Mathematics » Functions II » The Cosine Function

Functions of the Form y = cos (kθ)

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

We now consider cosine functions of the form \(y = \cos k\theta\) and the effects of parameter \(k\).

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

  1. Complete the following table for \(y_1 = \cos \theta\) for \(-\text{360}\text{°} \leq \theta \leq \text{360}\text{°}\):
    θ\(-\text{360}\)\(\text{°}\)\(-\text{300}\)\(\text{°}\)\(-\text{240}\)\(\text{°}\)\(-\text{180}\)\(\text{°}\)\(-\text{120}\)\(\text{°}\)\(-\text{60}\)\(\text{°}\)\(\text{0}\)\(\text{°}\)
    \(\cos \theta\)       
    θ\(\text{60}\)\(\text{°}\)\(\text{120}\)\(\text{°}\)\(\text{180}\)\(\text{°}\)\(\text{240}\)\(\text{°}\)\(\text{300}\)\(\text{°}\)\(\text{360}\)\(\text{°}\) 
    \(\cos \theta\)       
  2. Use the table of values to plot the graph of \(y_1 = \cos \theta\) for \(-\text{360}\text{°} \leq \theta \leq \text{360}\text{°}\).

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

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

     \(y_1\)\(y_2\)\(y_3\)\(y_4\)
    period    
    amplitude    
    domain    
    range    
    maximum turning points    
    minimum turning points    
    \(y\)-intercept(s)    
    \(x\)-intercept(s)    
    effect of \(k\)    
  5. What do you notice about \(y_1 = \cos \theta\) and \(y_2 = \cos (-\theta)\)?

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

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

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

The value of \(k\) affects the period of the cosine function.

  • For \(k > 0\):

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

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

  • For \(k < 0\):

    For \(-1 < k < 0\), the period increases.

    For \(k < -1\), the period decreases.

Negative angles: \[\cos (-\theta) = \cos \theta\] Notice that for negative values of \(\theta\), the graph is not reflected about the \(x\)-axis.

Calculating the period:

To determine the period of \(y = \cos k\theta\) we use, \[\text{Period } = \cfrac{\text{360}\text{°}}{|k|}\] where \(|k|\) is the absolute value of \(k\).

\(0 < k < 1\)

\(-1 < k < 0\)

58356ab256b4b9fa0156a64409f04592.pngddd3310bebaf8519b910d5fea228456d.png

\(k > 1\)

\(k < -1\)

984b092fb28a2d1de95c0e3dff84ceac.pngd8f9c98e34b1613992c75002fd8a56be.png

Example

Question

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

    1. Period
    2. Amplitude
    3. Domain and range
    4. \(x\)- and \(y\)-intercepts
    5. Maximum and minimum turning points

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

Notice that for \(y_2 = \cos \cfrac{\theta}{2}\), \(k < 1\) therefore the period of the graph increases.

Complete a table of values

θ\(-\text{180}\)\(\text{°}\)\(-\text{135}\)\(\text{°}\)\(-\text{90}\)\(\text{°}\)\(-\text{45}\)\(\text{°}\)\(\text{0}\)\(\text{°}\)\(\text{45}\)\(\text{°}\)\(\text{90}\)\(\text{°}\)\(\text{135}\)\(\text{°}\)\(\text{180}\)\(\text{°}\)
\(\cos \theta\)\(-\text{1}\)\(-\text{0.71}\)\(\text{0}\)\(\text{0.71}\)\(\text{1}\)\(\text{0.71}\)\(\text{0}\)\(-\text{0.71}\)\(-\text{1}\)
\(\cos \cfrac{\theta}{2}\)\(\text{0}\)\(\text{0.38}\)\(\text{0.71}\)\(\text{0.92}\)\(\text{1}\)\(\text{0.92}\)\(\text{0.71}\)\(\text{0.38}\)\(\text{0}\)

Sketch the cosine graphs

02547eb34be12eaa2b97e43d7a3a9d15.png

Complete the table

 \(y_1 = \cos \theta\)\(y_2 = \cos \cfrac{\theta}{2}\)
period\(\text{360}\text{°}\)\(\text{720}\text{°}\)
amplitude\(\text{1}\)\(\text{1}\)
domain\([-\text{180}\text{°};\text{180}\text{°}]\)\([-\text{180}\text{°};\text{180}\text{°}]\)
range\([-1;1]\)\([0;1]\)
maximum turning points\((\text{0}\text{°};1)\)\((\text{0}\text{°};1)\)
minimum turning points\((-\text{180}\text{°};-1) \text{ and } (\text{180}\text{°};-1)\)none
\(y\)-intercept(s)\((\text{0}\text{°};1)\)\((\text{0}\text{°};1)\)
\(x\)-intercept(s)\((-\text{90}\text{°};0) \text{ and } (\text{90}\text{°};0)\)\((-\text{180}\text{°};0) \text{ and } (\text{180}\text{°};0)\)

Discovering the characteristics

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

Domain and range

The domain is \(\{ \theta: \theta \in \mathbb{R} \}\) because there is no value for \(\theta\) for which \(f(\theta)\) is undefined.

The range is \(\{ f(\theta): -1 \leq f(\theta) \leq 1, f(\theta) \in \mathbb{R} \}\) or \([-1;1]\).

Intercepts

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

The \(y\)-intercept is calculated by letting \(\theta = \text{0}\text{°}\) and solving for \(f(\theta)\). \begin{align*} y &= \cos k\theta \\ &= \cos \text{0}\text{°} \\ &= 1 \end{align*} This gives the point \((\text{0}\text{°};1)\).


Got questions about this content? Get access to an AI-Powered Study Help/Tutor you can chat with as you learn! Continue Learning With Ulearngo


[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