Mathematics » Functions II » The Tangent Function

Functions of the Form y = tan (kθ)

Functions of the form \(y=\tan (k\theta)\)

Optional Investigation: The effects of \(k\) on a tangent graph

  1. Complete the following table for \(y_1 = \tan \theta\) for \(-\text{360}\text{°} \leq \theta \leq \text{360}\text{°}\):
    \(\tan \theta\)       
    \(\tan \theta\)       
  2. Use the table of values to plot the graph of \(y_1 = \tan \theta\) for \(-\text{360}\text{°} \leq \theta \leq \text{360}\text{°}\).

  3. On the same system of axes, plot the following graphs:

    1. \(y_2 = \tan (-\theta)\)
    2. \(y_3 = \tan 3\theta\)
    3. \(y_4 = \tan \cfrac{\theta}{2}\)
  4. Use your sketches of the functions above to complete the following table:

    effect of \(k\)    
  5. What do you notice about \(y_1 = \tan \theta\) and \(y_2 = \tan (-\theta)\)?

  6. Is \(\tan (-\theta) = -\tan \theta\) a true statement? Explain your answer.

  7. Can you deduce a formula for determining the period of \(y = \tan k\theta\)?

The effect of the parameter on \(y = \tan k\theta\)

The value of \(k\) affects the period of the tangent function. If \(k\) is negative, then the graph is reflected about the \(y\)-axis.

  • For \(k > 0\):

    For \(k > 1\), the period of the tangent function decreases.

    For \(0 < k < 1\), the period of the tangent function increases.

  • For \(k < 0\):

    For \(-1 < k < 0\), the graph is reflected about the \(y\)-axis and the period increases.

    For \(k < -1\), the graph is reflected about the \(y\)-axis and the period decreases.

Negative angles: \[\tan (-\theta) = -\tan \theta\]

Calculating the period:

To determine the period of \(y = \tan k\theta\) we use, \[\text{Period} = \cfrac{\text{180}\text{°}}{|k|}\] where \(|k|\) is the absolute value of \(k\).

\(k > 0\)

\(k < 0\)

Functions of the Form <em>y = tan (kθ)</em>Functions of the Form <em>y = tan (kθ)</em>



  1. Sketch the following functions on the same set of axes for \(-\text{180}\text{°} \leq \theta \leq \text{180}\text{°}\).
    1. \(y_1 = \tan \theta\)
    2. \(y_2 = \tan \cfrac{3\theta}{2}\)
  2. For each function determine the following:

    • Period
    • Domain and range
    • \(x\)- and \(y\)-intercepts
    • Asymptotes

Examine the equations of the form \(y = \tan k\theta\)

Notice that \(k > 1\) for \(y_2 = \tan \cfrac{3\theta}{2}\), therefore the period of the graph decreases.

Complete a table of values

\(\tan \theta\)\(\text{0}\)\(\text{1}\)undef\(-\text{1}\)\(\text{0}\)\(\text{1}\)undef\(-\text{1}\)\(\text{0}\)
\(\tan \cfrac{3\theta}{2}\)undef\(-\text{0.41}\)\(\text{1}\)\(-\text{2.41}\)\(\text{0}\)\(\text{2.41}\)\(-\text{1}\)\(\text{0.41}\)undef

Sketch the tangent graphs

Functions of the Form <em>y = tan (kθ)</em>

Complete the table

 \(y_1 = \tan \theta\)\(y_2 = \tan \cfrac{3\theta}{2}\)
domain\(\{\theta: -\text{180}\text{°} \leq \theta \leq \text{180}\text{°}, \theta \ne -\text{90}\text{°}; \text{90}\text{°}\}\)\(\{\theta: -\text{180}\text{°} < \theta < \text{180}\text{°}, \theta \ne -\text{60}\text{°}; \text{60}\text{°}\}\)
range\(\{f(\theta): f(\theta) \in \mathbb{R}\}\)\(\{f(\theta): f(\theta) \in \mathbb{R}\}\)
\(x\)-intercept(s)\((-\text{180}\text{°};0)\), \((\text{0}\text{°};0)\) and \((\text{180}\text{°};0)\)\((-\text{120}\text{°};0)\), \((\text{0}\text{°};0)\) and \((\text{120}\text{°};0)\)
asymptotes\(\theta = -\text{90}\text{°}\) and \(\theta = \text{90}\text{°}\)\(\theta = -\text{180}\text{°}\); \(-\text{60}\text{°}\) and \(\text{180}\text{°}\)

Discovering the characteristics

For functions of the general form: \(f(\theta) = y =\tan k\theta\):

Domain and range

The domain of one branch is \(\{ \theta: -\cfrac{\text{90}\text{°}}{k} < \theta < \cfrac{\text{90}\text{°}}{k}, \theta \in \mathbb{R}\}\) because \(f(\theta)\) is undefined for \(\theta = -\cfrac{\text{90}\text{°}}{k}\) and \(\theta = \cfrac{\text{90}\text{°}}{k}\).

The range is \(\{ f(\theta): f(\theta) \in \mathbb{R} \}\) or \((-\infty; \infty)\).


The \(x\)-intercepts are determined by letting \(f(\theta) = 0\) and solving for \(\theta\).

The \(y\)-intercept is calculated by letting \(\theta = 0\) and solving for \(f(\theta)\). \begin{align*} y &= \tan k\theta \\ &= \tan \text{0}\text{°} \\ &= 0 \end{align*} This gives the point \((\text{0}\text{°};0)\).


These are the values of \(k\theta\) for which \(\tan k\theta\) is undefined.

Do you want to suggest a correction or an addition to this content? Leave Contribution

[Attributions and Licenses]

This is a lesson from the tutorial, Functions II and you are encouraged to log in or register, so that you can track your progress.

Log In

Share Thoughts

Do NOT follow this link or you will be banned from the site!