Free worked solutions · 2026 Edition

IB Maths AA SL Chapter 6: Trigonometric functions and equations worked solutions

Worked solutions to the practice questions in Chapter 6 of The Ultimate Guide for the IB: Mathematics: Analysis and Approaches SL (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 unit circle meets the coordinate axes at four points.

(a) State the four coordinate pairs for the unit-circle angles 0, \dfrac{\pi}{2}, \pi and \dfrac{3\pi}{2}.

[2]

(b) Hence write down the value of

(i) \sin\pi (ii) \cos\dfrac{3\pi}{2} (iii) \cos\pi (iv) \sin\dfrac{3\pi}{2} (v) \tan\pi (vi) \cos 2\pi

[3]

(c) Explain why \tan\dfrac{\pi}{2} is not defined.

[1]

Show solutionHide solution

(a) 0: (1,0); \dfrac{\pi}{2}: (0,1); \pi: (-1,0); \dfrac{3\pi}{2}: (0,-1)

Note: award A1 for two or three correct points.

(b) cosine is the x-coordinate and sine is the y-coordinate

(i) \sin\pi=0 and (ii) \cos\dfrac{3\pi}{2}=0

(iii) \cos\pi=-1 and (iv) \sin\dfrac{3\pi}{2}=-1

(v) \tan\pi=\dfrac{0}{-1}=0 and (vi) \cos 2\pi=1 (same point as angle 0)

Note: award each A1 only when both values in the pair are correct.

(c) \tan\dfrac{\pi}{2}=\dfrac{\sin(\pi/2)}{\cos(\pi/2)}=\dfrac{1}{0}, and division by zero is not defined

Final answer

(a) (1,0),(0,1),(-1,0),(0,-1)
(b) 0,0,-1,-1,0,1
(c) division by zero.

Question 2

[6 marks]

Find the exact value of

(a) \sin\dfrac{5\pi}{6} (b) \cos\dfrac{4\pi}{3} (c) \tan\dfrac{7\pi}{4} (d) \sin\dfrac{5\pi}{4} (e) \cos150^\circ (f) \tan240^\circ

[6]

Show solutionHide solution

reference angle (acute angle to the x-axis) gives the size; the quadrant gives the sign

(a) second quadrant, reference angle \dfrac{\pi}{6}, \sin>0: \sin\dfrac{5\pi}{6}=\dfrac{1}{2}

(b) third quadrant, reference angle \dfrac{\pi}{3}, \cos<0: \cos\dfrac{4\pi}{3}=-\dfrac{1}{2}

(c) fourth quadrant, reference angle \dfrac{\pi}{4}, \tan<0: \tan\dfrac{7\pi}{4}=-1

(d) third quadrant, reference angle \dfrac{\pi}{4}, \sin<0: \sin\dfrac{5\pi}{4}=-\dfrac{\sqrt{2}}{2}

(e) second quadrant, reference angle 30^\circ, \cos<0: \cos150^\circ=-\dfrac{\sqrt{3}}{2}

(f) third quadrant, reference angle 60^\circ, \tan>0: \tan240^\circ=\sqrt{3}

Note: accept equivalent exact forms such as -\dfrac{1}{\sqrt{2}}.

Final answer

\frac12,-\frac12,-1,-\frac{\sqrt2}{2},-\frac{\sqrt3}{2},\sqrt3.

Question 3

[6 marks]

The angle x is acute and \cos x=k, where 0<k<1.

(a) Find \sin x in terms of k.

[2]

(b) Find \tan x in terms of k.

[1]

(c) Given also that \tan x=2\sqrt2, find the value of k.

[3]

Show solutionHide solution

(a) \sin^2x=1-k^2

\sin x=\sqrt{1-k^2} (positive because x is acute)

(b) \tan x=\dfrac{\sqrt{1-k^2}}{k}

(c) \dfrac{\sqrt{1-k^2}}{k}=2\sqrt2

1-k^2=8k^2 (squaring both sides)

k^2=\dfrac19, and k>0, so k=\dfrac{1}{3}

Note: FT from (b). Do not award the final A1 for k=\pm\dfrac13.

Final answer

\sqrt{1-k^2}; \sqrt{1-k^2}/k; k=\frac13.

Question 4

[9 marks]

Simplify each expression fully. You may assume that no denominator is zero.

(a) \dfrac{\sin2x}{\sin x}

[2]

(b) \dfrac{1-\cos2x}{\sin x}

[2]

(c) 6\sin3x\cos3x

[2]

(d) \cos^22x-\sin^22x

[1]

(e) \dfrac{\cos2x}{\cos x-\sin x}

[2]

Show solutionHide solution

(a) \dfrac{2\sin x\cos x}{\sin x}

=2\cos x

(b) \dfrac{2\sin^2x}{\sin x}

=2\sin x

(c) 3(2\sin3x\cos3x)

=3\sin6x

(d) \cos4x

(e) \dfrac{(\cos x-\sin x)(\cos x+\sin x)}{\cos x-\sin x}

=\cos x+\sin x

Note: in (b) the key choice is the form \cos2x=1-2\sin^2x; in (e) it is \cos2x=\cos^2x-\sin^2x factorised as a difference of two squares.

Final answer

2\cos x; 2\sin x; 3\sin6x; \cos4x; \cos x+\sin x.

Question 5

[7 marks]

Let f(x)=7-4\cos 2x, for 0 \le x \le 2\pi.

(a) Write down the maximum value and the minimum value of f.

[2]

(b) Find the values of x at which f takes its maximum value.

[2]

(c) Find the range of f(x) for 0 \le x \le \dfrac{\pi}{6}.

[3]

Show solutionHide solution

(a) maximum 11

minimum 3

(b) maximum when \cos 2x=-1, i.e. 2x=\pi or 2x=3\pi (for 0 \le 2x \le 4\pi)

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

(c) for 0 \le x \le \dfrac{\pi}{6}, 0 \le 2x \le \dfrac{\pi}{3}, where \cos 2x decreases from 1 to \dfrac{1}{2}, so f increases

f(0)=7-4=3 and f\left(\dfrac{\pi}{6}\right)=7-4\cos\dfrac{\pi}{3}=5

3 \le f(x) \le 5

Final answer

(a) 11,3
(b) \frac\pi2,\frac{3\pi}2
(c) [3,5].

Question 6

[5 marks]

The graph of y=\cos x is mapped onto the graph of y=3\cos 2x-5 by a sequence of three transformations.

(a) Describe fully each of the three transformations.

[3]

(b) The point (\pi,-1) lies on the graph of y=\cos x. Find the coordinates of its image on the graph of y=3\cos 2x-5.

[2]

Show solutionHide solution

(a) vertical stretch with scale factor 3

horizontal stretch with scale factor \dfrac{1}{2}

translation by the vector \begin{pmatrix}0\\-5\end{pmatrix}

Note: the vertical stretch must come before the vertical translation; the horizontal stretch may be in any position. Award A0 for “stretch by 2” in the x-direction.

(b) x-coordinate halved: \dfrac{\pi}{2}

y-coordinate: 3(-1)-5=-8, image \left(\dfrac{\pi}{2},-8\right)

Final answer

Vertical stretch 3, horizontal stretch \frac12, then translate 5 down; (\frac\pi2,-8).

Question 7

[7 marks]

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

(a) \cos 2x=-\dfrac{\sqrt2}{2}, for 0\le x\le 2\pi.

[4]

(b) \sin\dfrac{x}{2}=\dfrac{1}{2}, for 0\le x\le 4\pi.

[3]

Show solutionHide solution

(a) 0\le 2x\le 4\pi

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

2x=\dfrac{11\pi}{4},\ \dfrac{13\pi}{4} (adding 2\pi)

x=\dfrac{3\pi}{8},\ \dfrac{5\pi}{8},\ \dfrac{11\pi}{8},\ \dfrac{13\pi}{8}

(b) 0\le\dfrac{x}{2}\le 2\pi

\dfrac{x}{2}=\dfrac{\pi}{6},\ \dfrac{5\pi}{6}

x=\dfrac{\pi}{3},\ \dfrac{5\pi}{3}

Note: a candidate who solves \cos 2x=-\dfrac{\sqrt2}{2} only for 0\le2x\le2\pi finds two of the four solutions: award M0A1A0A0.

Final answer

(a) \frac{3\pi}8,\frac{5\pi}8,\frac{11\pi}8,\frac{13\pi}8
(b) \frac\pi3,\frac{5\pi}3.

Question 8

[7 marks]

Consider the equation \tan 3x=-\sqrt3, for 0\le x\le 2\pi.

(a) Write down the interval in which 3x lies.

[1]

(b) Solve the equation, giving your answers as exact multiples of \pi.

[4]

(c) Find the sum of all the solutions.

[2]

Show solutionHide solution

(a) 0\le 3x\le 6\pi

(b) first value in the interval 3x=\dfrac{2\pi}{3}; \tan has period \pi, so keep adding \pi while 3x\le6\pi

3x=\dfrac{2\pi}{3},\ \dfrac{5\pi}{3},\ \dfrac{8\pi}{3},\ \dfrac{11\pi}{3},\ \dfrac{14\pi}{3},\ \dfrac{17\pi}{3}

dividing by 3 (M1)

x=\dfrac{2\pi}{9},\ \dfrac{5\pi}{9},\ \dfrac{8\pi}{9},\ \dfrac{11\pi}{9},\ \dfrac{14\pi}{9},\ \dfrac{17\pi}{9}

(c) \dfrac{\pi}{9}(2+5+8+11+14+17)

=\dfrac{57\pi}{9}=\dfrac{19\pi}{3}

Note: in (b) \tan^{-1}(-\sqrt3)=-\dfrac{\pi}{3} is not in the interval, and \dfrac{20\pi}{3}>6\pi, so there are exactly six solutions. In (c) accept the sum as an arithmetic series, \dfrac{6}{2}\left(\dfrac{2\pi}{9}+\dfrac{17\pi}{9}\right).

Final answer

(a) [0,6\pi]
(b) \frac\pi9(2,5,8,11,14,17)
(c) \frac{19\pi}3.

Tutorial

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

Question 9

[5 marks]

(a) Find the exact value of \sin^2(2\pi/3)-\cos(4\pi/3)\tan(3\pi/4).

[3]

(b) Write down the exact value of \cos(13\pi/6).

[1]

(c) Write down the exact value of \tan(10\pi/3).

[1]

Show solutionHide solution

(a) \sin\dfrac{2\pi}{3} = \dfrac{\sqrt3}{2}, so \sin^2\dfrac{2\pi}{3} = \dfrac34.

[1]

\cos\dfrac{4\pi}{3} = -\dfrac12 and \tan\dfrac{3\pi}{4} = -1.

[1]

Expression = \dfrac34 - \left(-\dfrac12\right)(-1) = \dfrac34 - \dfrac12 = \dfrac14.

[1]

(b) \dfrac{13\pi}{6} = 2\pi + \dfrac{\pi}{6}, and cosine has period 2\pi, so \cos\dfrac{13\pi}{6} = \cos\dfrac{\pi}{6} = \dfrac{\sqrt3}{2}.

[1]

(c) \dfrac{10\pi}{3} = 3\pi + \dfrac{\pi}{3}, and tangent has period \pi, so \tan\dfrac{10\pi}{3} = \tan\dfrac{\pi}{3} = \sqrt3.

[1]

Final answer

\frac14,\frac{\sqrt3}{2},\sqrt3.

Question 10

[5 marks]

The angle \theta is obtuse and \sin\theta = \dfrac{5}{13}.

(a) Find the exact value of \cos\theta.

[2]

(b) Find the exact value of \tan\theta.

[1]

(c) Find the exact value of \sin 2\theta.

[2]

Show solutionHide solution

(a) \cos^2\theta = 1 - \sin^2\theta = 1 - \dfrac{25}{169} = \dfrac{144}{169}.

[1]

\theta is obtuse, so cosine is negative: \cos\theta = -\dfrac{12}{13}.

[1]

(b) \tan\theta = \dfrac{\sin\theta}{\cos\theta} = -\dfrac{5}{12}.

[1]

(c) \sin 2\theta = 2\sin\theta\cos\theta = 2\left(\dfrac{5}{13}\right)\left(-\dfrac{12}{13}\right)

[1]

= -\dfrac{120}{169}.

[1]

Final answer

-\frac{12}{13}; -\frac5{12}; -\frac{120}{169}.

Question 11

[5 marks]

(a) Write \cos^2 3x - \sin^2 3x as a single trigonometric function.

[1]

(b) Write 8\sin(x/2)\cos(x/2) in the form k\sin x, stating the value of k.

[1]

(c) Show that \dfrac{1 + \cos 2x}{2\cos x} \equiv \cos x, for \cos x \neq 0.

[3]

Show solutionHide solution

(a) \cos^2 3x - \sin^2 3x = \cos 6x (double angle with \theta = 3x).

[1]

(b) 8\sin\dfrac{x}{2}\cos\dfrac{x}{2} = 4\left(2\sin\dfrac{x}{2}\cos\dfrac{x}{2}\right) = 4\sin x, so k = 4.

[1]

(c) Numerator: 1 + \cos 2x = 1 + (2\cos^2 x - 1) = 2\cos^2 x.

[1]

\dfrac{1 + \cos 2x}{2\cos x} = \dfrac{2\cos^2 x}{2\cos x}.

[1]

= \cos x (as \cos x \neq 0).

[1]

Final answer

\cos6x; k=4; proved by 1+\cos2x=2\cos^2x.

Question 12

[6 marks]

(a) Show that \cos^6 x + \sin^6 x \equiv 1 - \tfrac34\sin^2 2x.

[3]

(b) Hence find the exact value of \cos^6\dfrac{\pi}{12} + \sin^6\dfrac{\pi}{12}.

[2]

(c) Write down the minimum value of \cos^6 x + \sin^6 x.

[1]

Show solutionHide solution

(a) Using a^3 + b^3 = (a + b)^3 - 3ab(a + b) with a = \cos^2 x, b = \sin^2 x: \cos^6 x + \sin^6 x = (\cos^2 x + \sin^2 x)^3 - 3\sin^2 x\cos^2 x(\cos^2 x + \sin^2 x).

[1]

= 1 - 3\sin^2 x\cos^2 x.

[1]

Since \sin 2x = 2\sin x\cos x, we have \sin^2 2x = 4\sin^2 x\cos^2 x, so 3\sin^2 x\cos^2 x = \tfrac34\sin^2 2x and the expression is 1 - \tfrac34\sin^2 2x.

[1]

(b) With x = \dfrac{\pi}{12}: 1 - \tfrac34\sin^2\dfrac{\pi}{6}.

[1]

= 1 - \tfrac34\left(\dfrac14\right) = \dfrac{13}{16}.

[1]

(c) The minimum occurs when \sin^2 2x = 1: 1 - \tfrac34 = \dfrac14.

[1]

Final answer

(a) proved
(b) \frac{13}{16}
(c) \frac14.

Question 13

[4 marks]

Consider the function f(x) = 3 - 5\cos\dfrac{x}{2}.

(a) Write down the amplitude.

[1]

(b) Calculate the period.

[1]

(c) State the principal-axis equation (the line midway between the maximum and minimum).

[1]

(d) Write down the range of f.

[1]

Show solutionHide solution

(a) Amplitude = |{-5}| = 5.

[1]

(b) Period = \dfrac{2\pi}{1/2} = 4\pi.

[1]

(c) y = 3.

[1]

(d) Maximum 3 + 5 = 8, minimum 3 - 5 = -2, so -2 \le f(x) \le 8.

[1]

Final answer

5,4\pi,y=3,[-2,8].

Question 14

[5 marks]

The graph of g(x) = 3\sin\bigl(2(x - \tfrac{\pi}{6})\bigr) comes from y=\sin x after three transformations in sequence.

(a) Describe the three transformations.

[3]

(b) Write down the maximum value of g, and the smallest positive value of x at which it occurs.

[2]

Show solutionHide solution

(a) Vertical stretch, scale factor 3.

[1]

Horizontal stretch, scale factor \dfrac12 (from \sin x to \sin 2x).

[1]

Translation by the vector \begin{pmatrix}\pi/6\\ 0\end{pmatrix} (applied last).

[1]

(b) Maximum value = 3.

[1]

It occurs when 2\left(x - \dfrac{\pi}{6}\right) = \dfrac{\pi}{2}, so x = \dfrac{\pi}{6} + \dfrac{\pi}{4} = \dfrac{5\pi}{12}.

[1]

Final answer

Vertical stretch 3, horizontal stretch \frac12, then right \frac\pi6; max 3 at \frac{5\pi}{12}.

Question 15

[5 marks]

The height h metres of a buoy above the sea bed at time t seconds is modelled by h(t) = 4 - 1.5\cos\left(\dfrac{\pi t}{3}\right), for t \ge 0.

(a) Find the maximum and minimum heights of the buoy.

[2]

(b) Find the period of the motion.

[1]

(c) Find the height of the buoy when t = 1.

[1]

(d) Write down the first time after t = 0 at which the buoy is at its maximum height.

[1]

Show solutionHide solution

(a) Maximum = 4 + 1.5 = 5.5 m.

[1]

Minimum = 4 - 1.5 = 2.5 m.

[1]

(b) Period = \dfrac{2\pi}{\pi/3} = 6 s.

[1]

(c) h(1) = 4 - 1.5\cos\dfrac{\pi}{3} = 4 - 0.75 = 3.25 m.

[1]

(d) The maximum occurs when \cos\dfrac{\pi t}{3} = -1, i.e.; half a period after t = 0: t = 3 s.

[1]

Final answer

5.5 m, 2.5 m; 6 s; 3.25 m; 3 s.

Question 16

[5 marks]

Consider the equation \sin\left(3x - \dfrac{\pi}{4}\right) = \dfrac12.

(a) Find the smallest positive value of x that satisfies the equation.

[3]

(b) Find the next positive value of x.

[2]

Show solutionHide solution

(a) For x > 0 we have 3x - \dfrac{\pi}{4} > -\dfrac{\pi}{4}, and the first angle in this range with sine \dfrac12 is \dfrac{\pi}{6}.

[1]

3x = \dfrac{\pi}{6} + \dfrac{\pi}{4} = \dfrac{5\pi}{12}.

[1]

x = \dfrac{5\pi}{36}.

[1]

(b) The next angle with sine \dfrac12 is \pi - \dfrac{\pi}{6} = \dfrac{5\pi}{6}, so 3x = \dfrac{5\pi}{6} + \dfrac{\pi}{4} = \dfrac{13\pi}{12}.

[1]

x = \dfrac{13\pi}{36}.

[1]

Final answer

\frac{5\pi}{36}; \frac{13\pi}{36}.

Question 17

[5 marks]

Solve \sin 2x + \sin x = 0 for 0 \le x \le 2\pi.

Show solutionHide solution

Expand the double angle: 2\sin x\cos x + \sin x = 0.

[1]

Factorise (rather than divide by \sin x): \sin x\,(2\cos x + 1) = 0.

[1]

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

[1]

\cos x = -\dfrac12: x = \dfrac{2\pi}{3},\ \dfrac{4\pi}{3}.

[1]

All five solutions: x = 0,\ \dfrac{2\pi}{3},\ \pi,\ \dfrac{4\pi}{3},\ 2\pi.

[1]

Final answer

0,\frac{2\pi}3,\pi,\frac{4\pi}3,2\pi.

Mastery

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

Question 18

[6 marks]

The function g is defined by g(x) = p\cos(qx) + r, where p > 0 and q > 0. The period of g is \pi, g(0) = 8 and g\left(\dfrac{\pi}{3}\right) = 2.

(a) Write down the value of q.

[1]

(b) Find the values of p and r.

[3]

(c) Find the exact value of g\left(\dfrac{\pi}{8}\right).

[1]

(d) Write down the number of solutions of g(x) = 3 for 0 \le x \le 2\pi.

[1]

Show solutionHide solution

(a) q = \dfrac{2\pi}{\pi} = 2.

[1]

(b) g(0) = p + r = 8 and g\left(\dfrac{\pi}{3}\right) = p\cos\dfrac{2\pi}{3} + r = r - \dfrac{p}{2} = 2.

[1]

Subtract: \dfrac{3p}{2} = 6, so p = 4.

[1]

r = 8 - 4 = 4.

[1]

(c) g\left(\dfrac{\pi}{8}\right) = 4\cos\dfrac{\pi}{4} + 4 = 2\sqrt2 + 4.

[1]

(d) g has range 0 \le g \le 8 and period \pi, so [0, 2\pi] contains two periods. In each period the line y = 3 (between the minimum 0 and the principal axis 4) is crossed twice: 4 solutions.

[1]

Final answer

q=2, p=4, r=4; 4+2\sqrt2; 4 solutions.

Question 19

[12 marks]

Let f(x) = \cos 2x + 2\cos x, for 0 \le x \le 2\pi.

(a) Show that f(x) = 2\cos^2 x + 2\cos x - 1.

[2]

(b) Hence solve f(x) = -1.

[4]

(c) Show that f(x) = 2\left(\cos x + \tfrac12\right)^2 - \tfrac32.

[2]

(d) Hence write down the minimum value of f(x), and the exact values of x at which it occurs.

[2]

(e) Find the range of f.

[2]

Show solutionHide solution

(a) f(x) = (2\cos^2 x - 1) + 2\cos x

[1]

= 2\cos^2 x + 2\cos x - 1.

[1]

(b) 2\cos^2 x + 2\cos x - 1 = -1 \Rightarrow 2\cos^2 x + 2\cos x = 0.

[1]

2\cos x\,(\cos x + 1) = 0.

[1]

\cos x = 0: x = \dfrac{\pi}{2} and \dfrac{3\pi}{2}.

[1]

\cos x = -1: x = \pi.

[1]

(c) 2\left(\cos x + \tfrac12\right)^2 - \tfrac32 = 2\left(\cos^2 x + \cos x + \tfrac14\right) - \tfrac32

[1]

= 2\cos^2 x + 2\cos x - 1 = f(x).

[1]

(d) The square is never negative, so the minimum value is -\dfrac32, when \cos x = -\dfrac12.

[1]

x = \dfrac{2\pi}{3} and x = \dfrac{4\pi}{3}.

[1]

(e) The maximum occurs when \cos x = 1 (the square is largest there): f = 2\left(\dfrac32\right)^2 - \dfrac32 = 3.

[1]

Range: -\dfrac32 \le f(x) \le 3.

[1]

Final answer

(a),(c) proved
(b) \frac\pi2,\pi,\frac{3\pi}2
(d) -\frac32 at \frac{2\pi}3,\frac{4\pi}3
(e) [-\frac32,3].

Question 20

[8 marks]

(a) Show that (1 + \sin\theta + \cos\theta)^2 \equiv 2(1 + \sin\theta)(1 + \cos\theta).

[3]

(b) Hence solve (1 + \sin\theta + \cos\theta)^2 = 2(1 + \sin\theta) for 0 \le \theta \le 2\pi.

[5]

Show solutionHide solution

(a) Expand: (1 + \sin\theta + \cos\theta)^2 = 1 + \sin^2\theta + \cos^2\theta + 2\sin\theta + 2\cos\theta + 2\sin\theta\cos\theta.

[2]

Use \sin^2\theta + \cos^2\theta = 1: = 2 + 2\sin\theta + 2\cos\theta + 2\sin\theta\cos\theta.

[1]

= 2(1 + \sin\theta) + 2\cos\theta(1 + \sin\theta) = 2(1 + \sin\theta)(1 + \cos\theta).

(b) By part (a): 2(1 + \sin\theta)(1 + \cos\theta) = 2(1 + \sin\theta).

[1]

Do not divide by 1 + \sin\theta (it can be zero). Collect: 2(1 + \sin\theta)\bigl[(1 + \cos\theta) - 1\bigr] = 0.

[1]

2(1 + \sin\theta)\cos\theta = 0.

[1]

\cos\theta = 0: \theta = \dfrac{\pi}{2} or \dfrac{3\pi}{2}.

[1]

\sin\theta = -1: \theta = \dfrac{3\pi}{2}, already found. Solutions: \theta = \dfrac{\pi}{2},\ \dfrac{3\pi}{2}.

[1]

Final answer

(a) proved
(b) \frac\pi2,\frac{3\pi}2.

Question 21

[6 marks]

(a) Show that \dfrac{1}{1 + \cos x} + \dfrac{1}{1 - \cos x} \equiv \dfrac{2}{\sin^2 x}.

[3]

(b) Hence solve \dfrac{1}{1 + \cos x} + \dfrac{1}{1 - \cos x} = 8 for 0 < x < \pi.

[3]

Show solutionHide solution

(a) \dfrac{1}{1 + \cos x} + \dfrac{1}{1 - \cos x} = \dfrac{(1 - \cos x) + (1 + \cos x)}{(1 + \cos x)(1 - \cos x)}

[1]

= \dfrac{2}{1 - \cos^2 x}

[1]

= \dfrac{2}{\sin^2 x}.

Uses 1 - \cos^2 x = \sin^2 x

[1]

(b) Using part (a): \dfrac{2}{\sin^2 x} = 8, so \sin^2 x = \dfrac14.

[1]

For 0 < x < \pi, \sin x > 0, so \sin x = \dfrac12.

[1]

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

[1]

Final answer

(a) proved
(b) \frac\pi6,\frac{5\pi}6.

Question 22

[7 marks]

(a) Show that 4\sin x\cos x\cos 2x \equiv \sin 4x.

[3]

(b) Hence solve 4\sin x\cos x\cos 2x = \dfrac12 for 0 \le x \le \pi.

[4]

Show solutionHide solution

(a) 4\sin x\cos x\cos 2x = 2(2\sin x\cos x)\cos 2x = 2\sin 2x\cos 2x

[1]

= \sin 4x (using \sin 2\alpha = 2\sin\alpha\cos\alpha with \alpha = 2x).

[1]

Both double-angle steps shown correctly, with the result stated

[1]

(b) Using part (a): \sin 4x = \dfrac12.

[1]

The interval 0 \le x \le \pi becomes 0 \le 4x \le 4\pi.

[1]

4x = \dfrac{\pi}{6},\ \dfrac{5\pi}{6},\ \dfrac{13\pi}{6},\ \dfrac{17\pi}{6}.

[1]

x = \dfrac{\pi}{24},\ \dfrac{5\pi}{24},\ \dfrac{13\pi}{24},\ \dfrac{17\pi}{24}.

[1]

Final answer

(a) proved
(b) \frac\pi{24},\frac{5\pi}{24},\frac{13\pi}{24},\frac{17\pi}{24}.

Question 23

[18 marks]

An illuminated sculpture repeats a brightness cycle every 12 s. Its sensor records a minimum of 30 units and a maximum of 66 units. At t=1 s the reading crosses the midline while increasing. Model the reading by L(t)=a\sin(b(t-c))+d, with a,b>0 and 0\le c<12.

(a) Find the four model parameters.

[4]

(b) Sketch one cycle on 0\le t\le12, marking its maximum and minimum.

[3]

(c) Find the reading at t=0.

[2]

(d) A camera records only when L(t)>57. Find the recording time within the first cycle.

[3]

(e) The sculpture runs for 31 s from t=0. Find the total recording time.

[3]

(f) A second sensor measures J(t)=L(t)+6\cos(\pi(t-1)/6). Use your GDC to find the maximum value of J and the first time at which it occurs.

[3]

Show solutionHide solution

a=(66-30)/2=18, d=(66+30)/2=48 A1A1 Take half the difference and the mean.

b=2\pi/12=\pi/6, c=1 A1A1 Use the period and the rising midline time.

Plot the midline L=48, maximum (4,66) and minimum (10,30) A1A1 The extrema are quarter-cycle steps from the rising midline.

A smooth cycle passes through (1,48) and (7,48) A1 Join with a sinusoid and stop at the given domain.

L(0)=48+18\sin(-\pi/6) M1 Substitute in radian mode.

L(0)=39.0 units A1 Use the whole stored model.

L(t)=57 means \sin(\pi(t-1)/6)=1/2 M1 Equality gives the boundary times.

Crossings t=2.00,6.00; L>57 between them A1 Check which side of the level the curve occupies.

First-cycle recording time 4.00 s A1 Subtract the two crossing times.

High intervals are (2,6), (14,18) and (26,30) M1A1 Repeat by twelve seconds and clip at 31.

Total recording time 12.0 s A1 All three intervals are complete.

Plot J(t) in radian mode M1 Enter the complete sum.

Maximum 67.0 units, first maximum at t=3.39 s A1A1 Read the coordinates of the first maximum.

Final answer

(a) 18,\pi/6,1,48
(b) max (4,66), min (10,30)
(c) 39.0
(d) 4.00 s
(e) 12.0 s
(f) 67.0 units, 3.39 s.

Going further

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

Worked example

[6 marks]

The equal sides of a triangle are 1 unit and its apex angle is 36^\circ. Its base has length x. Bisect one base angle; the resulting similar triangles give x^2+x=1.

(a) Find the positive value of x.

[2]

(b) Use the cosine rule at the apex to find \cos36^\circ.

[2]

(c) Deduce \cos72^\circ.

[2]

Show solutionHide solution

x=(-1\pm\sqrt5)/2 M1 Apply the quadratic formula.

x=(\sqrt5-1)/2 A1 A length is positive.

x^2=2-2\cos36^\circ M1 The sides enclosing the apex both have length one.

\cos36^\circ=1-x^2/2=(1+\sqrt5)/4 A1 Use the verified value of the base.

\cos72^\circ=2\cos^236^\circ-1 M1 Use a double angle.

\cos72^\circ=(\sqrt5-1)/4 A1 Expand the square before simplifying.

Final answer

x=(\sqrt5-1)/2; \cos36^\circ=(1+\sqrt5)/4; \cos72^\circ=(\sqrt5-1)/4.

Investigation

[10 marks]

For the triangle above, keep each equal side at 1 but let the apex angle be \alpha and the base be u.

(a) Show that u^2=2-2\cos\alpha.

[2]

(b) Find u^2 for \alpha=60^\circ, 90^\circ and 120^\circ.

[3]

(c) Given that u^2+u=1, explain why exactly one positive value of u is possible, and find the corresponding apex angle.

[3]

(d) For a regular pentagon of side 1, use \cos36^\circ to find the length of a diagonal.

[2]

Show solutionHide solution

u^2=1^2+1^2-2\cos\alpha M1A1 Cosine rule at the apex.

Pairs (60^\circ,1),(90^\circ,2),(120^\circ,3) A1A1A1 Use the standard exact cosines.

u=(-1\pm\sqrt5)/2; exactly one root is positive M1R1 Reject the negative length.

\alpha=36^\circ A1 The preceding worked example identifies this cosine; a triangle angle lies between 0 and 180 degrees.

A perpendicular halves a diagonal in the 36^\circ,36^\circ,108^\circ triangle M1 Two adjacent sides enclose the pentagon interior angle.

Diagonal =2\cos36^\circ=(1+\sqrt5)/2 A1 The two half-diagonals are equal.

Final answer

(a) 2-2\cos\alpha
(b) 1,2,3
(c) u=(\sqrt5-1)/2, 36^\circ
(d) (1+\sqrt5)/2.

Worked example

[6 marks]

A scanner triggers when \sin(3x)=\frac12, for 0\le x\le2\pi.

(a) Determine all trigger positions exactly.

[4]

(b) Count the trigger positions when \sin(8x)=\frac12, without solving again.

[2]

Show solutionHide solution

0\le3x\le6\pi M1 There are three full cycles of the inner angle.

3x=\pi/6,5\pi/6,13\pi/6,17\pi/6,25\pi/6,29\pi/6 A1 Two positive-sine crossings per cycle.

x=\pi/18,5\pi/18,13\pi/18,17\pi/18,25\pi/18,29\pi/18 A1 Divide every inner angle by three.

There are exactly six positions; the endpoints do not satisfy the equation R1 A cycle count checks completeness.

Eight complete cycles, each with two crossings M1 The level lies strictly between minus one and one.

16 positions A1 The interval endpoints are not crossings.

Final answer

(a) \frac\pi{18},\frac{5\pi}{18},\frac{13\pi}{18},\frac{17\pi}{18},\frac{25\pi}{18},\frac{29\pi}{18}
(b) 16.

Investigation

[13 marks]

Let N(b,k) count the distinct solutions of \sin(bx)=k on the closed interval 0\le x\le2\pi, where b is a positive integer.

(a) For b=1, 2 and 4, find N(b,k) for k=\frac12, 1, 0, -1 and \frac32. Show your results in a table.

[5]

(b) State and justify a rule for each of these five levels for any positive integer b.

[5]

(c) On 0\le x\le\pi instead, count the solutions of \sin(3x)=\frac12 and of \sin(4x)=\frac12. Explain why the two counts agree.

[3]

Show solutionHide solution

For b=1: 2,1,3,1,0; for b=2: 4,2,5,2,0; for b=4: 8,4,9,4,0 A1A1A1A1A1 Each column corresponds to the stated level.

At k=1/2: 2b; at k=1 or -1: b; at k=0: 2b+1; at k=3/2: 0 A1A1A1A1 Two crossings, one extremum, half-cycle boundaries, or an unattainable height.

The 2b half-cycles have 2b+1 distinct boundary points, including both endpoints R1 Do not count a shared zero twice.

For b=3 and b=4 on [0,\pi], the counts are both 4 A1A1 Each interval contains two positive half-cycles.

The extra half-cycle for b=4 is negative and adds no positive-level crossing R1 A partial-cycle count depends on which half-cycle is added.

Final answer

(a) rows 2,1,3,1,0; 4,2,5,2,0; 8,4,9,4,0
(b) 2b,b,2b+1,b,0
(c) 4,4.

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 12.

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 SL

Chapter 6 teaches trigonometric functions and equations 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