Math Core

Sine and cosine give the coordinates of a point on the unit circle. The third basic trigonometric function, tangent, measures something just as geometric: the slope of the terminal side. Its graph looks nothing like a wave. It has breaks, climbs without bound, and repeats twice as often as sine and cosine.

Tangent on the unit circle

For an angle $\theta$ whose terminal side meets the unit circle at $(x, y)$ ,

\tan\theta = \frac{y}{x} = \frac{\sin\theta}{\cos\theta}.

The terminal side passes through the origin and through $(x, y)$ , so its slope is $\dfrac{y - 0}{x - 0} = \dfrac{y}{x}$ . In other words, $\tan\theta$ is the slope of the terminal side of $\theta$ .

Tangent

\tan\theta = \frac{\sin\theta}{\cos\theta}, \qquad \text{defined whenever } \cos\theta \ne 0.

It equals the slope of the terminal side. Tangent is positive in Quadrants I and III and negative in Quadrants II and IV.

The sign pattern follows from the slope picture: a terminal side in Quadrant I or III rises from left to right, and one in Quadrant II or IV falls.

Worked example: Exact values of tangent

Find the exact value of each expression.

$\tan\dfrac{\pi}{3}$
$\tan\dfrac{3\pi}{4}$
$\tan\dfrac{7\pi}{6}$

Solutions.

$\tan\dfrac{\pi}{3} = \dfrac{\sqrt{3}/2}{1/2} = \sqrt{3}$ .
The point at $\dfrac{3\pi}{4}$ is $\left(-\dfrac{\sqrt{2}}{2}, \dfrac{\sqrt{2}}{2}\right)$ , so $\tan\dfrac{3\pi}{4} = -1$ . The terminal side is the line $y = -x$ , with slope $-1$ .
The point at $\dfrac{7\pi}{6}$ is $\left(-\dfrac{\sqrt{3}}{2}, -\dfrac{1}{2}\right)$ , so $\tan\dfrac{7\pi}{6} = \dfrac{-1/2}{-\sqrt{3}/2} = \dfrac{1}{\sqrt{3}} = \dfrac{\sqrt{3}}{3}$ .

Here are the Quadrant I values. Every other value comes from a reference angle and the sign rule.

$\theta$	$0$	$\dfrac{\pi}{6}$	$\dfrac{\pi}{4}$	$\dfrac{\pi}{3}$	$\dfrac{\pi}{2}$
$\tan\theta$	$0$	$\dfrac{\sqrt{3}}{3}$	$1$	$\sqrt{3}$	undefined

Where tangent is undefined

At $\theta = \dfrac{\pi}{2}$ , the terminal side is the vertical $y$ -axis. A vertical line has no slope, and in the formula $\cos\dfrac{\pi}{2} = 0$ puts a zero in the denominator. The same thing happens at $\dfrac{3\pi}{2}$ , $-\dfrac{\pi}{2}$ , and every angle whose terminal side is vertical:

\tan x \text{ is undefined at } x = \frac{\pi}{2} + k\pi \quad (k \text{ any integer}).

Think about what happens as $x$ approaches $\dfrac{\pi}{2}$ from the left. The terminal side gets steeper and steeper, so $\tan x$ grows without bound. Just past $\dfrac{\pi}{2}$ , the terminal side is in Quadrant II and very steep with a negative slope, so $\tan x$ is a huge negative number. The graph has a vertical asymptote there.

The graph of y = tan x

y = tan x with its vertical asymptotes at x = ±π/2 and x = ±3π/2 dashed.Open in grapher →

Between two neighboring asymptotes, the graph rises from $-\infty$ to $+\infty$ , crossing the $x$ -axis halfway. Then the pattern repeats.

Why is the period $\pi$ and not $2\pi$ ? Adding $\pi$ to an angle sends its point to the opposite side of the circle, $(-x, -y)$ . The slope $\dfrac{-y}{-x}$ equals $\dfrac{y}{x}$ , so the tangent doesn't change. A half turn is enough to repeat.

Key features of y = tan x

feature	$y = \tan x$
domain	all real numbers except $x = \dfrac{\pi}{2} + k\pi$
range	all real numbers
period	$\pi$
$x$ -intercepts	$x = k\pi$
vertical asymptotes	$x = \dfrac{\pi}{2} + k\pi$
symmetry	odd: $\tan(-x) = -\tan x$

Common mistake

Tangent has no amplitude. Its graph has no maximum or minimum, because it climbs forever near each asymptote. In $y = 3\tan x$ , the $3$ is a vertical stretch (the graph passes through $\left(\dfrac{\pi}{4}, 3\right)$ instead of $\left(\dfrac{\pi}{4}, 1\right)$ ), but don't call it an amplitude.

Transforming the tangent graph

The same transformations apply, with one change: since the basic period is $\pi$ , the period of $y = a\tan(bx)$ is

\text{period} = \frac{\pi}{|b|}.

To find the asymptotes, set the inside equal to where tangent is undefined: $bx = \dfrac{\pi}{2} + k\pi$ .

Worked example: Period and asymptotes

Find the period and the asymptotes of $y = 2\tan(3x)$ .

The period is $\dfrac{\pi}{3}$ . For the asymptotes, solve

3x = \frac{\pi}{2} + k\pi \quad\Longrightarrow\quad x = \frac{\pi}{6} + \frac{k\pi}{3}.

So there are asymptotes at $x = \pm\dfrac{\pi}{6}$ , $\pm\dfrac{\pi}{2}$ , $\pm\dfrac{5\pi}{6}$ , and so on, spaced $\dfrac{\pi}{3}$ apart, exactly one period.

y = 2tan(3x). Its branches repeat every π/3.Open in grapher →

Solving with tangent

Because tangent has period $\pi$ , solutions of $\tan x = c$ repeat every $\pi$ . On the interval $0 \le x < 2\pi$ , you get two solutions, one from each half turn.

Worked example: Solving an equation

Solve $\tan x = \sqrt{3}$ on the interval $0 \le x \le 2\pi$ .

From the table, $\tan\dfrac{\pi}{3} = \sqrt{3}$ . Add the period $\pi$ to get the other solution: $\dfrac{\pi}{3} + \pi = \dfrac{4\pi}{3}$ . Adding $\pi$ again leaves the interval, so the solutions are $x = \dfrac{\pi}{3}$ and $x = \dfrac{4\pi}{3}$ .

Check with signs: $\dfrac{4\pi}{3}$ is in Quadrant III, where tangent is positive.