Mathematics » Functions II » The Cosine Function

Sketching Cosine Graphs

Sketching cosine graphs

Example

Question

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

θ\(\text{0}\)\(\text{°}\)\(\text{45}\)\(\text{°}\)\(\text{90}\)\(\text{°}\)\(\text{135}\)\(\text{°}\)\(\text{180}\)\(\text{°}\)\(\text{225}\)\(\text{°}\)\(\text{270}\)\(\text{°}\)\(\text{315}\)\(\text{°}\)\(\text{360}\)\(\text{°}\)
\(f(\theta)\)\(-\text{1}\)\(\text{0.71}\)\(\text{0}\)\(-\text{0.71}\)\(\text{1}\)\(-\text{0.71}\)\(\text{0}\)\(\text{0.71}\)\(-\text{1}\)

Plot the points and join with a smooth curve

Sketching Cosine Graphs

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 \((\text{330}\text{°};0)\)

Example

Question

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

Sketching Cosine Graphs

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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