Mathematics » Functions II » The Cosine Function

Sketching Cosine Graphs

Sketching cosine graphs



Sketch the graph of \(f(\theta) = \cos (\text{180}\text{°} – 3\theta)\) for \(\text{0}\text{°} \leq \theta \leq \text{360}\text{°}\).

Examine the form of the equation

Write the equation in the form \(y = \cos k(\theta + p)\). \begin{align*} f(\theta) &= \cos (\text{180}\text{°} – 3\theta)\\ &= \cos (-3\theta + \text{180}\text{°}) \\ &= \cos ( -3(\theta – \text{60}\text{°}) ) \\ &= \cos 3(\theta – \text{60}\text{°}) \end{align*}

To draw a graph of the above function, the standard cosine graph, \(y = \cos \theta\), must be changed in the following ways:

  • decrease the period by a factor of \(\text{3}\)
  • shift to the right by \(\text{60}\text{°}\).

Complete a table of values


Plot the points and join with a smooth curve


Period: \(\text{120}\)\(\text{°}\)

Amplitude: \(\text{1}\)

Domain: \([\text{0}\text{°};\text{360}\text{°}]\)

Range: \([-1;1]\)

Maximum turning point: \((\text{60}\text{°};1)\), \((\text{180}\text{°};1)\) and \((\text{300}\text{°};1)\)

Minimum turning point: \((\text{0}\text{°}; -1)\), \((\text{120}\text{°};-1)\), \((\text{240}\text{°};-1)\) and \((\text{360}\text{°};-1)\)

\(y\)-intercepts: \((\text{0}\text{°};-1)\)

\(x\)-intercept: \((\text{30}\text{°};0)\), \((\text{90}\text{°};0)\), \((\text{150}\text{°};0)\), \((\text{210}\text{°};0)\), \((\text{270}\text{°};0)\) and \((\text{330}\text{°};0)\)



Given the graph of \(y = a \cos (k\theta + p)\), determine the values of \(a\), \(k\), \(p\) and the minimum turning point.


Determine the value of \(k\)

From the sketch we see that the period of the graph is \(\text{360}\text{°}\), therefore \(k = 1\).

\[y = a \cos ( \theta + p)\]

Determine the value of \(a\)

From the sketch we see that the maximum turning point is \((\text{45}\text{°};2)\), so we know that the amplitude of the graph is \(\text{2}\) and therefore \(a = 2\).

\[y = 2 \cos ( \theta + p)\]

Determine the value of \(p\)

Compare the given graph with the standard cosine function \(y = \cos \theta\) and notice the difference in the maximum turning points. We see that the given function has been shifted to the right by \(\text{45}\)\(\text{°}\), therefore \(p = \text{45}\text{°}\).

\[y = 2 \cos ( \theta – \text{45}\text{°})\]

Determine the minimum turning point

At the minimum turning point, \(y = -2\):

\begin{align*} y &= 2 \cos ( \theta – \text{45}\text{°}) \\ -2 &= 2 \cos ( \theta – \text{45}\text{°}) \\ -1 &= \cos ( \theta – \text{45}\text{°}) \\ \cos^{-1}(-1) &= \theta – \text{45}\text{°} \\ \text{180}\text{°} &= \theta – \text{45}\text{°} \\ \text{225}\text{°} &= \theta \end{align*}

This gives the point \((\text{225}\text{°};-2)\).

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