3\ldots

[1]

D(t)>E(t) for 65.8<t<317.3.

[1]

(f) Maximum of y=D(t)-E(t) on the GDC: greatest value 2.23 (that is, D-E=2.2290\ldots)

[1]

at t=191.5.

[1]

Final answer

(a) a=3.6, b=\frac{2\pi}{365}, c=172, d=12.2
(b) 13.4 hours
(c) 93.8\le t\le250.2
(d) t=80.75
(e) 65.8<t<317.3
(f) 2.23 hours at t=191.5

Question 27

[16 marks]

A farmer has 60 m of fencing to enclose a paddock in the shape of a sector of a circle, with radius r metres and angle \theta radians, where 0 < \theta < 2\pi. All of the fencing is used for the straight edges and the curved edge of the sector.

(a) Show that r = \dfrac{60}{2 + \theta}.

[2]

(b) Show that the area A of the paddock is given by A = \dfrac{1800\theta}{(2 + \theta)^2}.

[2]

(c) For \theta = 0.8, find the radius r and the area A.

[3]

(d) The farmer wants an area of 150 m^2. Find the value of \theta that achieves this.

[4]

(e) Use your GDC to find the greatest possible area of the paddock, and the value of \theta at which it occurs.

[2]

(f) Show that A \le 225 for all \theta > 0.

[3]

Show solutionHide solution

(a) The perimeter is two radii plus the arc: 2r + r\theta = 60.

[1]

r(2 + \theta) = 60

[1]

so r = \dfrac{60}{2 + \theta}.

(b) A = \tfrac12 r^2\theta = \tfrac12\cdot\dfrac{3600}{(2 + \theta)^2}\cdot\theta

[1]

= \dfrac{1800\theta}{(2 + \theta)^2}.

Substitution of r shown

[1]

(c) r = \dfrac{60}{2.8} = 21.428\ldots \approx 21.4 m.

[1]

A = \dfrac{1800(0.8)}{(2.8)^2} = 183.67\ldots \approx 184 m^2.

[2]

(d) \dfrac{1800\theta}{(2 + \theta)^2} = 150, so 12\theta = (2 + \theta)^2.

[1]

\theta^2 - 8\theta + 4 = 0.

[1]

\theta = 4 \pm 2\sqrt3 = 0.536\ldots or 7.46\ldots

[1]

Since \theta < 2\pi = 6.28\ldots, \theta = 4 - 2\sqrt3 = 0.536 radians.

[1]

(e) Graph A(\theta) = \dfrac{1800\theta}{(2 + \theta)^2} and use the Maximum tool: the greatest area is 225 m^2

[1]

at \theta = 2 radians.

[1]

(f) 225 - A = \dfrac{225(2 + \theta)^2 - 1800\theta}{(2 + \theta)^2}

[1]

= \dfrac{225\left(\theta^2 + 4\theta + 4 - 8\theta\right)}{(2 + \theta)^2} = \dfrac{225(\theta - 2)^2}{(2 + \theta)^2}.

[1]

This is non-negative for all \theta, so A \le 225, with equality only at \theta = 2.

[1]

Final answer

(a), (b) shown
(c) r=21.4 m, A=184 m^2
(d) \theta=4-2\sqrt3\approx0.536 (4+2\sqrt3>2\pi is rejected)
(e) 225 m^2 at \theta=2
(f) shown

Going further

Beyond the syllabus: the worked examples and investigations of the chapter's Going further pages.

G1 Area from three sides

G1 Worked example

[10 marks]

A triangle has sides 7, 8 and x, and area A.

(a) Use Heron's formula to show that 16A^2=(x^2-1)(225-x^2).

[3]

(b) State the condition on x for such a triangle to exist.

[1]

(c) Find the value of x for which A is greatest, and this greatest area. Show that the triangle is then right-angled.

[3]

(d) Find the two values of x for which A=14\sqrt3. For each, find the angle between the sides of length 7 and 8.

[3]

Show solutionHide solution

(a) s=\dfrac{15+x}{2}, so s-7=\dfrac{x+1}{2}, s-8=\dfrac{x-1}{2}, s-x=\dfrac{15-x}{2}

A^2=\dfrac{(15+x)(x+1)(x-1)(15-x)}{16}

pair the brackets as differences of squares: 16A^2=(x^2-1)(225-x^2)

A1AG

(b) 1<x<15

Note: both brackets must be positive; this is the triangle inequality.

(c) with u=x^2, 16A^2=-u^2+226u-225, a quadratic with its vertex at u=113

x=\sqrt{113}, 16A^2=112\times112, so A=28

x^2=113=7^2+8^2, so by the cosine rule the angle between the sides 7 and 8 has cosine 0: it is a right angle

Note: the check A=\tfrac12(7)(8)\sin C\le28, with equality only at C=90^\circ, gives the same answer.

(d) 16\times588=9408=(u-1)(225-u), so u^2-226u+9633=0

u=113\pm56: x=\sqrt{57} or x=13

\cos C=\dfrac{113-u}{112}=\dfrac12 or -\dfrac12: C=60^\circ (when x=\sqrt{57}) and C=120^\circ (when x=13)

Final answer

(b) 1<x<15
(c) x=\sqrt{113}, A=28
(d) x=\sqrt{57}, 60^\circ
x=13, 120^\circ

G1 Investigation

[11 marks]

A triangle is called Heronian if its three sides and its area are all integers. (This is not on the syllabus; a Paper 3 question would define it.)

(a) Use Heron's formula to find the areas of the triangles with sides 13,14,15; 5,5,6; and 5,5,8.

[2]

(b) A triangle has sides a, b, c and s=\frac12(a+b+c). Let p=s-a, q=s-b and r=s-c. Show that p, q and r are positive, that p+q+r=s, and that the area is \sqrt{s\,pqr}.

[2]

(c) Find every Heronian triangle with perimeter 32.

[4]

(d) Prove that every Heronian triangle has an even perimeter.

[3]

Show solutionHide solution

(a) s=21: \sqrt{21\times8\times7\times6}=\sqrt{7056}=84; ; s=8: \sqrt{8\times3\times3\times2}=12; ; s=9: \sqrt{9\times4\times4\times1}=12

A1A1

Note: one mark for 84; one mark for the two areas of 12.

(b) p=\dfrac{b+c-a}{2}>0 by the triangle inequality, and the same for q and r; p+q+r=3s-(a+b+c)=3s-2s=s, and Heron's formula reads \sqrt{s\,pqr}

A1A1

(c) s=16, so p,q,r are positive integers with p+q+r=16, and A=\sqrt{16pqr}=4\sqrt{pqr}

A is an integer only if pqr is a perfect square

list p\le q\le r with p+q+r=16 (21 triples) and test pqr: only 1\times3\times12=36 and 4\times6\times6=144 are squares

sides 16-p, 16-q, 16-r: the triangle 15,13,4 (area 24) and the triangle 12,10,10 (area 48)

(d) suppose the perimeter P=a+b+c is odd. Heron's formula in the form 16A^2=P(P-2a)(P-2b)(P-2c)

each of the four factors is odd, so 16A^2 is odd

but if A is an integer then 16A^2 is even: a contradiction, so P is even

Final answer

(a) 84, 12, 12
(c) 4,13,15 (area 24) and 10,10,12 (area 48)
(d) proof

G2 The triangle inside the pentagon

G2 Worked example

[8 marks]

Let \theta=36^\circ.

(a) Explain why \sin3\theta=\sin2\theta.

[1]

(b) Given that \sin3\theta=3\sin\theta-4\sin^3\theta, show that 4\cos^2\theta-2\cos\theta-1=0.

[3]

(c) Hence find the exact value of \cos36^\circ.

[2]

(d) A regular pentagon has sides of length 1. Show that each diagonal has length \dfrac{1+\sqrt5}{2}.

[2]

Show solutionHide solution

(a) 3\theta=108^\circ=180^\circ-72^\circ=180^\circ-2\theta, and \sin(180^\circ-\alpha)=\sin\alpha

(b) 3\sin\theta-4\sin^3\theta=2\sin\theta\cos\theta

\sin\theta\ne0, so 3-4\sin^2\theta=2\cos\theta

3-4(1-\cos^2\theta)=2\cos\theta, so 4\cos^2\theta-2\cos\theta-1=0

A1AG

(c) \cos\theta=\dfrac{2\pm\sqrt{4+16}}{8}=\dfrac{1\pm\sqrt5}{4}

\cos36^\circ>0, so \cos36^\circ=\dfrac{1+\sqrt5}{4}

(d) a diagonal and two adjacent sides form an isosceles triangle with apex angle 108^\circ and base angles 36^\circ

the perpendicular from the apex bisects the diagonal, so the diagonal is 2\times1\times\cos36^\circ=\dfrac{1+\sqrt5}{2}

A1AG

Final answer

G2 Investigation

[10 marks]

(a) Use the exact values of \cos36^\circ and \cos72^\circ to show that \cos36^\circ-\cos72^\circ=\frac12 and \cos36^\circ\cos72^\circ=\frac14.

[2]

(b) Use your GDC to evaluate \cos\frac{\pi}{7}-\cos\frac{2\pi}{7}+\cos\frac{3\pi}{7}.

[1]

(c) For an odd integer n=2m+1\ge3, let T_n=\cos\frac{\pi}{n}-\cos\frac{2\pi}{n}+\cos\frac{3\pi}{n}-\cdots+(-1)^{m+1}\cos\frac{m\pi}{n}, a sum of m terms with alternating signs. Find T_9 with your GDC, and write down a conjecture for T_n.

[2]

(d) By expanding \cos(A+B) and \cos(A-B), show that 2\cos A\cos B=\cos(A+B)+\cos(A-B).

[1]

(e) Let \theta=\frac{\pi}{n}. Use (d) with A=k\theta and B=\frac{\theta}{2} to write 2\cos\frac{\theta}{2}\,T_n as a sum in which almost every term cancels. Hence prove your conjecture.

[4]

Show solutionHide solution

(a) \dfrac{1+\sqrt5}{4}-\dfrac{\sqrt5-1}{4}=\dfrac24=\dfrac12 and \dfrac{(\sqrt5+1)(\sqrt5-1)}{16}=\dfrac{4}{16}=\dfrac14

A1A1

(b) 0.5

(c) T_9=\cos20^\circ-\cos40^\circ+\cos60^\circ-\cos80^\circ=0.5

conjecture: T_n=\dfrac12 for every odd n\ge3

Note: T_3=\cos60^\circ=\frac12 and T_5=\frac12 by (a) also fit.

(d) \cos(A\pm B)=\cos A\cos B\mp\sin A\sin B. Adding the two expansions, the sine terms cancel and the cosine terms double

A1AG

(e) 2\cos\frac{\theta}{2}\cos k\theta=\cos\left(k+\frac12\right)\theta+\cos\left(k-\frac12\right)\theta

with the signs of T_n: \left(\cos\frac{3\theta}{2}+\cos\frac{\theta}{2}\right)-\left(\cos\frac{5\theta}{2}+\cos\frac{3\theta}{2}\right)+\left(\cos\frac{7\theta}{2}+\cos\frac{5\theta}{2}\right)-\cdots; each term after the first cancels with the next bracket

2\cos\frac{\theta}{2}\,T_n=\cos\frac{\theta}{2}+(-1)^{m+1}\cos\left(m+\frac12\right)\theta, and \left(m+\frac12\right)\theta=\dfrac{(2m+1)\pi}{2n}=\dfrac{\pi}{2}, so the last term is 0

\cos\frac{\theta}{2}\ne0, so T_n=\dfrac12

Final answer

(b) 0.5
(c) T_9=0.5, conjecture T_n=\frac12
(e) proof

TI-84 Plus CE keystrokes

The GDC panels in the book give keystrokes for the TI-Nspire CX II and the Casio fx-CG50. These are the TI-84 Plus CE keystrokes for the same calculator jobs. The full list of TI-84 keystrokes is in Chapter 19.

Job	TI-84 Plus CE keystrokes
Radian mode	`MODE, Radian, ENTER`
Intersections	`Graph screen: Y1 and Y2, 2nd TRACE (CALC), 5:intersect, three ENTERs per crossing`
Solve an equation	`MATH, Numeric Solver (MATH, 0:Solver on older OS versions): enter the equation, then SOLVE. Or graph both sides and use 2nd TRACE (CALC), 5:intersect`

From the book

The Ultimate Guide for the IB: Mathematics: Analysis and Approaches HL

Chapter 10 teaches trigonometry from the basics to full exam questions: how the IB asks it, every question type with a marked example, and these practice questions in three steps.

About the book