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The Tangent Function (College Board AP® Precalculus): Study Guide

Written by: Roger B

Reviewed by: Mark Curtis

Updated on 15 April 2026

The tangent function & the unit circle

How is the tangent function defined using the unit circle?

Given an angle $θ$ in standard position and a unit circle centered at the origin
- the terminal ray intersects the circle at a point $P$
The value of the tangent function, $f (θ) = \tan θ$
- is given by the slope of the terminal ray
This connects directly to the unit circle coordinates of point $P$
- $P$ has coordinates $(\cos θ, \sin θ)$
- So the slope of a line through the origin $(0, 0)$ and the point $P$ is
  - $slope = \frac{\sin θ - 0}{\cos θ - 0} = \frac{\sin θ}{\cos θ}$
Therefore
- $\tan θ = \frac{\sin θ}{\cos θ}, where \cos θ \neq 0$
This slope interpretation means
- When the terminal ray is steep and rising
  - $\tan θ$ is a large positive number
- When the terminal ray is horizontal (along the $x$ -axis)
  - $\tan θ = 0$
- When the terminal ray is steep and falling
  - $\tan θ$ is a large negative number
- When the terminal ray is vertical
  - the slope is undefined
  - so $\tan θ$ is undefined

What are the exact values of tangent for key angles?

Since $\tan θ = \frac{\sin θ}{\cos θ}$ , the exact values can be calculated from the sine and cosine values

$θ$	$\sin θ$	$\cos θ$	$\tan θ$
$- \frac{π}{2}$	$- 1$	$0$	undefined
$- \frac{π}{3}$	$- \frac{\sqrt{3}}{2}$	$\frac{1}{2}$	$- \sqrt{3}$
$- \frac{π}{4}$	$- \frac{\sqrt{2}}{2}$	$\frac{\sqrt{2}}{2}$	$- 1$
$- \frac{π}{6}$	$- \frac{1}{2}$	$\frac{\sqrt{3}}{2}$	$- \frac{1}{\sqrt{3}} = - \frac{\sqrt{3}}{3}$
$0$	$0$	$1$	$0$
$\frac{π}{6}$	$\frac{1}{2}$	$\frac{\sqrt{3}}{2}$	$\frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$
$\frac{π}{4}$	$\frac{\sqrt{2}}{2}$	$\frac{\sqrt{2}}{2}$	$1$
$\frac{π}{3}$	$\frac{\sqrt{3}}{2}$	$\frac{1}{2}$	$\sqrt{3}$
$\frac{π}{2}$	$1$	$0$	undefined

Because tangent has a period of $π$ , the values repeat every $π$ radians
- E.g. $\tan \frac{7 π}{6} = \tan \frac{π}{6} = \frac{\sqrt{3}}{3}$

Key characteristics of the tangent function

What is the period of the tangent function?

The slope values of the terminal ray repeat every half revolution of the unit circle
- For example, the terminal rays at $θ$ and $θ + π$ point in opposite directions
  - but they lie on the same line
  - and therefore have the same slope
Therefore the tangent function has a period of $π$ :

$\tan (θ + π) = \tan θ$

This is different from sine and cosine
- which both have a period of $2 π$

Where does the tangent function have asymptotes?

The tangent function is undefined when $\cos θ = 0$ , which occurs at

$θ = \frac{π}{2} + k π, for integer values of k$

At these input values, the terminal ray is vertical (pointing straight up or straight down)
- so its slope does not exist
The graph of the tangent function has vertical asymptotes at each of these values
- E.g. as $θ$ approaches $\frac{π}{2}$ from the left, $\tan θ$ increases without bound
- and as $θ$ approaches $\frac{π}{2}$ from the right, $\tan θ$ decreases without bound

The tangent function therefore demonstrates periodic asymptotic behavior
- the pattern of asymptotes repeats with the same period as the function

What is the behavior of the tangent function between consecutive asymptotes?

Between any two consecutive asymptotes
- the tangent function is always increasing
The graph changes from concave down to concave up between consecutive asymptotes
- Specifically, the graph is
  - concave down on the first half of the interval
  - and concave up on the second half
- The changeover occurs at the point of inflection
  - which is where $\tan θ = 0$
  - i.e. at integer multiples of $π$
Between consecutive asymptotes, the tangent function takes all real number output values, from $- \infty$ to $+ \infty$

Graph showing the tangent function y = tan θ with vertical asymptotes at odd multiples of π/2, between -2π and 2π, with curves rising and falling. — **Graph of y=tanθ**

Transformations of the tangent function

How do additive transformations affect the tangent function?

A vertical translation of the tangent function is given by $g (θ) = \tan θ + d$
- This shifts the graph vertically by $d$ units
- The line containing the points of inflection is also shifted up by $d$ units
  - from $y = 0$ to $y = d$
- The asymptotes are not affected by a vertical translation

A horizontal translation (phase shift) is given by $g (θ) = \tan (θ + c)$
- This shifts the graph horizontally by $- c$ units
  - to the left if $c > 0$ , to the right if $c < 0$
- The asymptotes shift by $- c$ units as well

How do multiplicative transformations affect the tangent function?

A vertical dilation is given by $g (θ) = a \tan θ$
- This stretches the graph vertically by a factor of $| a |$
- If $a < 0$ , the graph is also reflected over the $x$ -axis
  - this causes the function to be decreasing (instead of increasing) between consecutive asymptotes
  - and reverses the concavity pattern
- The asymptotes and period are not affected

A horizontal dilation is given by $g (θ) = \tan (b θ)$
- This changes the period of the function by a factor of $\frac{1}{| b |}$
  - The new period is $\frac{π}{| b |}$
- If $b < 0$ , the graph is also reflected over the $y$ -axis
  - this causes the function to be decreasing (instead of increasing) between consecutive asymptotes
  - and reverses the concavity pattern
- The asymptotes are compressed or stretched accordingly

How are all transformations combined?

The general form combining all transformations is $f (θ) = a \tan (b (θ + c)) + d$

The parameters have the following effects
- $| a |$ : vertical dilation factor (with reflection over the $x$ -axis if $a < 0$ )
- $\frac{π}{| b |}$ : the period of the function
- $- c$ : the phase shift (horizontal translation)
- $d$ : the vertical shift (the new midline of the points of inflection is $y = d$ )
For example, $g (θ) = 3 \tan (2 (θ - \frac{π}{4})) + 1$ has:
- Vertical dilation factor $3$
- Period $= \frac{π}{2}$
- Phase shift of $\frac{π}{4}$ to the right (since $c = - \frac{π}{4}$ )
- Vertical shift of $1$ unit up

Examiner Tips and Tricks

When identifying the period of a transformed tangent function, remember that the base period is $π$ , not $2 π$ . A common mistake is to use the sine/cosine period formula $\frac{2 π}{| b |}$ instead of the correct tangent period formula $\frac{π}{| b |}$ .

Worked Example

The function $g$ is defined by $g (θ) = - 2 \tan (\frac{θ}{3}) + 4$ .

(a) Find the period of $g$ .

Answer:

The function has the form $a \tan (b θ) + d$ where $b = \frac{1}{3}$

So the period of the tangent function is

$\frac{π}{| b |} = \frac{π}{\frac{1}{3}} = 3 π$

$period = 3 π$

(b) Find the equations of two consecutive vertical asymptotes of the graph of $g$ .

Answer:

The asymptotes of $y = \tan θ$ occur at $θ = \frac{π}{2} + k π$

For $g (θ) = - 2 \tan (\frac{θ}{3}) + 4$ , asymptotes occur when

$\frac{θ}{3} = \frac{π}{2} + k π$

i.e. when

$θ = \frac{3 π}{2} + 3 k π$

So a possible set of two consecutive asymptotes (using $k = 0$ and $k = 1$ ) is

$θ = \frac{3 π}{2} + 3 (0) π = \frac{3 π}{2}$

$and θ = \frac{3 π}{2} + 3 (1) π = \frac{9 π}{2}$

$θ = \frac{3 π}{2} and θ = \frac{9 π}{2}$

(c) Find the coordinates of the point of inflection of the graph of $g$ that lies between the two asymptotes found in part (b).

Answer:

The point of inflection lies midway between the two consecutive asymptotes

$θ = \frac{1}{2} (\frac{3 π}{2} + \frac{9 π}{2}) = \frac{1}{2} \times \frac{12 π}{2} = 3 π$

At a point of inflection of a transformed tangent function, the tangent part equals zero, so

$g (3 π) = - 2 \tan (\frac{3 π}{3}) + 4 = - 2 \tan (π) + 4 = - 2 (0) + 4 = 4$

Therefore

The point of inflection is $(3 π, 4)$

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The Tangent Function (College Board AP® Precalculus): Study Guide

The tangent function & the unit circle

How is the tangent function defined using the unit circle?

What are the exact values of tangent for key angles?

Key characteristics of the tangent function

What is the period of the tangent function?

Where does the tangent function have asymptotes?

What is the behavior of the tangent function between consecutive asymptotes?

Transformations of the tangent function

How do additive transformations affect the tangent function?

How do multiplicative transformations affect the tangent function?

How are all transformations combined?

Examiner Tips and Tricks

Worked Example

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

Author: Roger B

Reviewer: Mark Curtis