Free AP Precalculus Trigonometric and Polar Functions

AP Precalculus · Trigonometric and Polar Functions · free lesson

Sine and Cosine Graphs; Sinusoidal Functions

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 graph

As an angle $\theta$ grows, the point $P = (\cos\theta, \sin\theta)$ goes round the unit circle. Its height above the $x$ -axis is $\sin\theta$ . Plot that height against $\theta$ and you get the graph of $y = \sin\theta$ . Watch it being built: the dashed line carries the height of $P$ across to the graph.

One turn of $P$ draws one cycle of the graph, so $\sin$ is periodic with period $2\pi$ . In the same way, the horizontal position of $P$ , $\cos\theta$ , plotted against $\theta$ gives the graph of $y = \cos\theta$ . Both are drawn below.

Feature	$y = \sin x$	$y = \cos x$
period	$2\pi$	$2\pi$
frequency (cycles per unit of $x$ )	$\frac{1}{2\pi}$	$\frac{1}{2\pi}$
amplitude	1	1
midline	$y = 0$	$y = 0$
range	$[-1, 1]$	$[-1, 1]$
value at $x = 0$	$0$	$1$
zeros in $[0, 2\pi]$	$[0, \pi, 2\pi]$	$[\frac{\pi}{2}, \frac{3\pi}{2}]$

Period, frequency, amplitude and midline

These four describe any sine-shaped graph, a sinusoid.

The period is the input length of one cycle; the frequency is its reciprocal, the number of cycles in one unit of input. A period of 4 seconds is a frequency of $\frac{1}{4}$ cycle per second.
The midline is the horizontal line halfway between the greatest and least values: $y = \frac{\max + \min}{2}$ .
The amplitude is the distance from the midline to a greatest value: $\frac{\max - \min}{2}$ . It is half the height of the graph, not the whole height.

The sinusoid below rises to $5$ and falls to $1$ , and its peaks are $4$ units apart. So its amplitude is $\frac{ {5} - {1} }{2} = {2}$ , its midline is $y = \frac{ {5} + {1} }{2} = {3}$ , its period is $4$ and its frequency is $\frac{1}{4}$ .

Sine is odd, cosine is even

The angle $-\theta$ turns clockwise as far as $\theta$ turns counterclockwise, so its point on the unit circle is the reflection of $P$ in the $x$ -axis: the same $x$ , the opposite $y$ .

Rule.

$\sin(-\theta) = -\sin\theta$ : sine is an odd function, its graph symmetric about the origin. $\cos(-\theta) = \cos\theta$ : cosine is an even function, its graph symmetric about the $y$ -axis.

For example, $\sin\left(-\frac{\pi}{6}\right) = {-\frac{1}{2}}$ and $\cos\left(-\frac{\pi}{3}\right) = {\frac{1}{2}}$ .

Cosine is a shifted sine

The two graphs have the same shape; the cosine graph is the sine graph moved $\frac{\pi}{2}$ to the left: $\cos x = \sin\left(x + \frac{\pi}{2}\right)$ . Move the slider to slide the blue graph of $y = \sin(x + h)$ to the left until it lies on the red cosine graph.

The graphs first agree at $h = \frac{\pi}{2}$ ; they agree again whenever $h = \frac{\pi}{2} + 2\pi n$ . The other way round, $\sin x = \cos\left(x - \frac{\pi}{2}\right)$ : sine is cosine moved $\frac{\pi}{2}$ to the right.

Where the graph rises, falls and bends

On one cycle of $y = \sin x$ , $0 \le x \le 2\pi$ , read the behavior from the unit circle: $\sin x$ rises while $P$ climbs, fastest as $P$ crosses the $x$ -axis and slowest near the top and bottom.

interval	$\left(0, \frac{\pi}{2}\right)$	$\left(\frac{\pi}{2}, \pi\right)$	$\left(\pi, \frac{3\pi}{2}\right)$	$\left(\frac{3\pi}{2}, 2\pi\right)$
$\sin x$ is	increasing	decreasing	decreasing	increasing
its rate of change is	decreasing	decreasing	increasing	increasing
the graph is	concave down	concave down	concave up	concave up

The graph is concave down above the midline and concave up below it; it changes concavity, at a point of inflection, each time it crosses the midline. So on $\left(0, \frac{\pi}{2}\right)$ sine is increasing at a decreasing rate, and on $\left(\frac{3\pi}{2}, 2\pi\right)$ it is increasing at an increasing rate.

Warning.

The usual slips: the amplitude taken as the greatest value minus the least (that is twice the amplitude); the midline taken as $y = 0$ for a graph that has been moved up or down; period and frequency mixed up; cosine shifted to sine the wrong way; sine called even; and "concave down" read as "decreasing".

Worked examples

Each example is one kind of question the AP exam asks about the graphs of sine and cosine and about sinusoids. Every number in the working is found by the engine.

Example.

1. Reading a sinusoid's graph. The graph of a sinusoidal function $g$ is shown. Find its amplitude, midline, period and frequency.

Read the greatest and least values, $2$ and $-4$ , and the distance between consecutive peaks, at $x = 0$ and $x = 6$ :

\begin{aligned} \text{amplitude} &= \frac{ {2} - ({-4}) }{2} = {3} \\ \text{midline: } y &= \frac{ {2} + ({-4}) }{2} = {-1} \\ \text{period} &= 6 - 0 = 6 \\ \text{frequency} &= \frac{1}{ {6} } = {\frac{1}{6}} \end{aligned}

The graph is $3$ above and below the line $y = -1$ , not the $x$ -axis; its whole height, $6$ , is twice the amplitude.

Example.

2. Features from a description. The depth of water at a pier, in feet, is sinusoidal in time. The greatest depth, $11$ feet, comes at $t = 4$ hours and again at $t = 16$ hours, with no greater depth between them; the least depth is $3$ feet. Find the amplitude, the midline, the period and the frequency, and the time of the least depth between $t = 4$ and $t = 16$ .

\begin{aligned} \text{amplitude} &= \frac{ {11} - {3} }{2} = {4} \\ \text{midline: } y &= \frac{ {11} + {3} }{2} = {7} \\ \text{period} &= 16 - 4 = 12 \\ \text{frequency} &= \frac{1}{ {12} } = {\frac{1}{12}} \end{aligned}

The period is $12$ hours and the frequency $\frac{1}{12}$ cycle per hour. A least value comes halfway between two greatest values, at $t = {4} + \frac{ {12} }{2} = {10}$ hours.

Example.

3. Odd and even. Which of these statements are true for every $x$ ? I. $\sin(-x) = \sin x$ . II. $\cos(-x) = \cos x$ . III. $\sin(-x) = -\sin x$ .

Test with one value, $x = \frac{\pi}{3}$ , then explain with the unit circle:

\begin{aligned} \sin\left(-\frac{\pi}{3}\right) &= {-\frac{\sqrt{3}}{2}}, \quad \sin\frac{\pi}{3} = {\frac{\sqrt{3}}{2}} \\ \cos\left(-\frac{\pi}{3}\right) &= {\frac{1}{2}}, \quad \cos\frac{\pi}{3} = {\frac{1}{2}} \end{aligned}

One value can only disprove. Statement I fails at $x = \frac{\pi}{3}$ . For II and III, the point of $-x$ on the unit circle is the reflection of the point of $x$ in the $x$ -axis: same $x$ -coordinate (cosine), opposite $y$ -coordinate (sine), for every $x$ .

Result.

II and III are true: cosine is even and sine is odd.

Example.

4. Cosine as a shifted sine. Find the number $h$ with $0 < h < 2\pi$ for which $\sin(x + h) = \cos x$ for every $x$ , and check it at two values of $x$ .

The cosine graph reaches its peak at $x = 0$ ; the sine graph reaches its peak at $x = \frac{\pi}{2}$ , which is $\frac{\pi}{2}$ later. So the sine graph must be moved $\frac{\pi}{2}$ to the left: $h = \frac{\pi}{2}$ .

\begin{aligned} \sin\left(\frac{\pi}{6} + \frac{\pi}{2}\right) &= \sin\frac{2\pi}{3} = {\frac{\sqrt{3}}{2}}, \quad \cos\frac{\pi}{6} = {\frac{\sqrt{3}}{2}} \\ \sin\left(\pi + \frac{\pi}{2}\right) &= \sin\frac{3\pi}{2} = {-1}, \quad \cos\pi = {-1} \end{aligned}

Moving to the right instead, $\sin\left(x - \frac{\pi}{2}\right)$ , gives $-\cos x$ , the cosine graph upside down.

Example.

5. Rising, falling and bending: the cosine graph. On $0 < x < 2\pi$ , where is $y = \cos x$ decreasing with its graph concave down? Where is it increasing at an increasing rate?

Follow $P$ round the unit circle: $\cos x$ is its $x$ -coordinate. From $x = 0$ to $x = \pi$ , $P$ moves left, so cosine decreases; from $\pi$ to $2\pi$ it moves right, so cosine increases. The graph is concave down above its midline $y = 0$ , on $\left(0, \frac{\pi}{2}\right)$ and $\left(\frac{3\pi}{2}, 2\pi\right)$ , and concave up below it.

The average rates of change over $\left[0, \frac{\pi}{4}\right]$ and $\left[\frac{\pi}{4}, \frac{\pi}{2}\right]$ are about $-0.373$ and $-0.900$ : negative and decreasing, which confirms decreasing and concave down on $\left(0, \frac{\pi}{2}\right)$ .

Result.

$\cos x$ is decreasing and concave down on $\left(0, \frac{\pi}{2}\right)$ , and increasing at an increasing rate (concave up) on $\left(\pi, \frac{3\pi}{2}\right)$ .

Practice

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

Practise sine and cosine graphs; sinusoidal functions with new numbers