The Tangent Function Revision Notes for VCE SSCE Mathematical Methods

The Tangent Function

Introduction to the tangent function

The tangent of an angle is defined in terms of sine and cosine:

\tan \theta = \frac{\sin \theta}{\cos \theta} \quad \text{for } \cos \theta \neq 0

This definition is only valid when $\cos \theta \neq 0$ , as division by zero is undefined. This restriction creates vertical asymptotes in the graph of the tangent function.

chatImportant

The tangent function is undefined wherever the cosine equals zero. This fundamental restriction determines the location of all vertical asymptotes in tangent graphs.

The basic tangent function $y = \tan \theta$

Table of values

The following table shows key values of the tangent function:

Notice that the function is undefined (ud) at $\theta = -\frac{\pi}{2}, \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}$ , which is where $\cos \theta = 0$ .

Graph of $y = \tan \theta$

The graph shows several important features:

Vertical asymptotes: These occur at $\theta = -\frac{\pi}{2}, \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}$ , and generally at all odd multiples of $\frac{\pi}{2}$ .

Repeating pattern: The graph has a periodic structure, repeating every $\pi$ units along the $\theta$ -axis.

Unbounded values: Unlike sine and cosine, the tangent function can take any real value, approaching positive or negative infinity near the asymptotes.

infoNote

Unlike sine and cosine functions which are bounded between -1 and 1, the tangent function has no upper or lower bounds. This is why it can approach infinity near its asymptotes.

Key properties of $y = \tan \theta$

Period: The tangent function repeats every $\pi$ units. This is half the period of sine and cosine functions.

Range: The range is all real numbers, $\mathbb{R}$ . The function can take any value from negative infinity to positive infinity.

Vertical asymptotes: These occur at $\theta = \frac{(2k + 1)\pi}{2}$ where $k \in \mathbb{Z}$ (i.e., $k$ is any integer). This formula represents all odd multiples of $\frac{\pi}{2}$ .

Axis intercepts: The graph crosses the $\theta$ -axis at $\theta = k\pi$ where $k \in \mathbb{Z}$ (i.e., at integer multiples of $\pi$ ).

Transformations: the function $y = a \tan(nt)$

When we apply transformations to the basic tangent function, we obtain $y = a \tan(nt)$ where $a$ and $n$ are positive numbers.

How to obtain the graph

The graph of $y = a \tan(nt)$ is created from the graph of $y = \tan t$ by applying:

A dilation of factor $a$ from the $t$ -axis (vertical stretch)
A dilation of factor $\frac{1}{n}$ from the $y$ -axis (horizontal compression)

Properties of $f(t) = a \tan(nt)$

Period: $\frac{\pi}{n}$

The coefficient $n$ affects the period. A larger value of $n$ means a shorter period (more cycles in the same interval).

Range: $\mathbb{R}$

The range remains all real numbers regardless of the transformations.

Vertical asymptotes: $t = \frac{(2k + 1)\pi}{2n}$ where $k \in \mathbb{Z}$

The asymptotes become closer together as $n$ increases.

Axis intercepts: $t = \frac{k\pi}{n}$ where $k \in \mathbb{Z}$

The zeros become closer together as $n$ increases.

infoNote

When the coefficient $n$ increases, both the asymptotes and zeros become more frequent (closer together), while the period decreases. The vertical stretch factor $a$ affects the steepness of the curve but does not change the period or the locations of asymptotes.

Worked example: sketching transformed tangent graphs

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Worked Example: Sketch $y = 3 \tan(2x)$ for $x \in [-\pi, \pi]$

Step 1: Identify the period

Period $= \frac{\pi}{n} = \frac{\pi}{2}$

Step 2: Find the asymptotes

Asymptotes occur at $x = \frac{(2k + 1)\pi}{4}$ where $k \in \mathbb{Z}$

For the given interval, asymptotes are at $x = -\frac{3\pi}{4}, -\frac{\pi}{4}, \frac{\pi}{4}, \frac{3\pi}{4}$

Step 3: Find the axis intercepts

Axis intercepts occur at $x = \frac{k\pi}{2}$ where $k \in \mathbb{Z}$

For the given interval, intercepts are at $x = -\pi, -\frac{\pi}{2}, 0, \frac{\pi}{2}, \pi$

Step 4: Sketch the graph

The factor of 3 stretches the graph vertically, making it steeper between asymptotes.

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Worked Example: Sketch $y = -2 \tan(3x)$ for $x \in [-\pi, \pi]$

Step 1: Identify the period

Period $= \frac{\pi}{n} = \frac{\pi}{3}$

Step 2: Find the asymptotes

Asymptotes occur at $x = \frac{(2k + 1)\pi}{6}$ where $k \in \mathbb{Z}$

Step 3: Find the axis intercepts

Axis intercepts occur at $x = \frac{k\pi}{3}$ where $k \in \mathbb{Z}$

Step 4: Note the reflection

The negative sign reflects the graph in the $x$ -axis, so the curves slope downward instead of upward.

Horizontal translations

When the tangent function includes a horizontal shift, we write it as $y = a \tan(n(x - c))$ or equivalently $y = a \tan(nx - d)$ where $d = nc$ .

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Worked Example: Sketch $y = 3 \tan\left(2x - \frac{\pi}{3}\right)$ for $\frac{\pi}{6} \leq x \leq \frac{13\pi}{6}$

Step 1: Rewrite in factored form

y = 3 \tan\left(2\left(x - \frac{\pi}{6}\right)\right)

Step 2: Identify the transformations

The graph of $y = \tan x$ undergoes:

A dilation of factor 3 from the $x$ -axis
A dilation of factor $\frac{1}{2}$ from the $y$ -axis
A translation of $\frac{\pi}{6}$ units in the positive $x$ -direction

Step 3: Calculate the period

Period $= \frac{\pi}{2}$

Step 4: Find the asymptotes

Asymptotes: $x = \frac{(2k + 1)\pi}{4} + \frac{\pi}{6} = \frac{(6k + 5)\pi}{12}$ where $k \in \mathbb{Z}$

Step 5: Find the axis intercepts

Axis intercepts: $x = \frac{k\pi}{2} + \frac{\pi}{6} = \frac{(3k + 1)\pi}{6}$ where $k \in \mathbb{Z}$

Vertical translations

When we add a constant to the tangent function, we shift it vertically.

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Worked Example: Sketch $y = 3 \tan\left(2x - \frac{\pi}{3}\right) + \sqrt{3}$ for $\frac{\pi}{6} \leq x \leq \frac{13\pi}{6}$

Step 1: Rewrite in factored form

y = 3 \tan\left(2\left(x - \frac{\pi}{6}\right)\right) + \sqrt{3}

Step 2: Identify the transformations

Starting from $y = \tan x$ , apply:

A dilation of factor 3 from the $x$ -axis
A dilation of factor $\frac{1}{2}$ from the $y$ -axis
A translation of $\frac{\pi}{6}$ units in the positive $x$ -direction
A translation of $\sqrt{3}$ units in the positive $y$ -direction

Step 3: Note about asymptotes

The vertical translation does NOT affect the positions of the vertical asymptotes. They remain at the same $x$ -values.

Step 4: Find the $x$ -axis intercepts

To find where the graph crosses the $x$ -axis, solve:

3 \tan\left(2x - \frac{\pi}{3}\right) + \sqrt{3} = 0

\tan\left(2x - \frac{\pi}{3}\right) = -\frac{\sqrt{3}}{3} = -\frac{1}{\sqrt{3}}

2x - \frac{\pi}{3} = \frac{5\pi}{6}, \frac{11\pi}{6}, \frac{17\pi}{6}, \frac{23\pi}{6}

2x = \frac{7\pi}{6}, \frac{13\pi}{6}, \frac{19\pi}{6}, \frac{25\pi}{6}

x = \frac{7\pi}{12}, \frac{13\pi}{12}, \frac{19\pi}{12}, \frac{25\pi}{12}

The $x$ -axis intercepts are at $\frac{7\pi}{12}, \frac{13\pi}{12}, \frac{19\pi}{12}$ , and $\frac{25\pi}{12}$ .

chatImportant

Common Mistake Alert

Vertical translations shift the entire graph up or down but do NOT change the positions of the vertical asymptotes. Only horizontal transformations (scaling or translations) affect where the asymptotes occur.

Solving equations involving the tangent function

Exact values to remember

These fundamental values are essential for solving tangent equations:

\tan 0 = 0

\tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}

\tan\left(\frac{\pi}{4}\right) = 1

\tan\left(\frac{\pi}{3}\right) = \sqrt{3}

infoNote

Memorizing these exact values will make solving tangent equations much faster. They correspond to the special angles that appear frequently in trigonometry problems.

Symmetry properties

Periodic property: $\tan(\pi + \theta) = \tan \theta$

This reflects the fact that the tangent function has period $\pi$ .

Odd function property: $\tan(-\theta) = -\tan \theta$

The tangent function is an odd function, meaning its graph is symmetric about the origin.

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Worked Example: Solve $3 \tan(2x) = \sqrt{3}$ for $x \in (0, 2\pi)$

Step 1: Isolate the tangent

\tan(2x) = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}

Step 2: Find the reference solution

Since $\tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}$ , we have $2x = \frac{\pi}{6}$ as one solution.

Step 3: Find all solutions for $2x$

We need solutions for $x$ in $(0, 2\pi)$ , which means $2x$ is in $(0, 4\pi)$ .

Using the period of $\pi$ , we add multiples of $\pi$ :

2x = \frac{\pi}{6}, \frac{\pi}{6} + \pi, \frac{\pi}{6} + 2\pi, \frac{\pi}{6} + 3\pi

2x = \frac{\pi}{6}, \frac{7\pi}{6}, \frac{13\pi}{6}, \frac{19\pi}{6}

Step 4: Solve for $x$

x = \frac{\pi}{12}, \frac{7\pi}{12}, \frac{13\pi}{12}, \frac{19\pi}{12}

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Worked Example: Solve $\tan\left(\frac{1}{2}\left(x - \frac{\pi}{4}\right)\right) = -1$ for $x \in [-2\pi, 2\pi]$

Step 1: Find the reference solution

Since $\tan\left(\frac{\pi}{4}\right) = 1$ , and we need $\tan = -1$ , we use the odd function property:

\tan\left(-\frac{\pi}{4}\right) = -1

Also, $\tan\left(\frac{3\pi}{4}\right) = -1$

Step 2: Set up the equation

\frac{1}{2}\left(x - \frac{\pi}{4}\right) = -\frac{\pi}{4} \text{ or } \frac{3\pi}{4}

Step 3: Solve for the transformed variable

x - \frac{\pi}{4} = -\frac{\pi}{2} \text{ or } \frac{3\pi}{2}

Step 4: Solve for $x$

x = -\frac{\pi}{4} \text{ or } \frac{7\pi}{4}

Note: We need to check which solutions fall within the given domain. For $x \in [-2\pi, 2\pi]$ , this means $x - \frac{\pi}{4} \in \left[-\frac{9\pi}{4}, \frac{7\pi}{4}\right]$ , which means $\frac{1}{2}\left(x - \frac{\pi}{4}\right) \in \left[-\frac{9\pi}{8}, \frac{7\pi}{8}\right]$ .

Solving equations of the form $\sin(nx) = k \cos(nx)$

When solving equations where sine equals a constant times cosine, we can use a useful technique.

Key principle

If $\sin(nx) = k \cos(nx)$ , then $\tan(nx) = k$ .

This is obtained by dividing both sides by $\cos(nx)$ (provided $\cos(nx) \neq 0$ ).

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This technique allows us to find the intersection points of sine and cosine curves by converting to a tangent equation. This conversion simplifies the problem significantly.

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Worked Example: Find where $y = \sin x$ and $y = \cos x$ intersect for $x \in [0, 2\pi]$

Step 1: Set up the equation

\sin x = \cos x

Step 2: Convert to tangent

\tan x = 1

Step 3: Solve

Since $\tan\left(\frac{\pi}{4}\right) = 1$ , and using the period of $\pi$ :

x = \frac{\pi}{4} \text{ or } \frac{5\pi}{4}

Step 4: Find the coordinates

When $x = \frac{\pi}{4}$ : $y = \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$

When $x = \frac{5\pi}{4}$ : $y = \sin\left(\frac{5\pi}{4}\right) = -\frac{1}{\sqrt{2}}$

The points of intersection are $\left(\frac{\pi}{4}, \frac{1}{\sqrt{2}}\right)$ and $\left(\frac{5\pi}{4}, -\frac{1}{\sqrt{2}}\right)$ .

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Worked Example: Solve $\sin(2x) = \cos(2x)$ for $x \in [0, 2\pi]$

Step 1: Convert to tangent

\tan(2x) = 1

Step 2: Find all solutions for $2x$

Since $x$ is in $[0, 2\pi]$ , we need $2x$ in $[0, 4\pi]$ .

2x = \frac{\pi}{4}, \frac{5\pi}{4}, \frac{9\pi}{4}, \frac{13\pi}{4}

Step 3: Solve for $x$

x = \frac{\pi}{8}, \frac{5\pi}{8}, \frac{9\pi}{8}, \frac{13\pi}{8}

The graph shows the four intersection points labeled A, B, C, and D occurring at these $x$ -values.

Remember!

bookmarkSummary

Key Points to Remember:

The tangent function is defined as $\tan \theta = \frac{\sin \theta}{\cos \theta}$ where $\cos \theta \neq 0$ .
The basic tangent function has period $\pi$ , range $\mathbb{R}$ , vertical asymptotes at $\theta = \frac{(2k + 1)\pi}{2}$ (odd multiples of $\frac{\pi}{2}$ ), and axis intercepts at $\theta = k\pi$ (integer multiples of $\pi$ ).
For the transformed function $y = a \tan(nt)$ , the period is $\frac{\pi}{n}$ , asymptotes occur at $t = \frac{(2k + 1)\pi}{2n}$ , and intercepts occur at $t = \frac{k\pi}{n}$ .
Key symmetry properties: $\tan(\pi + \theta) = \tan \theta$ (periodic) and $\tan(-\theta) = -\tan \theta$ (odd function).
When solving equations of the form $\sin(nx) = k \cos(nx)$ , convert to $\tan(nx) = k$ for easier solution.

The Tangent Function (VCE SSCE Mathematical Methods): Revision Notes