Mathematics » Functions II » The Sine Function

Functions of the Form y = sin kθ

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

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

  1. Complete the following table for \(y_1 = \sin \theta\) for \(-\text{360}\text{°} \leq \theta \leq \text{360}\text{°}\):
    \(θ\)\(-\text{360}\)\(\text{°}\)\(-\text{270}\)\(\text{°}\)\(-\text{180}\)\(\text{°}\)\(-\text{90}\)\(\text{°}\)\(\text{0}\)\(\text{°}\)\(\text{90}\)\(\text{°}\)\(\text{180}\)\(\text{°}\)\(\text{270}\)\(\text{°}\)\(\text{360}\)\(\text{°}\)
    \(\sin \theta\)         
  2. Use the table of values to plot the graph of \(y_1 = \sin \theta\) for \(-\text{360}\text{°} \leq \theta \leq \text{360}\text{°}\).

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

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

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

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

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

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

The value of \(k\) affects the period of the sine function. If \(k\) is negative, then the graph is reflected about the \(y\)-axis.

  • For \(k > 0\):

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

    For \(0 < k < 1\), the period of the sine 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: \[\sin (-\theta) = -\sin \theta\]

Calculating the period:

To determine the period of \(y = \sin k\theta\) we use, \[\text{Period } = \cfrac{\text{360}\text{°}}{|k|}\] where \(|k|\) is the absolute value of \(k\) (this means that \(k\) is always considered to be positive).

\(0 < k < 1\)

\(-1 < k < 0\)

1b39c5955de4ff4769778a00dc6f9bec.png9489eeaa93dfc8a3ea08de3bbb12e149.png

\(k > 1\)

\(k < -1\)

aed1a16b6d01b9250823e94de6445009.png4c3ec29fdca6d6878e134da51a3c64d7.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. \(y_1 = \sin \theta\)
    2. \(y_2 = \sin \cfrac{3\theta}{2}\)
  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 \(y = \sin k\theta\)

Notice that \(k > 1\) for \(y_2 = \sin \cfrac{3\theta}{2}\), therefore the period of the graph decreases.

Complete a table of values

\(θ\)\(-\text{180}\)\(\text{°}\)\(-\text{135}\)\(\text{°}\)\(-\text{90}\)\(\text{°}\)\(-\text{45}\)\(\text{°}\)\(\text{0}\)\(\text{°}\)\(\text{45}\)\(\text{°}\)\(\text{90}\)\(\text{°}\)\(\text{135}\)\(\text{°}\)\(\text{180}\)\(\text{°}\)
\(\sin \theta\)\(\text{0}\)\(-\text{0.71}\)\(-\text{1}\)\(-\text{0.71}\)\(\text{0}\)\(\text{0.71}\)\(\text{1}\)\(\text{0.71}\)\(\text{0}\)
\(\sin \cfrac{3\theta}{2}\)\(\text{1}\)\(\text{0.38}\)\(-\text{0.71}\)\(-\text{0.92}\)\(\text{0}\)\(\text{0.92}\)\(\text{0.71}\)\(-\text{0.38}\)\(-\text{1}\)

Sketch the sine graphs

11d6bcc079858c9218c92936fe08f725.png

Complete the table

 \(y_1 = \sin \theta\)\(y_2 = \sin \cfrac{3\theta}{2}\)
period\(\text{360}\)\(\text{°}\)\(\text{240}\)\(\text{°}\)
amplitude\(\text{1}\)\(\text{1}\)
domain\([-\text{180}\text{°};\text{180}\text{°}]\)\([-\text{180}\text{°};\text{180}\text{°}]\)
range\([-1;1]\)\([-1;1]\)
maximum turning points\((\text{90}\text{°};1)\)\((-\text{180}\text{°};1)\) and \((\text{60}\text{°};1)\)
minimum turning points\((-\text{90}\text{°};-1)\)\((-\text{60}\text{°};-1) \text{ and } (\text{180}\text{°};1)\)
\(y\)-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 \((\text{120}\text{°};0)\)

Discovering the characteristics

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

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 &= \sin k\theta \\ &= \sin \text{0}\text{°} \\ &= 0 \end{align*} This gives the point \((\text{0}\text{°};0)\).

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