The Tangent Function

Sine and cosine give smooth waves. The tangent ratio gives a very different graph: it repeats, but it shoots off to infinity, with vertical asymptotes. Seeing why comes straight from the unit circle and the fact that $\tan x = \dfrac{\sin x}{\cos x}$. All angles on this page are in radians.

Key ideas

From the tangent ratio to a function

For an angle $x$ in standard position, the terminal arm meets the unit circle at the point $(\cos x, \sin x)$. The tangent ratio is the $y$-coordinate of that point divided by its $x$-coordinate:

$$\tan x = \frac{\sin x}{\cos x}$$

Treating each angle as an input and its tangent as the output gives the tangent function, $f(x) = \tan x$. The tangent is also the slope of the terminal arm (rise over run), which is a handy way to picture it.

Here are its values for $-\tfrac{\pi}{2} \le x \le \tfrac{\pi}{2}$, from the special angles:

$x$	$-\tfrac{\pi}{2}$	$-\tfrac{\pi}{3}$	$-\tfrac{\pi}{4}$	$-\tfrac{\pi}{6}$	$0$	$\tfrac{\pi}{6}$	$\tfrac{\pi}{4}$	$\tfrac{\pi}{3}$	$\tfrac{\pi}{2}$
$\tan x$	undefined	$-\sqrt{3} \approx -1.73$	$-1$	$-\tfrac{\sqrt{3}}{3} \approx -0.58$	$0$	$\tfrac{\sqrt{3}}{3} \approx 0.58$	$1$	$\sqrt{3} \approx 1.73$	undefined

Why there are asymptotes

$\tan x = \dfrac{\sin x}{\cos x}$ is undefined wherever $\cos x = 0$: at $x = \tfrac{\pi}{2}$, $\tfrac{3\pi}{2}$, $-\tfrac{\pi}{2}$, and so on. Near those angles, the denominator is tiny, so the fraction is huge.

• As $x$ gets close to $\tfrac{\pi}{2}$ from the left, $\sin x$ is close to $1$ and $\cos x$ is a tiny positive number, so $\tan x$ becomes a very large positive number.
• Just to the right of $\tfrac{\pi}{2}$, $\cos x$ is a tiny negative number, so $\tan x$ is a very large negative number.

That’s a vertical asymptote, the same behaviour you see in reciprocal functions. In slope terms: as the terminal arm turns toward vertical, its slope grows without limit, and a vertical line has no slope at all.

$y = \tan x$: the same branch repeats every $\pi$, between asymptotes at $x = \tfrac{\pi}{2} + n\pi$.

Key properties of y = tan x

Property	$y = \tan x$
Period	$\pi$
Domain	$\left\{x \in \mathbb{R} \mid x \ne \tfrac{\pi}{2} + n\pi,\ n \in \mathbb{Z}\right\}$
Range	$\{y \in \mathbb{R}\}$
Vertical asymptotes	$x = \tfrac{\pi}{2} + n\pi$, $n \in \mathbb{Z}$
Zeros	$x = n\pi$, $n \in \mathbb{Z}$
$y$-intercept	$0$
Amplitude	none (no maximum or minimum)
Behaviour	increasing on every interval between asymptotes

($\mathbb{Z}$ is the set of integers, so $n$ can be $\ldots, -2, -1, 0, 1, 2, \ldots$)

• Period $\pi$, not $2\pi$. Turning the terminal arm half a turn ($\pi$) takes the point $(\cos x, \sin x)$ to the opposite point $(-\cos x, -\sin x)$. Both coordinates change sign, so their ratio doesn’t change: $\tan(x + \pi) = \tan x$.
• Zeros. $\tan x = 0$ exactly where $\sin x = 0$.
• No amplitude. The range is all real numbers, so there’s no maximum or minimum, and amplitude doesn’t apply.
• Symmetry. $\tan(-x) = -\tan x$: the graph is symmetric about the origin.

Worked examples

Example 1: Close to an asymptote

Use a calculator (radian mode) to evaluate $\tan x$ for $x = 1.5$, $1.55$, $1.57$, $1.58$, and $1.6$, to $3$ decimal places. What do the values show?

Solution.

$x$	$1.5$	$1.55$	$1.57$	$1.58$	$1.6$
$\tan x$	$14.101$	$48.078$	$1255.766$	$-108.649$	$-34.233$

Since $\tfrac{\pi}{2} \approx 1.5708$, the first three inputs are just to the left of $\tfrac{\pi}{2}$, and the tangent grows very quickly through large positive values. The last two are just to the right, and the tangent is large and negative. The graph jumps from the top of the screen to the bottom across the vertical asymptote $x = \tfrac{\pi}{2}$.

Example 2: Properties on an interval

For $-2\pi \le x \le 2\pi$, list the zeros and the equations of the vertical asymptotes of $y = \tan x$.

Solution. Zeros are at multiples of $\pi$:

$$x = -2\pi,\ -\pi,\ 0,\ \pi,\ 2\pi$$

Asymptotes are halfway between the zeros, at odd multiples of $\tfrac{\pi}{2}$:

$$x = -\frac{3\pi}{2},\quad x = -\frac{\pi}{2},\quad x = \frac{\pi}{2},\quad x = \frac{3\pi}{2}$$

That’s three full branches between the asymptotes, plus half a branch at each end of the interval.

Example 3: Using the period

Given $\tan\dfrac{\pi}{3} = \sqrt{3}$, find $\tan\dfrac{4\pi}{3}$, $\tan\left(-\dfrac{2\pi}{3}\right)$, and $\tan\dfrac{7\pi}{3}$ without a calculator. Then explain why $\tan\dfrac{2\pi}{3}$ is different.

Solution. The period is $\pi$, so adding or subtracting any multiple of $\pi$ doesn’t change the tangent:

$$\begin{aligned} \tan\frac{4\pi}{3} &= \tan\left(\frac{\pi}{3} + \pi\right) = \sqrt{3} \\ \tan\left(-\frac{2\pi}{3}\right) &= \tan\left(\frac{\pi}{3} - \pi\right) = \sqrt{3} \\ \tan\frac{7\pi}{3} &= \tan\left(\frac{\pi}{3} + 2\pi\right) = \sqrt{3} \end{aligned}$$

But $\tfrac{2\pi}{3}$ is not a multiple of $\pi$ away from $\tfrac{\pi}{3}$ (the difference is $\tfrac{\pi}{3}$). It’s in quadrant II, where tangent is negative: $\tan\tfrac{2\pi}{3} = -\sqrt{3}$.

Example 4: Reading the graph

For $0 \le x \le 2\pi$:

• (a) Where is $\tan x = -1$?
• (b) Where is $\tan x$ undefined?
• (c) On which intervals is $\tan x \gt 0$?

Solution.

(a) The related angle is $\tfrac{\pi}{4}$, and tangent is negative in quadrants II and IV: $x = \tfrac{3\pi}{4}$ and $x = \tfrac{7\pi}{4}$. These are one period apart, as they should be.

(b) Where $\cos x = 0$: $x = \tfrac{\pi}{2}$ and $x = \tfrac{3\pi}{2}$.

(c) The graph is above the $x$-axis just after each zero, up to the next asymptote: $0 \lt x \lt \tfrac{\pi}{2}$ and $\pi \lt x \lt \tfrac{3\pi}{2}$. These are quadrants I and III, matching CAST. ✓

Common mistakes

Saying the period is 2π. Sine and cosine repeat every $2\pi$, but tangent repeats every $\pi$. Look at the graph: a whole branch fits between two asymptotes $\pi$ apart.

Drawing the graph through the asymptote. The branches never touch or cross the dashed lines. Near an asymptote, the graph gets steeper and steeper on each side, heading in opposite directions.

Putting asymptotes at the zeros of sin x. $\tan x = \tfrac{\sin x}{\cos x}$ is undefined where the denominator, $\cos x$, is $0$. Where $\sin x = 0$, the tangent is $0$.

Giving tan x an amplitude. The range is all real numbers. There’s no maximum or minimum, so there’s no amplitude.

Leaving the asymptotes out of the domain statement. The domain isn’t all real numbers. Write $\left\{x \in \mathbb{R} \mid x \ne \tfrac{\pi}{2} + n\pi,\ n \in \mathbb{Z}\right\}$.

Practice

1. (Warm-up) State the period, domain, and range of $y = \tan x$.

Solution

Period $\pi$. Domain $\left\{x \in \mathbb{R} \mid x \ne \tfrac{\pi}{2} + n\pi,\ n \in \mathbb{Z}\right\}$. Range $\{y \in \mathbb{R}\}$.

2. (Warm-up) Find the exact value: (a) $\tan\dfrac{\pi}{4}$ (b) $\tan\dfrac{5\pi}{4}$ (c) $\tan\left(-\dfrac{\pi}{4}\right)$

Solution

(a) $1$

(b) $\tfrac{5\pi}{4} = \tfrac{\pi}{4} + \pi$, and the period is $\pi$, so $\tan\tfrac{5\pi}{4} = 1$.

(c) Tangent is symmetric about the origin: $\tan\left(-\tfrac{\pi}{4}\right) = -\tan\tfrac{\pi}{4} = -1$.

3. (Warm-up) Write the equations of the vertical asymptotes of $y = \tan x$ for $0 \le x \le 2\pi$.

Solution

$x = \dfrac{\pi}{2}$ and $x = \dfrac{3\pi}{2}$.

4. (Core) Evaluate $\tan 1$, $\tan 1.4$, $\tan 1.5$, and $\tan 1.7$ to $3$ decimal places. Explain the change in sign between the last two.

Solution

$\tan 1 \approx 1.557$, $\tan 1.4 \approx 5.798$, $\tan 1.5 \approx 14.101$, $\tan 1.7 \approx -7.697$.

The values grow quickly as $x$ approaches $\tfrac{\pi}{2} \approx 1.571$. Between $1.5$ and $1.7$ the input passes the vertical asymptote at $x = \tfrac{\pi}{2}$, where $\cos x$ changes from positive to negative while $\sin x$ stays positive, so the tangent changes from large positive to negative.

5. (Core) Explain why it makes sense to talk about the amplitude of $y = \sin x$ but not of $y = \tan x$.

Solution

Amplitude is half the distance between the maximum and minimum values. $y = \sin x$ has a maximum of $1$ and a minimum of $-1$, so its amplitude is $1$. $y = \tan x$ takes every real value (its range is $\{y \in \mathbb{R}\}$) and has no maximum or minimum, so amplitude isn’t defined for it.

6. (Core) Find all $x$ with $-\pi \le x \le \pi$ such that $\tan x = \sqrt{3}$.

Solution

$\tan\tfrac{\pi}{3} = \sqrt{3}$, so $x = \tfrac{\pi}{3}$ is one answer. The period is $\pi$, so the others are $\tfrac{\pi}{3} + n\pi$. In the interval, $\tfrac{\pi}{3} - \pi = -\tfrac{2\pi}{3}$ also works ($\tfrac{\pi}{3} + \pi = \tfrac{4\pi}{3}$ is too big).

$$x = -\frac{2\pi}{3} \qquad\text{or}\qquad x = \frac{\pi}{3}$$

7. (Core) Using $\tan x = \dfrac{\sin x}{\cos x}$, explain (a) why $\tan x = 0$ at exactly the same $x$-values where $\sin x = 0$, and (b) why $\tan x$ is positive in quadrants I and III.

Solution

(a) A fraction is $0$ exactly when its numerator is $0$ (and the denominator isn’t). Where $\sin x = 0$ ($x = n\pi$), $\cos x = \pm 1$, so $\tan x = 0$.

(b) In quadrant I, $\sin x$ and $\cos x$ are both positive; in quadrant III, both are negative. Either way, the quotient is positive. In quadrants II and IV they have opposite signs, so the tangent is negative.

8. (Challenge) Find the period of $y = \tan(2x)$, and give its zeros and vertical asymptotes for $0 \le x \le \pi$.

Solution

A horizontal compression by a factor of $\tfrac{1}{2}$ halves the period: $\tfrac{\pi}{2}$.

Zeros: $\tan(2x) = 0$ when $2x = n\pi$, so $x = \tfrac{n\pi}{2}$: $x = 0, \tfrac{\pi}{2}, \pi$.

Asymptotes: $\tan(2x)$ is undefined when $2x = \tfrac{\pi}{2} + n\pi$, so $x = \tfrac{\pi}{4} + \tfrac{n\pi}{2}$: $x = \tfrac{\pi}{4}$ and $x = \tfrac{3\pi}{4}$.

9. (Challenge) A line through the origin makes an angle of $1.2$ rad with the positive $x$-axis.

• (a) Find its slope to $3$ decimal places.
• (b) What happens to the slope as the angle increases toward $\tfrac{\pi}{2}$? How does this connect to the graph of $y = \tan x$?

Solution

(a) The slope is $\tan 1.2 \approx 2.572$.

(b) The line gets steeper, and its slope increases without limit. At exactly $\tfrac{\pi}{2}$ the line is vertical and its slope is undefined. On the graph of $y = \tan x$, that’s the branch rising toward the vertical asymptote at $x = \tfrac{\pi}{2}$.