Trig graphs · Trigonometric graphs

A trig graph question looks like the quadratic one: complete a table, plot, draw, then read answers. The reports say the same thing about both: most candidates complete the table, fewer draw the graph, and many don’t answer the reading questions at all. Those last parts carry many of the marks.

What the curves look like

$y = \sin x$ and $y = \cos x$ are smooth waves that go between $-1$ and $1$ and repeat every $360^\circ$ (see the graphs in Sine and cosine beyond 90°). Exam relations stretch and shift them:

$y = 2\sin x + 1$ goes between $-1$ and $3$ (twice as tall, moved up 1);
$y = 3\sin 2x$ repeats every $180^\circ$ instead of $360^\circ$ ;
$y = 3\sin x + 2\cos x$ is still one smooth wave, just shifted sideways.

More: the shape of the sine curve

Try it

Trig table and graphFill in the table

x	0°	30°	60°	90°	120°	150°	180°	210°	240°	270°	300°	330°	360°
y

Work out each value with your calculator in degree mode, to 1 decimal place. Each right value is plotted straight away.

Reading the graph

Solving an equation. Rearrange it so one side is exactly the relation you drew, then draw the horizontal line for the other side, and read down from each crossing.

Worked example · WAEC 2014 Paper 2, Q7(c)

WAEC 2014 · Paper 2 · Q7 (c)

Use the graph to find the values of $x$ for which $\sin x = \frac14$ .

Match the equation to the graph
The graph is of $2\sin x + 1$ , but the equation is about $\sin x$ on its own. Change the equation so it contains $2\sin x + 1$ .
Think first. The graph is of $y = 2\sin x + 1$ . What is $2\sin x + 1$ when $\sin x = \frac14$ ?
Rearrange
Multiply both sides of $\sin x = \frac14$ by 2 and add 1:
$2\sin x + 1 = 2 \times \tfrac14 + 1 = 1.5$
Draw the line and read
Draw the horizontal line $y = 1.5$ . It crosses the curve twice. Reading down: $x \approx 15^\circ$ and $x \approx 165^\circ$ .

Greatest and least values. Read the top and bottom of the wave, and the $x$ -values where they happen. They are usually between plotted points, so draw the curve smoothly through the top rather than flattening it at a plotted value.

Your turn

WAEC 2019 · Paper 2 · Q9

Trigonometric graphs

(a)
Copy and complete the table of values for $y = 2\cos x + 3\sin x$ for $0^\circ \le x \le 360^\circ$ .

$x$ $0^\circ$ $60^\circ$ $120^\circ$ $180^\circ$ $240^\circ$ $300^\circ$ $360^\circ$

$y$ $2.0$ $-3.6$

Model answer

$x$ 0° 60° 120° 180° 240° 300° 360°

$y$ 2.0 3.6 1.6 −2.0 −3.6 −1.6 2.0

For example, at $x = 60^\circ$ : $y = 2(0.5) + 3(0.866) = 3.6$ (1 d.p.).
(b)
Using a scale of 2 cm to $60^\circ$ on the $x$ -axis and 2 cm to 1 unit on the $y$ -axis, draw the graph of $y = 2\cos x + 3\sin x$ for $0^\circ \le x \le 360^\circ$ .
Model answer
Plot every point from the table, then join them with one smooth curve (not straight lines between points). With points only every $60^\circ$ , draw a smooth wave: it peaks at about $3.6$ near $56^\circ$ and dips to about $-3.6$ near $236^\circ$ .

For (c): (i) draw $y = -1$ : $x \approx 162.4^\circ$ and $310.2^\circ$ . (ii) At $x = 342^\circ$ , $y \approx 1.0$ .
(c)(i)
Using the graph, solve $2\cos x + 3\sin x = -1$ .
Separate values with commas, e.g. 3, −2
(c)(ii)
Using the graph, find, correct to one decimal place, the value of $y$ when $x = 342^\circ$ .

$x$	0°	60°	120°	180°	240°	300°	360°
$y$	2.0	3.6	1.6	−2.0	−3.6	−1.6	2.0

Try it on a graph

x in degrees. The line y = −1 gives the solutions of (c)(i).

Worked solution (try it first)

(a)

In degree mode, to 1 decimal place.
For example, $x = 60^\circ$ : $2(0.5) + 3(0.866) = 1 + 2.598 = 3.6$ .
$x = 300^\circ$ : $1 - 2.598 = -1.6$ .
The full row is $2.0, 3.6, 1.6, -2.0, -3.6, -1.6, 2.0$ .

(b)

Plot the points with the scales given and join them with a smooth curve.

(c)(i)

Draw the line $y = -1$ and read down from where it meets the curve: $x \approx 162^\circ$ and $x \approx 310^\circ$ .

(ii)

Read up from $x = 342^\circ$ to the curve and across: $y \approx 1.0$ .
(By calculation, $2\cos 342^\circ + 3\sin 342^\circ \approx 1.90 - 0.93$
$= 0.97$ .)

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