Mathematics » Functions II » The Sine Function

Sketching Sine Graphs

Sketching Sine Graphs

Example

Question

Sketch the graph of \(f(\theta) = \sin (\text{45}\text{°} – \theta)\) for \(\text{0}\text{°} \leq \theta \leq \text{360}\text{°}\).

Examine the form of the equation

Write the equation in the form \(y = \sin (\theta + p)\).

\begin{align*} f(\theta) &= \sin (\text{45}\text{°} – \theta)\\ &= \sin (-\theta + \text{45}\text{°}) \\ &= \sin ( -(\theta – \text{45}\text{°}) ) \\ &= -\sin (\theta – \text{45}\text{°}) \end{align*}

To draw a graph of the above function, we know that the standard sine graph, \(y = \sin\theta\), must:

  • be reflected about the \(x\)-axis
  • be shifted to the right by \(\text{45}\text{°}\)

Complete a table of values

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

Plot the points and join with a smooth curve

c48bdaef4f3955e156bcc39c2c65dd9e.png

Period: \(\text{360}\text{°}\)

Amplitude: \(\text{1}\)

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

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

Maximum turning point: \((\text{315}\text{°};1)\)

Minimum turning point: \((\text{135}\text{°};-1)\)

\(y\)-intercepts: \((\text{0}\text{°};\text{0.71})\)

\(x\)-intercept: \((\text{45}\text{°};0) \text{ and } (\text{225}\text{°};0)\)

Example

Question

Sketch the graph of \(f(\theta) = \sin (3\theta + \text{60}\text{°})\) for \(\text{0}\text{°} \leq \theta \leq \text{180}\text{°}\).

Examine the form of the equation

Write the equation in the form \(y = \sin k(\theta + p)\).

\begin{align*} f(\theta) &= \sin (3\theta + \text{60}\text{°})\\ &= \sin 3(\theta + \text{20}\text{°}) \end{align*}

To draw a graph of the above equation, the standard sine graph, \(y = \sin\theta\), must be changed in the following ways:

  • decrease the period by a factor of \(\text{3}\);
  • shift to the left by \(\text{20}\text{°}\).

Complete a table of values

θ\(\text{0}\)\(\text{°}\)\(\text{30}\)\(\text{°}\)\(\text{60}\)\(\text{°}\)\(\text{90}\)\(\text{°}\)\(\text{120}\)\(\text{°}\)\(\text{150}\)\(\text{°}\)\(\text{180}\)\(\text{°}\)
\(f(\theta)\)\(\text{0.87}\)\(\text{0.5}\)\(-\text{0.87}\)\(-\text{0.5}\)\(\text{0.87}\)\(\text{0.5}\)\(-\text{0.87}\)

Plot the points and join with a smooth curve

df7384a16d5d9bb9e2dcbb588d9ab470.png

Period: \(\text{120}\text{°}\)

Amplitude: \(\text{1}\)

Domain: \([\text{0}\text{°}; \text{180}\text{°}]\)

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

Maximum turning point: \((\text{10}\text{°}; 1) \text{ and } (\text{130}\text{°}; 1)\)

Minimum turning point: \((\text{70}\text{°}; -1)\)

\(y\)-intercept: \((\text{0}\text{°}; \text{0.87})\)

\(x\)-intercepts: \((\text{40}\text{°}; 0)\), \((\text{100}\text{°}; 0)\) and \((\text{160}\text{°}; 0)\)

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