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 \(-\text{360}\text{°} \leq \theta \leq \text{360}\text{°}\):
    θ\(-\text{360}\)\(\text{°}\)\(-\text{300}\)\(\text{°}\)\(-\text{240}\)\(\text{°}\)\(-\text{180}\)\(\text{°}\)\(-\text{120}\)\(\text{°}\)\(-\text{60}\)\(\text{°}\)\(\text{0}\)\(\text{°}\)
    \(\tan \theta\)       
    θ\(\text{60}\)\(\text{°}\)\(\text{120}\)\(\text{°}\)\(\text{180}\)\(\text{°}\)\(\text{240}\)\(\text{°}\)\(\text{300}\)\(\text{°}\)\(\text{360}\)\(\text{°}\) 
    \(\tan \theta\)       
  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:

     \(y_1\)\(y_2\)\(y_3\)\(y_4\)
    period    
    domain    
    range    
    \(y\)-intercept(s)    
    \(x\)-intercept(s)    
    asymptotes    
    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>

Example

Question

  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

θ\(-\text{180}\)\(\text{°}\)\(-\text{135}\)\(\text{°}\)\(-\text{90}\)\(\text{°}\)\(-\text{45}\)\(\text{°}\)\(\text{0}\)\(\text{°}\)\(\text{45}\)\(\text{°}\)\(\text{90}\)\(\text{°}\)\(\text{135}\)\(\text{°}\)\(\text{180}\)\(\text{°}\)
\(\tan \theta\)\(\text{0}\)\(\text{1}\)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

 \(y_1 = \tan \theta\)\(y_2 = \tan \cfrac{3\theta}{2}\)
period\(\text{180}\)\(\text{°}\)\(\text{120}\)\(\text{°}\)
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\(\{f(\theta): f(\theta) \in \mathbb{R}\}\)\(\{f(\theta): f(\theta) \in \mathbb{R}\}\)
\(y\)-intercept(s)\((\text{0}\text{°};0)\)\((\text{0}\text{°};0)\)
\(x\)-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 \((-\infty; \infty)\).

Intercepts

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 \((\text{0}\text{°};0)\).

Asymptotes

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

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