Topic 3.8

The Tangent Function

Sine and cosine gave the coordinates of a point on the unit circle. Tangent answers a different question: how steep is the terminal ray? That one idea explains everything about the tangent graph — why it repeats every $\pi$ instead of every $2\pi$ , and why it shoots off to infinity.

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01
For tangent: period $\pi / |b|$ , asymptotes where the input is $\pi /2 + k\pi$ , and quarter points where $\tan = \pm 1$ .

Practice 33 problems Next: Chapter 3.9 Inverse Trigonometric Functions

Tangent as the Slope of the Terminal Ray

Remember from 3.2B: for an angle $\theta$ in standard position, $\tan \theta = y/x$ , where $(x, y)$ is the point where the terminal ray meets the unit circle. Since that point is $(\cos \theta , \sin \theta )$ , and the ray starts at the origin, y/x is exactly rise over run from $(0, 0)$ to the point.

\tan\theta=\frac{\sin\theta}{\cos\theta}=\frac{y}{x}=\ \text{slope of the terminal ray}

CONCEPT

Tangent = slope

$\tan \theta$ is the slope of the terminal ray of $\theta$ , as long as $\cos \theta \ne 0$ .

When $\cos \theta = 0$ the ray is vertical ( $\theta = \pi /2$ , $3\pi /2$ , …), a vertical line has no slope, and $\tan \theta$ is undefined.

Signs follow from slope: the ray points up-right or down-left in Quadrants I and III (positive slope), and up-left or down-right in Quadrants II and IV (negative slope).

Worked example

Example 1. Find the slope of the terminal ray of $\theta = 2\pi /3$ . Then evaluate $\tan (7\pi /6)$ , $\tan (-\pi /4)$ , and $\tan (3\pi /2)$ .

01
$\theta = 2\pi /3$ has reference angle $\pi /3$ in Quadrant II, so the ray meets the unit circle at (−1/2, $\sqrt{3}/2$ ) (3.3). Slope = $(\sqrt{3}/2) \div (-1/2) = -\sqrt{3}$ . The ray points up and to the left, so a negative slope makes sense.
02
$7\pi /6$ is in Quadrant III with reference angle $\pi /6$ : the point is ( $-\sqrt{3}/2$ , −1/2). $\tan (7\pi /6) = (-1/2) \div (-\sqrt{3}/2) = 1/\sqrt{3} = \sqrt{3}/3$ .
03
$-\pi /4$ is in Quadrant IV: ( $\sqrt{2}/2$ , $-\sqrt{2}/2$ ). $\tan (-\pi /4) = -1$ .
04
$3\pi /2$ points straight down: $(0, -1)$ . Dividing by $x = 0$ is impossible, so $\tan (3\pi /2)$ is undefined.

Left: the slope of the ray for $2\pi /3$ . Right: $\pi /6$ and $7\pi /6$ point in opposite directions along the same line.

Look at the right-hand picture. Adding $\pi$ to an angle turns the ray around to point the opposite way, but it stays on the same line. Opposite rays on one line have the same slope, so $\tan (\theta + \pi ) = \tan \theta$ . That is why tangent repeats every $\pi$ , not every $2\pi$ .

COMMON MISTAKE

" $\tan (2\pi /3) = \sqrt{3}$ , because the reference angle is $\pi /3$ and $\tan (\pi /3) = \sqrt{3}$ ."

The reference angle gives the size of the value, but not its sign. In Quadrant II, $x$ is negative and $y$ is positive, so the ray slopes downward from left to right: $\tan (2\pi /3) = -\sqrt{3}$ .

Quick check: picture the ray. Up-left or down-right means negative.

The Graph of y = tan θ

Start at $\theta = 0$ and let the ray swing counterclockwise. Its slope begins at 0 and grows. As $\theta$ gets close to $\pi /2$ the ray becomes almost vertical, and the slope becomes huge. Just past $\pi /2$ the ray points up-left, and the slope is hugely negative. The table and graph show this numerically and graphically.

θ	−π/3	−π/4	−π/6	0	π/6	π/4	π/3	→ π/2
tan θ	−√3	−1	−√3/3	0	√3/3	1	√3	→ ∞

$y = \tan \theta$ : one branch between every pair of asymptotes, repeating every $\pi$ .

Why an asymptote at $\pi /2$ ? As $\theta$ approaches $\pi /2$ from the left, $\cos \theta$ shrinks toward 0 while $\sin \theta$ stays near 1. Dividing about 1 by a tiny positive number gives a huge positive number, so $\tan \theta$ → ∞. From the right, $\cos \theta$ is a tiny negative number, so $\tan \theta$ → −∞.

CONCEPT

Features of $y = \tan \theta$

Period $\pi$ . Vertical asymptotes at $\theta = \pi /2 + k\pi$ , where $k$ is any integer (wherever $\cos \theta = 0$ ).

Zeros at $\theta = k\pi$ (wherever $\sin \theta = 0$ ). Domain: all $\theta$ except the asymptotes. Range: all real numbers.

Increasing on every interval between consecutive asymptotes.

Concave down on $(-\pi /2, 0)$ and concave up on $(0, \pi /2)$ ; the zeros are inflection points.

You can see the concavity in the table (remember from Unit 1: concave up means the rate of change is increasing). From $\theta = 0$ to $\pi /6$ , $\tan \theta$ rises by $\sqrt{3}/3 \approx 0.577$ . Over the next $\pi /6$ , from $\pi /6$ to $\pi /3$ , it rises by $\sqrt{3} - \sqrt{3}/3 = 2\sqrt{3}/3 \approx 1.155$ , exactly twice as much. Equal steps in $\theta$ , bigger and bigger steps in $\tan \theta$ : concave up.

COMMON MISTAKE

"Tangent is built from sine and cosine, so its period is $2\pi$ too."

The period of $\sin \theta \div \cos \theta$ is shorter than either one. Every $\pi$ , both $\sin \theta$ and $\cos \theta$ change sign, and the two sign changes cancel in the quotient. On the unit circle this is the opposite-ray picture from Section 1.

Check on the graph: the branch between $-\pi /2$ and $\pi /2$ repeats between $\pi /2$ and $3\pi /2$ . That is a horizontal distance of $\pi$ .

Quick check

Where are the vertical asymptotes of $y = \tan\theta$ , and what is its period?

Transformations of Tangent

Tangent takes the same transformations as sine and cosine in 3.6A and 3.6B:

f(\theta)=a\,\tan\left(b(\theta+c)\right)+d

Constant	Transformation	Effect on y = tan θ
a	vertical dilation by \|a\|	stretches the branches; a < 0 reflects them (decreasing)
b	horizontal dilation	period π/\|b\|
c	horizontal translation	left c units if c > 0; right \|c\| units if c < 0
d	vertical translation	moves every point up d units

Tangent has no maximum or minimum, so it has no amplitude and no midline. Instead, keep track of three things for each branch: the center point (the inflection point, $(0, 0)$ for $y = \tan \theta$ ), the two asymptotes, and the two "quarter points" halfway between the center and each asymptote, where $\tan$ equals $\pm 1$ .

To find the asymptotes of $y = \tan (b\theta )$ with $b > 0$ , set the input equal to the places where $\tan$ is undefined:

b\theta=\frac{\pi}{2}+k\pi\quad\Rightarrow\quad\theta=\frac{\pi}{2b}+\frac{k\pi}{b}

Worked example

Example 2. Graph $g(\theta ) = -2 \tan (\theta /3) + 1$ . Give the period and the asymptotes.

01
Period: $b = 1/3$ , so the period is $\pi \div (1/3) = 3\pi$ . Asymptotes: $\theta /3 = \pi /2 + k\pi$ gives $\theta = 3\pi /2 + 3k\pi$ , so one branch lives between $-3\pi /2$ and $3\pi /2$ .
02
Center point: at $\theta = 0$ , $g(0) = -2 \tan 0 + 1 = 1$ . The center moves up to $(0, 1)$ .
03
Quarter points: halfway to each asymptote, $\theta = \pm 3\pi /4$ , the input $\theta /3 = \pm \pi /4$ and $\tan ( \pm \pi /4) = \pm 1$ . So $g(3\pi /4) = -2(1) + 1 = -1$ and $g(-3\pi /4) = -2(-1) + 1 = 3$ .
04
Because $a = -2 < 0$ , the branch is reflected: it falls from left to right, through $(-3\pi /4, 3)$ , $(0, 1)$ , and $(3\pi /4, -1)$ .

$g(\theta ) = -2 \tan (\theta /3) + 1$ is decreasing on each branch because $a < 0$ .

Worked example

Example 3. Find the center point of one branch of $h(\theta ) = \tan (2\theta - \pi /4) + 3$ , and list its asymptotes.

01
Remember from 3.6B: factor out $b$ before reading a shift. $2\theta - \pi /4 = 2(\theta - \pi /8)$ , so the graph shifts $\pi /8$ to the right.
02
Center point: $y = \tan \theta$ has center $(0, 0)$ . Shift right $\pi /8$ and up 3: the center is $(\pi /8, 3)$ . Check: $h(\pi /8) = \tan (0) + 3 = 3$ ✓.
03
The period is $\pi /2$ , and a branch reaches half a period, $\pi /4$ , on either side of its center. So the asymptotes of this branch are $\theta = \pi /8 \pm \pi /4$ : $\theta = -\pi /8$ and $\theta = 3\pi /8$ . The rest repeat every $\pi /2$ : $\theta = 3\pi /8 + k\pi /2$ .
04
Quarter points: $\theta = 0$ and $\theta = \pi /4$ , giving $(0, 2)$ and $(\pi /4, 4)$ .

$h(\theta ) = \tan (2\theta - \pi /4) + 3$ : centers at $\pi /8$ and $5\pi /8$ , asymptotes every $\pi /2$ .

COMMON MISTAKE

"The shift is $\pi /4$ to the right, so the center is at $(\pi /4, 3)$ ."

$\pi /4$ is the shift of the whole input $2\theta$ , not of $\theta$ . After factoring, $2(\theta - \pi /8)$ shows a shift of $\pi /8$ . Test the wrong answer: $h(\pi /4) = \tan (\pi /4) + 3 = 4$ , not 3, so $(\pi /4, 3)$ is not even on the graph. $(\pi /4, 4)$ is actually a quarter point.

Quick check

What is the period of $y = 2\tan(3\theta)$ ?

Writing an Equation from a Graph

Worked example

Example 4. The graph of $f(x) = a \tan(bx)$ is shown, with vertical asymptotes at $x = -2$ , 2, and 6. It passes through $(1, 3)$ . Find $a$ and $b$ .

01
Consecutive asymptotes are $2 - (-2) = 4$ apart, so the period is 4. Then $\pi / |b| = 4$ , and $b = \pi /4$ ( $b > 0$ is the simplest choice).
02
The point $(1, 3)$ is halfway between the center $(0, 0)$ and the asymptote $x = 2$ , so it is a quarter point, where $\tan(bx) = \tan (\pi /4) = 1$ .
03
So $f(1) = a \cdot 1 = 3$ , and $a = 3$ . The model is $f(x) = 3 \tan (\pi x/4)$ . Check another point: $f(5) = 3 \tan (5\pi /4) = 3(1) = 3$ ✓, matching the point on the next branch.

KEY RULE

For tangent: period $\pi / |b|$ , asymptotes where the input is $\pi /2 + k\pi$ , and quarter points where $\tan = \pm 1$ .

Common slips

" $\tan (2\pi /3) = \sqrt{3}$ , because the reference angle is $\pi /3$ and $\tan (\pi /3) = \sqrt{3}$ ."

The reference angle gives the size of the value, but not its sign. In Quadrant ii, $x$ is negative and $y$ is positive, so the ray slopes downward from left to right: $\tan (2\pi /3) = -\sqrt{3}$ .

Quick check: picture the ray. Up-left or down-right means negative.
"Tangent is built from sine and cosine, so its period is $2\pi$ too."

The period of $\sin \theta \div \cos \theta$ is shorter than either one. Every $\pi$ , both $\sin \theta$ and $\cos \theta$ change sign, and the two sign changes cancel in the quotient. On the unit circle this is the opposite-ray picture from Section 1.

Check on the graph: the branch between $-\pi /2$ and $\pi /2$ repeats between $\pi /2$ and $3\pi /2$ . That is a horizontal distance of $\pi$ .
"The shift is $\pi /4$ to the right, so the center is at $(\pi /4, 3)$ ."

$\pi /4$ is the shift of the whole input $2\theta$ , not of $\theta$ . After factoring, $2(\theta - \pi /8)$ shows a shift of $\pi /8$ . Test the wrong answer: $h(\pi /4) = \tan (\pi /4) + 3 = 4$ , not 3, so $(\pi /4, 3)$ is not even on the graph. $(\pi /4, 4)$ is actually a quarter point.

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Recap card

6 lines to re-read the night before.

01
$\tan \theta = \sin \theta \div \cos \theta$ is the slope of the terminal ray. It is undefined when the ray is vertical ( $\cos \theta = 0$ ).
02
Opposite rays lie on the same line, so $\tan (\theta + \pi ) = \tan \theta$ : the period of tangent is $\pi$ .
03
$y = \tan \theta$ has vertical asymptotes at $\theta = \pi /2 + k\pi$ , zeros at $k\pi$ , and is increasing on each branch, changing from concave down to concave up at each zero.
04
In $a \tan (b(\theta + c)) + d$ , the period is $\pi / |b|$ . Tangent has no amplitude: a stretches the branches, and $a < 0$ makes them decreasing.
05
Graph with three landmarks per branch: the center, the two asymptotes, and the two quarter points where $\tan = \pm 1$ .
06
Factor out $b$ before reading the shift, and check any center or quarter point by plugging it in.

Practice 33 problems Next: Chapter 3.9 Inverse Trigonometric Functions

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The Tangent Function

Tangent as the Slope of the Terminal Ray

The Graph of y = tan θ

Transformations of Tangent

Writing an Equation from a Graph

Recap card

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