ACT Math · Functions · free lesson

Graphs of Sine and Cosine

The idea, then worked examples solved step by step, then practice questions whose numbers change every time you try them.

The idea

From the unit circle to the sine curve

On the unit circle, the point reached by turning an angle $x$ (in radians) from the positive $x$ -axis is $(\cos x, \sin x)$ . So $\sin x$ is the height of that point and $\cos x$ is how far it is to the right. Now let the point go round and plot its height against the angle turned. The curve you get is the graph of $y = \sin x$ .

In the animation, $P$ goes once round the circle on the left. The grey line carries its height across to the right, where the red curve is drawn at the matching angle. One full turn, $2\pi$ radians, draws one full wave.

The height starts at 0, rises to 1 at a quarter turn, falls back to 0 at a half turn, reaches $-1$ at three quarters, and returns to 0. The engine gives the sine and the cosine at these five angles:

\begin{aligned} s &= \sin(x) \\ &= [0, 1, 0, -1, 0] \\ c &= \cos(x) \\ &= [1, 0, -1, 0, 1] \end{aligned}

$x$	$0$	$\frac{\pi}{2}$	$\pi$	$\frac{3\pi}{2}$	$2\pi$
$\sin x$	$0$	$1$	$0$	$-1$	$0$
$\cos x$	$1$	$0$	$-1$	$0$	$1$

These five points fix the shape of one wave. The two graphs below draw themselves over one turn. The cosine curve is the sine curve moved $\frac{\pi}{2}$ to the left: it starts at its top instead of in the middle.

Amplitude, midline and period

Every sine or cosine graph on the ACT is this wave stretched and moved. Write it as $y = A\sin(Bx) + D$ or $y = A\cos(Bx) + D$ with $B > 0$ .

Definition.

The midline is the horizontal line $y = D$ that the wave runs along. The amplitude is $\lvert A \rvert$ : how far the wave rises above the midline and falls below it. The period is $\frac{2\pi}{B}$ : the length along the $x$ -axis of one full wave. So the maximum value is $D + \lvert A \rvert$ and the minimum is $D - \lvert A \rvert$ .

Take $y = 3\sin(2x) + 1$ . The engine finds the amplitude $a$ , the period $T$ , and the greatest and least values, $M$ and $m$ :

\begin{aligned} a &= \left|A\right| \\ &= \left|3\right| \\ &= 3 \\ T &= \frac{2\pi}{B} \\ &= \frac{2\pi}{2} \\ &= \pi \approx 3.141592654 \\ M &= D + a \\ &= 1 + 3 \\ &= 4 \\ m &= D - a \\ &= 1 - 3 \\ &= -2 \end{aligned}

The wave runs along $y = 1$ , rising to $4$ and falling to $-2$ , and it repeats every $\pi$ , so two whole waves fit between 0 and $2\pi$ . Watch it draw itself, with the midline dashed:

Now change the numbers yourself. $A$ stretches the wave up and down, $B$ squeezes it sideways (a larger $B$ means a shorter period and more waves), and $D$ lifts the whole wave.

Set $A$ below zero and the wave turns upside down: it falls first instead of rising. The amplitude is still the positive number $\lvert A \rvert$ .

Warning.

The period is $\frac{2\pi}{B}$ , not $B$ and not $2\pi B$ . In $y = \sin(4x)$ the wave repeats every $\frac{\pi}{2}$ ; four whole waves fit in $2\pi$ . And the amplitude is half the distance from the lowest point to the highest, not the whole height.

Phase shift

Replacing $x$ by $x - C$ moves a graph $C$ to the right, as for any function. So $y = \sin\left(x - \frac{\pi}{3}\right)$ is the sine wave moved $\frac{\pi}{3}$ to the right; this sideways move is the phase shift. When $B$ is not 1, take $B$ out first: $y = \sin\left(2x - \frac{\pi}{2}\right) = \sin\left(2\left(x - C\right)\right)$ , where $C$ is the number with $2C = \frac{\pi}{2}$ :

\begin{aligned} 2C &= \frac{\pi}{2} \\ C &= \frac{\pi}{4} \approx 0.7853981634 \end{aligned}

The shift is $\frac{\pi}{4}$ to the right, not $\frac{\pi}{2}$ . The grey curve below is $y = \sin(2x)$ ; the red one, drawn after it, is $y = \sin\left(2x - \frac{\pi}{2}\right)$ , the same wave moved $\frac{\pi}{4}$ to the right.

Reading an equation from a graph

To go from a graph to an equation, read the highest value $M$ and the lowest value $m$ . The midline is halfway between them and the amplitude is half the distance between them. Then read the period, the length of one full wave, and find $B$ from it. For a wave that rises to 7, falls to 1, and repeats every 4 units:

\begin{aligned} D &= \frac{M + m}{2} \\ &= \frac{7 + 1}{2} \\ &= 4 \\ a &= \frac{M - m}{2} \\ &= \frac{7 - 1}{2} \\ &= 3 \\ B &= \frac{2\pi}{T} \\ &= \frac{2\pi}{4} \\ &= \frac{\pi}{2} \approx 1.570796327 \end{aligned}

So the midline is $y = 4$ , the amplitude is $3$ , and $B = \frac{\pi}{2}$ . If the graph starts at its highest point when $x = 0$ , a cosine fits with no shift: $y = {3}\cos\left({\frac{\pi}{2}}x\right) + {4}$ .

Note.

On the ACT a question on these graphs asks one of a few things: the amplitude, period or midline of an equation, the greatest or least value, which equation matches a graph, or an equation for a periodic situation such as a tide or a Ferris wheel. The calculator must be in radian mode when $x$ is in radians.

Worked examples

Each example is one of the kinds of question the ACT asks about the graphs of sine and cosine. The numbers in the working are found by the engine. Angles are in radians.

Example.

1. Amplitude, period and midline. For $y = -4\cos(3x) + 2$ , find the amplitude, the period, the midline, and the greatest and least values of $y$ .

Match the equation with $y = A\cos(Bx) + D$ : here $A = -4$ , $B = 3$ and $D = 2$ . The amplitude is $\lvert A \rvert$ , the period is $\frac{2\pi}{B}$ , and the wave goes as far above $D$ as below it.

\begin{aligned} a &= \left|A\right| \\ &= \left|-4\right| \\ &= 4 \\ T &= \frac{2\pi}{B} \\ &= \frac{2\pi}{3} \\ &= \frac{2\pi}{3} \approx 2.094395102 \\ M &= D + a \\ &= 2 + 4 \\ &= 6 \\ m &= D - a \\ &= 2 - 4 \\ &= -2 \end{aligned}

The amplitude is $4$ (not $-4$ ), the period is $\frac{2\pi}{3}$ , the midline is $y = 2$ , and $y$ runs from $-2$ to $6$ . Because $A$ is negative, the wave starts at its lowest point.

Example.

2. A long period. What is the period of $y = 5\sin\left(\frac{x}{2}\right) - 1$ , and what is its range?

Here $B = \frac{1}{2}$ . A value of $B$ less than 1 stretches the wave, so the period is longer than $2\pi$ :

\begin{aligned} T &= \frac{2\pi}{B} \\ &= \frac{2\pi}{\frac{1}{2}} \\ &= 4\pi \approx 12.56637061 \\ M &= D + \left|A\right| \\ &= -1 + \left|5\right| \\ &= 4 \\ m &= D - \left|A\right| \\ &= -1 - \left|5\right| \\ &= -6 \end{aligned}

The period is $4\pi$ , so one wave runs from 0 to $4\pi$ . The range is all $y$ from $-6$ to $4$ . Taking the period as $2\pi \cdot \frac{1}{2} = \pi$ is the usual slip.

Example.

3. Which equation matches the graph? The graph below passes through the origin going up, rises to 3, falls to $-3$ , and repeats every $\pi$ . Which equation is it: $y = 3\sin(2x)$ , $y = 3\cos(2x)$ , $y = 2\sin(3x)$ , or $y = 3\sin\left(\frac{x}{2}\right)$ ?

Read three things. The wave goes 3 above and 3 below the $x$ -axis, so the midline is $y = 0$ and the amplitude is 3. It starts in the middle going up, so it is a sine with a positive $A$ , not a cosine (a cosine starts at its top). Its period is $\pi$ , so $B$ comes from $\frac{2\pi}{B} = \pi$ :

\begin{aligned} \frac{2\pi}{B} &= \pi \\ B &= 2 \end{aligned}

So the graph is $y = 3\sin({2}x)$ . The equation $y = 3\sin\left(\frac{x}{2}\right)$ would have period $4\pi$ , and $y = 2\sin(3x)$ would rise only to 2.

Example.

4. Phase shift. By how much, and which way, is the graph of $y = 2\sin(3x - \pi)$ shifted from the graph of $y = 2\sin(3x)$ ?

Take the 3 out of the bracket: $3x - \pi = 3(x - C)$ , where $3C = \pi$ .

\begin{aligned} 3C &= \pi \\ C &= \frac{\pi}{3} \approx 1.047197551 \end{aligned}

So $y = 2\sin\left(3\left(x - {\frac{\pi}{3}}\right)\right)$ : the graph is shifted $\frac{\pi}{3}$ to the right. Reading the shift as $\pi$ , without dividing by 3, is the trap. The period, $\frac{2\pi}{3}$ , and the amplitude, 2, do not change.

Example.

5. A Ferris wheel. A Ferris wheel has a diameter of 40 meters, and its lowest point is 2 meters above the ground. A rider gets on at the bottom, and the wheel turns once every 8 minutes. Write the rider's height $h$ , in meters, $t$ minutes after getting on, as $h = -a\cos(Bt) + D$ .

The lowest height is 2 and the highest is 2 + 40. The midline is halfway between them, the amplitude is half the distance between them, and $B$ comes from the period of 8 minutes:

\begin{aligned} D &= \frac{M + m}{2} \\ &= \frac{42 + 2}{2} \\ &= 22 \\ a &= \frac{M - m}{2} \\ &= \frac{42 - 2}{2} \\ &= 20 \\ B &= \frac{2\pi}{T} \\ &= \frac{2\pi}{8} \\ &= \frac{\pi}{4} \approx 0.7853981634 \end{aligned}

So $h = -{20}\cos\left({\frac{\pi}{4}}t\right) + {22}$ . The minus sign is there because the rider starts at the bottom: $-\cos$ starts at its lowest value. Check the start, $s$ , and the top, $u$ , half a turn later:

\begin{aligned} h(t) &= -a\cos(Bt) + D \\ &= -20\cos(\frac{\pi}{4}t) + 22 \\ s &= h(0) \\ &= -20\cos(\frac{\pi}{4} \times 0) + 22 \\ &= 2 \\ u &= h(4) \\ &= -20\cos(\frac{\pi}{4} \times 4) + 22 \\ &= 42 \end{aligned}

Example.

6. A tide. The depth of water in a harbor, in feet, $t$ hours after midnight is $d(t) = 3\sin\left(\frac{\pi}{6}t\right) + 8$ . What are the greatest and least depths, and when is the water first deepest?

\begin{aligned} M &= D + \left|A\right| \\ &= 8 + \left|3\right| \\ &= 11 \\ m &= D - \left|A\right| \\ &= 8 - \left|3\right| \\ &= 5 \\ T &= \frac{2\pi}{B} \\ &= \frac{2\pi}{\frac{\pi}{6}} \\ &= 12 \end{aligned}

The depth runs between $5$ and $11$ feet and repeats every $12$ hours. The sine is first at its top, 1, when its angle is $\frac{\pi}{2}$ :

\begin{aligned} \frac{\pi t}{6} &= \frac{\pi}{2} \\ t &= 3 \end{aligned}

So the water is first deepest $3$ hours after midnight, and deepest again $12$ hours later.

Practice

Answer ACT-style questions on graphs of sine and cosine. Each answer is checked, with the full solution, and every try brings new numbers. Practice needs a free account: it keeps your progress.

Practise graphs of sine and cosine with new numbers