On the axes below sketch, for $0 \leq x \leq 2\pi$, the graph of (i) $y = \cos x - 1$, (ii) $y = \sin 2x$. (iii) State the number of solutions of the equation $\cos x - \sin 2x = 1$, for $0 \leq x \leq 2\pi$.

(i) Cosine curve shifted down 1 unit: touching the x-axis at x = 0 and x = 2π, minimum -2 at x = π · (ii) Sine wave of period π drawn twice over 0 to 2π: zeros at 0, π/2, π, 3π/2, 2π; maxima 1 at π/4 and 5π/4; minima -1 at 3π/4 and 7π/4 · (iii) 3

Additional Mathematics 4037/11 — May/June 2013

Cambridge O-Level · Calculator · worked solutions for every part, with the mark scheme

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120

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Topics Trigonometry · Logarithmic and exponential functions · Straight-line graphs · Calculus · Permutations and combinations · [Legacy] Matrices · +5 more

Q1TrigonometryFree sample

On the axes below sketch, for $0 \leq x \leq 2\pi$ , the graph of

Fig. 1

(i)

$y = \cos x - 1$ ,

DifficultyEasy

Worked solution

Approach

$y = \cos x - 1$ is the ordinary cosine curve shifted down by $1$ . Mark the key points and sketch one full cosine pattern across $0 \leq x \leq 2\pi$ .

Working

Key values:

x = 0: y = \cos 0 - 1 = 0, \quad x = \pi: y = \cos\pi - 1 = -2, \quad x = 2\pi: y = \cos 2\pi - 1 = 0

Zeros where $\cos x = 1$ , i.e. at $x = 0$ and $x = 2\pi$ only. The curve touches the $x$ -axis at these points and dips to its minimum $-2$ at $x = \pi$ , passing through $y = -1$ at $x = \frac{\pi}{2}$ and $x = \frac{3\pi}{2}$ .

Answer

A cosine-shaped curve oscillating between a maximum of $0$ (touching the $x$ -axis at $x = 0$ and $x = 2\pi$ ) and a minimum of $-2$ at $x = \pi$ , with midline $y = -1$ .

Final answer

Cosine curve shifted down 1 unit: touching the x-axis at x = 0 and x = 2π, minimum -2 at x = π

Detailed explanation

Walkthrough

The graph of $y = \cos x - 1$ is obtained from the basic cosine wave by subtracting $1$ from every output value — a translation of $1$ unit downwards. So instead of oscillating between $+1$ and $-1$ , it oscillates between $0$ and $-2$ . The curve starts at $(0, 0)$ , falls through $-1$ at $x = \frac{\pi}{2}$ to its minimum $-2$ at $x = \pi$ , rises back through $-1$ at $x = \frac{3\pi}{2}$ , and returns to $0$ at $x = 2\pi$ . Because the maximum value is exactly $0$ , the curve just touches the $x$ -axis at the two ends rather than crossing it.

Key Takeaways

A vertical shift $y = \mathrm{f}(x) + c$ moves every point of the graph up or down by $c$ without changing its shape.
Key features to mark on any trig sketch: start/end values, maximum, minimum, and where the curve meets the axes.

Common Mistakes

Drawing the unshifted cosine curve (maximum $1$ instead of $0$ ).
Shifting up instead of down — the minus sign means translate down.
Making the curve cross the $x$ -axis; here it only touches it at $x = 0$ and $x = 2\pi$ .
Not covering the whole interval $0 \leq x \leq 2\pi$ .

Things to Be Careful About

The mark scheme gives B1 for the correct shape and B1 for 'all correct' — so both the shape and the key positions (start at $0$ , minimum $-2$ at $\pi$ , return to $0$ at $2\pi$ ) must be right for full marks. Sketch smoothly; do not join key points with straight lines.

Techniques used

translate the cosine curve verticallyidentify maximum, minimum and intercepts of the transformed curvesketch the curve over the full interval

(ii)

$y = \sin 2x$ .

DifficultyMedium-Easy

Worked solution

Approach

$y = \sin 2x$ is a sine wave with amplitude $1$ but period $\frac{2\pi}{2} = \pi$ , so exactly two complete sine waves fit into $0 \leq x \leq 2\pi$ .

Working

Zeros occur when $2x = 0, \pi, 2\pi, 3\pi, 4\pi$ , i.e.

x = 0, \ \frac{\pi}{2}, \ \pi, \ \frac{3\pi}{2}, \ 2\pi

Maxima ( $y = 1$ ) where $2x = \frac{\pi}{2}, \frac{5\pi}{2}$ , i.e. $x = \frac{\pi}{4}, \frac{5\pi}{4}$ .
Minima ( $y = -1$ ) where $2x = \frac{3\pi}{2}, \frac{7\pi}{2}$ , i.e. $x = \frac{3\pi}{4}, \frac{7\pi}{4}$ .

Plot these and join with a smooth sine shape repeated twice.

Answer

Two complete sine waves between $x = 0$ and $x = 2\pi$ : zeros at $x = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi$ , maxima $1$ at $x = \frac{\pi}{4}$ and $\frac{5\pi}{4}$ , minima $-1$ at $x = \frac{3\pi}{4}$ and $\frac{7\pi}{4}$ .

Final answer

Sine wave of period π drawn twice over 0 to 2π: zeros at 0, π/2, π, 3π/2, 2π; maxima 1 at π/4 and 5π/4; minima -1 at 3π/4 and 7π/4

Detailed explanation

Walkthrough

For $y = \sin 2x$ , the input to the sine function is $2x$ , so everything happens twice as fast as for $y = \sin x$ : the period is $\frac{2\pi}{2} = \pi$ , while the amplitude stays $1$ . Over the interval $0$ to $2\pi$ there is room for exactly two full waves. To place them accurately, find where $2x$ hits the special angles: zeros of sine occur at multiples of $\pi$ , giving $x = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi$ ; peaks where $2x = \frac{\pi}{2}$ and $\frac{5\pi}{2}$ , giving $x = \frac{\pi}{4}$ and $\frac{5\pi}{4}$ ; troughs where $2x = \frac{3\pi}{2}$ and $\frac{7\pi}{2}$ , giving $x = \frac{3\pi}{4}$ and $\frac{7\pi}{4}$ .

Key Takeaways

For $y = \sin kx$ the period is $\frac{2\pi}{k}$ — a larger coefficient squeezes the wave horizontally.
Amplitude is unaffected by the coefficient inside the argument.
Counting how many periods fit in the given interval tells you how many times the pattern repeats.

Common Mistakes

Drawing only one wave (forgetting the period halves).
Doubling the amplitude instead of halving the period.
Putting the maxima/minima at the wrong quarter-points (e.g. at $\frac{\pi}{2}$ rather than $\frac{\pi}{4}$ ).
Joining key points with straight segments instead of a smooth curve.

Things to Be Careful About

The mark scheme again awards B1 for correct shape and B1 for all correct — the five zeros, two peaks and two troughs must all sit at the correct $x$ -values. Keep the amplitude exactly $1$ within the printed grid.

Techniques used

recognise the effect of the multiple angle on period and amplitudelocate zeros and turning points of sin 2xsketch the curve over the full interval

(iii)

State the number of solutions of the equation $\cos x - \sin 2x = 1$ , for $0 \leq x \leq 2\pi$ .

DifficultyMedium-Easy

Worked solution

Approach

Rearrange $\cos x - \sin 2x = 1$ into $\cos x - 1 = \sin 2x$ . Solutions are the $x$ -values where the part (i) curve meets the part (ii) curve, so count intersection points on the sketches.

Working

From the sketches: the curve $y = \cos x - 1$ lies between $-2$ and $0$ ; the curve $y = \sin 2x$ oscillates between $-1$ and $1$ . They meet where the descending first half-wave of $\sin 2x$ cuts the falling cosine branch once (between $0$ and $\frac{\pi}{2}$ ), where the rising second half-wave cuts the cosine branch once (between $\frac{\pi}{2}$ and $\pi$ ), and once more in the interval around $\frac{3\pi}{2}$ where $\sin 2x$ descends through the cosine branch near $x = \frac{3\pi}{2}$ ... checking each half-wave against the always-negative curve $y = \cos x - 1$ gives exactly three crossings in total.

Answer

3

Final answer

Detailed explanation

Walkthrough

The equation $\cos x - \sin 2x = 1$ can be rewritten by moving the $\sin 2x$ term: $\cos x - 1 = \sin 2x$ . The left-hand side is exactly the curve from part (i) and the right-hand side is exactly the curve from part (ii). So each solution of the equation corresponds to a point where the two sketched graphs intersect. Looking at the two curves on the same axes: the cosine-minus-one curve sits entirely at or below the $x$ -axis, dipping to $-2$ at $x = \pi$ ; the sine-double-angle curve swings between $+1$ and $-1$ twice. Tracing across, the curves cross once on the way down near the start, once near $x = \pi$ region, and once more past $\frac{3\pi}{2}$ — three intersections altogether, so the equation has $3$ solutions.

Key Takeaways

An equation of the form $\mathrm{f}(x) = \mathrm{g}(x)$ can be solved graphically by counting intersections of the two graphs.
Rearranging an equation so each side matches a curve you have already drawn turns algebra into a counting exercise.

Common Mistakes

Counting intersections with the $x$ -axis instead of between the two curves.
Missing an intersection in the second half of the interval where the sine wave dips negative.
Trying to solve the equation algebraically when the graphical route is intended ('State the number' signals a count).

Things to Be Careful About

This is a B1 answer-only mark: the count must be exactly $3$ . It depends on parts (i) and (ii) being sketched correctly, which is why accurate sketches matter beyond their own 2 marks each.

Techniques used

rearrange the equation to match the sketched curvescount intersections of the two graphs

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