We think you are located in South Africa. Is this correct?

Trigonometric Functions

6.6 Trigonometric functions (EMA52)

This section describes the graphs of trigonometric functions.

Sine function (EMA53)

Functions of the form \(y=\sin\theta\) (EMA54)

Worked example 16: Plotting a sine graph

\(y=f(\theta) = \sin\theta \quad \left[0° \le \theta \le 360°\right]\)

Use your calculator to complete the following table.

Choose an appropriate scale and plot the values of \(\theta\) on the \(x\)-axis and of \(\sin\theta\) on the \(y\)-axis. Round answers to \(\text{2}\) decimal places.

\(\theta\)

\(0°\)

\(30°\)

\(60°\)

\(90°\)

\(120°\)

\(150°\)

\(180°\)

\(210°\)

\(240°\)

\(270°\)

\(300°\)

\(330°\)

\(360°\)

\(\sin\theta\)

Substitute values for \(\theta\)

\(\theta\)

\(0°\)

\(30°\)

\(60°\)

\(90°\)

\(120°\)

\(150°\)

\(180°\)

\(210°\)

\(240°\)

\(270°\)

\(300°\)

\(330°\)

\(360°\)

\(\sin\theta\)

\(\text{0}\)

\(\text{0,5}\)

\(\text{0,87}\)

\(\text{1}\)

\(\text{0,87}\)

\(\text{0,5}\)

\(\text{0}\)

\(-\text{0,5}\)

\(-\text{0,87}\)

\(-\text{1}\)

\(-\text{0,87}\)

\(-\text{0,5}\)

\(\text{0}\)

Plot the points and join with a smooth curve

82b4b3a590b82131912f5974525519b7.png

Notice the wave shape of the graph. Each complete wave takes \(360°\) to complete. This is called the period. The height of the wave above and below the \(x\)-axis is called the graph's amplitude. The maximum value of \(y=\sin\theta\) is \(\text{1}\) and the minimum value is \(-\text{1}\).

Domain: \(\left[0°;360°\right]\)

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

\(x\)-intercepts: \((0°;0)\), \((180°;0)\), \((360°;0)\)

\(y\)-intercept: \((0°;0)\)

Maximum turning point: \((90°;1)\)

Minimum turning point: \((270°;-1)\)

Functions of the form \(y=a\sin\theta +q\) (EMA55)

The effects of \(a\) and \(q\) on a sine graph

In the equation, \(y=a\sin\theta +q\), \(a\) and \(q\) are constants and have different effects on the graph. On the same set of axes, plot the following graphs for \(0°\le \theta \le 360°\):

  1. \({y}_{1} = \sin\theta -2\)

  2. \({y}_{2} = \sin\theta -1\)

  3. \({y}_{3} = \sin\theta\)

  4. \({y}_{4} = \sin\theta +1\)

  5. \({y}_{5} = \sin\theta +2\)

Use your results to deduce the effect of \(q\).

On the same set of axes, plot the following graphs for \(0° \le \theta \le 360°\):

  1. \({y}_{6}=-2\sin\theta\)

  2. \({y}_{7}=-\sin\theta\)

  3. \({y}_{8}=\sin\theta\)

  4. \({y}_{9}=2\sin\theta\)

Use your results to deduce the effect of \(a\).

The effect of \(q\)

The effect of \(q\) is called a vertical shift because the whole sine graph shifts up or down by \(q\) units.

  • For \(q>0\), the graph is shifted vertically upwards by \(q\) units.

  • For \(q<0\), the graph is shifted vertically downwards by \(q\) units.

The effect of \(a\)

The value of \(a\) affects the amplitude of the graph; the height of the peaks and the depth of the troughs.

  • For \(a>1\), there is a vertical stretch and the amplitude increases.

    For \(0<a<1\), the amplitude decreases

  • For \(a<0\), there is a reflection about the \(x\)-axis.

    For \(-1<a<0\), there is a reflection about the \(x\)-axis and the amplitude decreases.

    For \(a<-1\), there is a reflection about the \(x\)-axis and the amplitude increases.

Note that amplitude is always positive.

Effect of \(a\)

\(a>1\): vertical stretch, amplitude increases

de1048894d08e9e5163835ec7264066b.png

\(a=1\): basic sine graph

\(0<a<1\): vertical contraction, amplitude decreases

\(-1<a<0\): reflection about \(x\)-axis of \(0<a<1\)

\(a<-1\): reflection about \(x\)-axis of \(a>1\)

The effect of \(a\) on a sine graph.

Effect of \(q\)

\(q>0\): vertical shift upwards by \(q\) units

9bb513ac1b84cc784c06cfcd329cb5fa.png

\(q=0\): basic sine graph

\(q<0\): vertical shift downwards by \(q\) units

The effect of \(q\) on a sine graph.

Discovering the characteristics (EMA56)

Domain and range

For \(f(\theta) = a\sin\theta +q\), the domain is \(\left[0°;360°\right]\)

The range of \(f(\theta) = a\sin\theta +q\) depends on the values of \(a\) and \(q\).

For \(a>0\):

\begin{align*} -1& \le \sin \theta \le 1 \\ -a& \le a\sin\theta \le a \\ -a+q& \le a\sin\theta +q \le a+q \\ -a+q& \le f(\theta) \le a+q \end{align*}

For all values of \(\theta\), \(f(\theta)\) is always between \(-a+q\) and \(a+q\).

Therefore for \(a>0\), the range of \(f(\theta) = a\sin\theta +q\) is \(\left\{f(\theta):f(\theta)\in \left[-a+q, a+q\right]\right\}\)

Similarly, for \(a<0\), the range of \(f(\theta)=a \sin\theta +q\) is \(\left\{f(\theta):f(\theta)\in \left[a+q, -a+q\right]\right\}\)

Period

The period of \(y=a\sin\theta +q\) is \(360°\). This means that one sine wave is completed in \(360°\).

Intercepts

The \(y\)-intercept of \(f(\theta)=a\sin\theta +q\) is simply the value of \(f(\theta)\) at \(\theta = 0°\)

\begin{align*} y& = f(0°) \\ & = a\sin 0° + q \\ & = a(0) + q \\ & = q \end{align*}

This gives the point \((0;q)\)

Important: when sketching trigonometric graphs, always start with the basic graph and then consider the effects of \(a\) and \(q\).

Worked example 17: Sketching a sine graph

Sketch the graph of \(f(\theta)=2\sin\theta +3\) for \(\theta \in \left[0°;360°\right]\).

Examine the standard form of the equation

From the equation we see that \(a>1\) so the graph is stretched vertically. We also see that \(q>0\) so the graph is shifted vertically upwards by \(\text{3}\) units.

Substitute values for \(\theta\)

\(\theta\)

\(0°\)

\(30°\)

\(60°\)

\(90°\)

\(120°\)

\(150°\)

\(180°\)

\(210°\)

\(240°\)

\(270°\)

\(300°\)

\(330°\)

\(360°\)

\(f(\theta)\)

\(\text{3}\)

\(\text{4}\)

\(\text{4,73}\)

\(\text{5}\)

\(\text{4,73}\)

\(\text{4}\)

\(\text{3}\)

\(\text{2}\)

\(\text{1,27}\)

\(\text{1}\)

\(\text{1,27}\)

\(\text{2}\)

\(\text{3}\)

Plot the points and join with a smooth curve

c7d72293c6319a806d8a4ed60cfda9d5.png

Domain: \(\left[0°;360°\right]\)

Range: \(\left[1;5\right]\)

\(x\)-intercepts: none

\(y\)-intercepts: \((0°;3)\)

Maximum turning point: \((90°;5)\)

Minimum turning point: \((270°;1)\)

Cosine function (EMA57)

Functions of the form \(y=\cos\theta\) (EMA58)

Worked example 18: Plotting a cosine graph

\(y=f(\theta)=\cos\theta \quad \left[0°\le \theta \le 360°\right]\)

Use your calculator to complete the following table.

Choose an appropriate scale and plot the values of \(\theta\) on the \(x\)-axis and \(\cos\theta\) on the \(y\)-axis. Round your answers to \(\text{2}\) decimal places.

\(\theta\)

\(0°\)

\(30°\)

\(60°\)

\(90°\)

\(120°\)

\(150°\)

\(180°\)

\(210°\)

\(240°\)

\(270°\)

\(300°\)

\(330°\)

\(360°\)

\(\cos\theta\)

Substitute values for \(\theta\)

\(\theta\)

\(0°\)

\(30°\)

\(60°\)

\(90°\)

\(120°\)

\(150°\)

\(180°\)

\(210°\)

\(240°\)

\(270°\)

\(300°\)

\(330°\)

\(360°\)

\(\cos\theta\)

\(\text{1}\)

\(\text{0,87}\)

\(\text{0,5}\)

\(\text{0}\)

\(-\text{0,5}\)

\(-\text{0,87}\)

\(-\text{1}\)

\(-\text{0,87}\)

\(-\text{0,5}\)

\(\text{0}\)

\(\text{0,5}\)

\(\text{0,87}\)

\(\text{1}\)

Plot the points and join with a smooth curve

0ed89912afde93a21b42a332494a5c5d.png

Notice the similar wave shape of the graph. The period is also \(360°\) and the amplitude is \(\text{1}\). The maximum value of \(y=\cos\theta\) is \(\text{1}\) and the minimum value is \(-\text{1}\).

Domain: \(\left[0°;360°\right]\)

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

\(x\)-intercepts: \((90°;0)\), \((270°;0)\)

\(y\): \((0°;1)\)

Maximum turning points: \((0°;1)\), \((360°;1)\)

Minimum turning point: \((180°;-1)\)

Functions of the form \(y=a\cos\theta +q\) (EMA59)

The effects of \(a\) and \(q\) on a cosine graph

In the equation, \(y=a\cos\theta +q\), \(a\) and \(q\) are constants and have different effects on the graph.

On the same set of axes, plot the following graphs for \(0° \le \theta \le 360°\):

  1. \({y}_{1}=\cos\theta -2\)

  2. \({y}_{2}=\cos\theta -1\)

  3. \({y}_{3}=\cos\theta\)

  4. \({y}_{4}=\cos\theta +1\)

  5. \({y}_{5}=\cos\theta +2\)

Use your results to deduce the effect of \(q\).

On the same set of axes, plot the following graphs for \(0° \le \theta \le 360°\):

  1. \({y}_{6}=-2\cos\theta\)

  2. \({y}_{7}=-\cos\theta\)

  3. \({y}_{8}=\cos\theta\)

  4. \({y}_{9}=2\cos\theta\)

Use your results to deduce the effect of \(a\).

The effect of \(q\)

The effect of \(q\) is called a vertical shift because the whole cosine graph shifts up or down by \(q\) units.

  • For \(q>0\), the graph is shifted vertically upwards by \(q\) units.

  • For \(q<0\), the graph is shifted vertically downwards by \(q\) units.

The effect of \(a\)

The value of \(a\) affects the amplitude of the graph; the height of the peaks and the depth of the troughs.

  • For \(a>0\), there is a vertical stretch and the amplitude increases.

    For \(0<a<1\), the amplitude decreases.

  • For \(a<0\), there is a reflection about the \(x\)-axis.

    For \(-1<a<0\), there is a reflection about the \(x\)-axis and the amplitude decreases.

    For \(a<-1\), there is a reflection about the \(x\)-axis and the amplitude increases.

Note that amplitude is always positive.

Effect of \(a\)

\(a>1\): vertical stretch, amplitude increases

22cf1465b407ea16b3f471e1ce924582.png

\(a=1\): basic cosine graph

\(0<a<1\): amplitude decreases

\(-1<a<0\): reflection about \(x\)-axis, amplitude decreases

\(a<-1\): reflection about \(x\)-axis, amplitude increases

The effect of \(a\) on a cosine graph.

Effect of \(q\)

\(q>0\): vertical shift upwards by \(q\) units

b97fe321a3683bf4b2f432fe060f76e8.png

\(q=0\): basic cosine graph

\(q<0\): vertical shift downwards by \(q\) units

The effect of \(q\) on a cosine graph.

Discovering the characteristics (EMA5B)

Domain and range

For \(f(\theta)=a\cos\theta +q\), the domain is \(\left[0°;360°\right]\)

It is easy to see that the range of \(f(\theta)\) will be the same as the range of \(a\sin\theta +q\). This is because the maximum and minimum values of \(a\cos \theta + q\) will be the same as the maximum and minimum values of \(a\sin\theta +q\).

For \(a>0\) the range of \(f(\theta)=a\cos\theta +q\) is \(\left\{f(\theta):f(\theta)\in \left[-a+q;a+q\right]\right\}\)

For \(a<0\) the range of \(f(\theta)=a \cos\theta +q\) is \(\left\{f(\theta):f(\theta)\in \left[a+q;-a+q\right]\right\}\)

Period

The period of \(y=a\cos\theta +q\) is \(360°\). This means that one cosine wave is completed in \(360°\).

Intercepts

The \(y\)-intercept of \(f(\theta)=a\cos\theta +q\) is calculated in the same way as for sine.

\begin{align*} y& = f(0°) \\ & = a\cos 0°+q \\ & = a(1)+q \\ & = a+q \end{align*}

This gives the point \((0°;a+q)\).

Worked example 19: Sketching a cosine graph

Sketch the graph of \(f(\theta)=2\cos\theta +3\) for \(\theta \in \left[0°;360°\right]\).

Examine the standard form of the equation

From the equation we see that \(a>1\) so the graph is stretched vertically. We also see that \(q>0\) so the graph is shifted vertically upwards by \(\text{3}\) units.

Substitute values for \(\theta\)

\(\theta\)

\(0°\)

\(30°\)

\(60°\)

\(90°\)

\(120°\)

\(150°\)

\(180°\)

\(210°\)

\(240°\)

\(270°\)

\(300°\)

\(330°\)

\(360°\)

\(f(\theta)\)

\(\text{5}\)

\(\text{4,73}\)

\(\text{4}\)

\(\text{3}\)

\(\text{2}\)

\(\text{1,27}\)

\(\text{1}\)

\(\text{1,27}\)

\(\text{2}\)

\(\text{3}\)

\(\text{4}\)

\(\text{4,73}\)

\(\text{5}\)

Plot the points and join with a smooth curve

e27dc190f54bf29c89c821dbbfc2f238.png

Domain: \(\left[0°;360°\right]\)

Range: \(\left[1;5\right]\)

\(x\)-intercepts: none

\(y\)-intercept: \((0°;5)\)

Maximum turning points: \((0°;5)\), \((360°;5)\)

Minimum turning point: \((180°;1)\)

Comparison of graphs of \(y=\sin\theta\) and \(y=\cos\theta\) (EMA5C)

be9311a809cc01f6010f1453c10d64e0.png

Notice that the two graphs look very similar. Both waves move up and down along the \(x\)-axis. The distances between the peaks for each graph is the same. The height of the peaks and the depths of the troughs are also the same.

If you shift the whole cosine graph to the right by \(90°\) it will overlap perfectly with the sine graph. If you shift the sine graph \(90°\) to the left it would overlap perfectly with the cosine graph. This means that:

\[\begin{array}{ccl} \sin \theta & = \cos \left(\theta -90°\right) & \left(\text{shift the cosine graph to the right}\right) \\ \cos \theta & =\sin \left(\theta +90°\right) & \left(\text{shift the sine graph to the left}\right) \end{array}\]

Tangent function (EMA5D)

Functions of the form \(y=\tan\theta\) (EMA5F)

Worked example 20: Plotting a tangent graph

\(y=f(\theta)=\tan\theta \quad \left[0°\le \theta \le 360°\right]\)

Use your calculator to complete the following table.

Choose an appropriate scale and plot the values with \(\theta\) on the \(x\)-axis and \(\tan\theta\) on the \(y\)-axis. Round your answers to \(\text{2}\) decimal places.

\(\theta\)

\(0°\)

\(30°\)

\(45°\)

\(60°\)

\(90°\)

\(120°\)

\(135°\)

\(150°\)

\(180°\)

\(\tan\theta\)

\(\theta\)

\(210°\)

\(235°\)

\(240°\)

\(270°\)

\(300°\)

\(315°\)

\(330°\)

\(360°\)

\(\tan\theta\)

Substitute values for \(\theta\)

\(\theta\)

\(0°\)

\(30°\)

\(45°\)

\(60°\)

\(90°\)

\(120°\)

\(135°\)

\(150°\)

\(180°\)

\(\tan\theta\)

\(\text{0}\)

\(\text{0,58}\)

\(\text{1}\)

\(\text{1,73}\)

undef

\(-\text{1,73}\)

\(-\text{1}\)

\(-\text{0,58}\)

\(\text{0}\)

\(\theta\)

\(210°\)

\(235°\)

\(240°\)

\(270°\)

\(300°\)

\(315°\)

\(330°\)

\(360°\)

\(\tan\theta\)

\(\text{0,58}\)

\(\text{1}\)

\(\text{1,73}\)

undef

\(-\text{1,73}\)

\(-\text{1}\)

\(-\text{0,58}\)

\(\text{0}\)

Plot the points and join with a smooth curve

88c7bb650e28ba58734590b60129ba37.png

There is an easy way to visualise the tangent graph. Consider our definitions of \(\sin\theta\) and \(\cos\theta\) for right-angled triangles:

\begin{align*} \frac{\sin\theta}{\cos\theta} & = \frac{\left(\frac{\text{opposite}}{\text{hypotenuse}}\right)}{\left(\frac{\text{adjacent}}{\text{hypotenuse}}\right)} \\ & = \frac{\text{opposite}}{\text{hypotenuse}} \times \frac{\text{hypotenuse}}{\text{adjacent}} \\ & = \frac{\text{opposite}}{\text{adjacent}} \\ & = \tan\theta \end{align*}

So for any value of \(\theta\):

\(\tan{\theta} =\dfrac{\sin\theta }{\cos\theta }\)

So we know that for values of \(\theta\) for which \(\sin\theta = 0\), we must also have \(\tan\theta =0\). Also if \(\cos\theta =0\) the value of \(\tan\theta\) is undefined as we cannot divide by \(\text{0}\). The dashed vertical lines are at the values of \(\theta\) where \(\tan\theta\) is not defined and are called the asymptotes.

Asymptotes: the lines \(\theta =90°\) and \(\theta =270°\)

Period: \(180°\)

Domain: \(\left\{\theta :0° \le \theta \le 360°, \theta \ne 90°; 270°\right\}\)

Range: \(\left\{f\left(\theta \right):f\left(\theta \right)\in \mathbb{R}\right\}\)

\(x\)-intercepts: \(\left(0°;0\right)\), \(\left(180°;0\right)\), \(\left(360°;0\right)\)

\(y\)-intercept: \(\left(0°;0\right)\)

Functions of the form \(y=a\tan\theta + q\) (EMA5G)

The effects of \(a\) and \(q\) on a tangent graph

On the same set of axes, plot the following graphs for \(0° \le \theta \le 360°\):

  1. \({y}_{1}=\tan\theta -2\)

  2. \({y}_{2}=\tan\theta -1\)

  3. \({y}_{3}=\tan\theta\)

  4. \({y}_{4}=\tan\theta +1\)

  5. \({y}_{5}=\tan\theta +2\)

Use your results to deduce the effect of \(q\).

On the same set of axes, plot the following graphs for \(0° \le \theta \le 360°\):

  1. \({y}_{6}=-2\tan\theta\)

  2. \({y}_{7}=-\tan\theta\)

  3. \({y}_{8}=\tan\theta\)

  4. \({y}_{9}=2\tan\theta\)

Use your results to deduce the effect of \(a\).

The effect of \(q\)

The effect of \(q\) is called a vertical shift because the whole tangent graph shifts up or down by \(q\) units.

  • For \(q>0\), the graph is shifted vertically upwards by \(q\) units.

  • For \(q<0\), the graph is shifted vertically downwards by \(q\) units.

The effect of \(a\)

The value of \(a\) affects the steepness of each of the branches of the graph. The greater the value of \(a\), the quicker the branches of the graph approach the asymptotes.

\(a<0\)

\(a>0\)

\(q>0\)

c812156802dfa7f738e678e1b714c21c.png184730346a5fb1f3112637819fd27779.png

\(q=0\)

90d600dd16a70d252f054062f2109b20.png69a14975cc4481bd967091df11984ac8.png

\(q<0\)

e14640ad701433dc73ed7dd927f91b5e.png5658ab22986e45a0a54d6f8bc5237291.png

Table 6.4: The effect of \(a\) and \(q\) on a tangent graph.

Discovering the characteristics (EMA5H)

Domain and range

From the graph we see that \(\tan\theta\) is undefined at \(\theta =90°\) and \(\theta =270°\).

Therefore the domain is \(\left\{\theta :0°\le \theta \le 360°, \theta \ne 90°; 270°\right\}\).

The range is \(\left\{f(\theta):f(\theta)\in \mathbb{R}\right\}\).

Period

The period of \(y=a\tan\theta +q\) is \(180°\). This means that one tangent cycle is completed in \(180°\).

Intercepts

The \(y\)-intercept of \(f(\theta)=a\tan\theta +q\) is simply the value of \(f(\theta)\) at \(\theta =0°\).

\begin{align*} y& = f(0°) \\ & = a\tan 0°+q \\ & = a(0)+q \\ & = q \end{align*}

This gives the point \((0°;q)\).

Asymptotes

The graph has asymptotes at \(\theta = 90°\) and \(\theta =270°\).

Worked example 21: Sketching a tangent graph

Sketch the graph of \(y=2\tan\theta +1\) for \(\theta \in \left[0°;360°\right]\).

Examine the standard form of the equation

We see that \(a>1\) so the branches of the curve will be steeper. We also see that \(q>0\) so the graph is shifted vertically upwards by \(\text{1}\) unit.

Substitute values for \(\theta\)

\(\theta\)

\(0°\)

\(30°\)

\(60°\)

\(90°\)

\(120°\)

\(150°\)

\(180°\)

\(210°\)

\(240°\)

\(270°\)

\(300°\)

\(330°\)

\(360°\)

\(y\)

\(\text{1}\)

\(\text{2,15}\)

\(\text{4,46}\)

\(-\text{2,46}\)

\(-\text{0,15}\)

\(\text{1}\)

\(\text{2,15}\)

\(\text{4,46}\)

\(-\text{2,46}\)

\(-\text{0,15}\)

\(\text{1}\)

Plot the points and join with a smooth curve

acfee5c2d58221dee105f9c314c89cfd.png

Domain: \(\left\{\theta :0°\le \theta \le 360°,\theta \ne 90°;270°\right\}\)

Range: \(\left\{f(\theta):f(\theta)\in \mathbb{R}\right\}\)

Exercise 6.6

Shown the following graph of the following form: \(y = a \sin{\theta} + q\) where Point A is at \((180^{\circ};\text{1,5})\), and Point B is at \((90^{\circ};3)\), find the values of \(a\) and \(q\).

9145f9010adc40a34c15cc1d8be2aa16.png

To find \(q\) we note that \(q\) shifts the graph up or down. To determine \(q\) we can look at any point on the graph. For instance point \(A\) is at \((180^{\circ};\text{1,5})\). For an unshifted sine graph point \(A\) would be at \((180^{\circ};0)\). For this graph we see that this point has been shifted up by \(\text{1,5}\) or \(\frac{3}{2}\) spaces. Therefore \(q = \frac{3}{2}\).

To find \(a\) we note that the \(y\)-value at the middle (point \(A\)) is \(\text{1,5}\), while the \(y\)-value at the top (point \(B\)) is 3. We can find the amplitude by working out the distance from the top of the graph to the middle of the graph: \(3 - \text{1,5} = \text{1,5}\). Therefore \(a = \frac{3}{2}\).

The complete equation for the graph shown in this question is \(y = \frac{3}{2}\sin \theta + \frac{3}{2}\).

Therefore \(a = \frac{3}{2} \text{ and } q = \frac{3}{2}\)

Shown the following graph of the following form: \(y = a \sin{\theta} + q\) where Point A is at \((270^{\circ};-\text{6})\), and Point B is at \((90^{\circ};\text{2})\), determine the values of \(a\) and \(q\).

8526cb4a6491bce5d14fbf8f056a52bd.png

To find \(a\) we note that the \(y\)-value at the bottom (point \(A\)) is \(-\text{6}\), while the \(y\)-value at the top (point \(B\)) is 2. We can find the amplitude by working out the distance from the top of the graph to the bottom of the graph and then dividing this by 2 since this distance is twice the amplitude: \(\frac{2 - (-6)}{2} = 4\). Therefore \(a = 4\).

To find \(q\) we note that \(q\) shifts the graph up or down. To determine \(q\) we can look at any point on the graph. For instance point \(B\) is at \((90^{\circ};2)\). For an unshifted sine graph with the same \(a\) value (i.e. \(4 \sin \theta\)) point \(B\) would be at \((90^{\circ};4)\). For this graph we see that this point has been shifted down by \(\text{2}\) spaces. Therefore \(q = 2\).

The complete equation for the graph shown in this question is \(y = 4\sin \theta - 2\).

Therefore \(a = 4 \text{ and } q = -2\)

The graph below shows a trigonometric equation of the following form: \(y = a \cos{\theta} + q\). Two points are shown on the graph: Point A at \((180^{\circ};-\text{1,5})\), and Point B: \((0^{\circ};-\text{0,5})\). Calculate the values of \(a\) (the amplitude of the graph) and \(q\) (the vertical shift of the graph).

556979b9616418b9c2e1f197e3e38994.png

To find \(a\) we note that the \(y\)-value at the bottom (point \(A\)) is \(-\text{1,5}\), while the \(y\)-value at the top (point \(B\)) is \(-\text{0,5}\). We can find the amplitude by working out the distance from the top of the graph to the bottom of the graph and then dividing this by 2 since this distance is twice the amplitude: \(\frac{-\text{0,5} - (-\text{1,5})}{2} = \frac{1}{2}\). Therefore \(a = \frac{1}{2}\).

To find \(q\) we note that \(q\) shifts the graph up or down. To determine \(q\) we can look at any point on the graph. For instance point \(B\) is at \((0^{\circ};-\text{0,5})\). For an unshifted cosine graph with the same \(a\) value (i.e. \(\frac{1}{2} \cos \theta\)) point \(B\) would be at \((0^{\circ};\text{0,5})\). For this graph we see that this point has been shifted down by \(\text{1}\) space. Therefore \(q = 1\).

The complete equation for the graph shown in this question is \(y = \frac{1}{2}\cos \theta - 1\).

Therefore \(a = \frac{1}{2}, \text{ and } q= -1\).

The graph below shows a trigonometric equation of the following form: \(y = a \cos{\theta} + q\). Two points are shown on the graph: Point A at \((90^{\circ};\text{0,0})\), and Point B: \((180^{\circ};-\text{0,5})\). Calculate the values of \(a\) (the amplitude of the graph) and \(q\) (the vertical shift of the graph).

a7764f4e7f7fa1390336f89afe30d242.png

To find \(a\) we note that the \(y\)-value at the bottom (point \(B\)) is \(-\text{0,5}\), while the \(y\)-value at the middle (point \(A\)) is \(\text{0}\). We can find the amplitude by working out the distance from the top of the graph to the middle of the graph: \(\text{0} - (-\text{0,5}) = \frac{1}{2}\). Therefore \(a = \frac{1}{2}\).

To find \(q\) we note that \(q\) shifts the graph up or down. To determine \(q\) we can look at any point on the graph. For instance point \(A\) is at \((90^{\circ};\text{0})\). For an unshifted cosine graph with the same \(a\) value (i.e. \(\frac{1}{2} \cos \theta\)) point \(B\) would be at \((0^{\circ};\text{0})\). For this graph we see that this point has not been shifted. Therefore \(q = 0\).

The complete equation for the graph shown in this question is \(y = \frac{1}{2}\cos \theta\).

Therefore \(a = \frac{1}{2}, \text{ and } q= 0\).

On the graph below you see a tangent curve of the following form: \(y = a \tan{\theta} + q\). Two points are labelled on the curve: Point A is at \(\left(0^{\circ}; \frac{1}{3} \right)\), and Point B is at \(\left(45^{\circ}; \frac{10}{3} \right)\).

Calculate, or otherwise determine, the values of \(a\) and \(q\).

8af9c2c31a5212730202b1275ef0b07c.png

To find \(q\) we note that \(q\) shifts the graph up or down. To determine \(q\) we can look at any point on the graph. For instance point \(A\) is at \(\left(0^{\circ};\frac{1}{3}\right)\). For an unshifted tangent graph point \(A\) would be at \((0^{\circ};\text{0})\). For this graph we see that this point been shifted upwards by a \(\frac{1}{3}\). Therefore \(q = \frac{1}{3}\).

To find \(a\) we can substitute point \(B\) into the equation for the tangent graph:

\begin{align*} y & = a \tan{\theta} + \frac{1}{3} \\ \left( \frac{10}{3} \right) & = a \tan{45^{\circ}} + \frac{1}{3} \\ \frac{10}{3} & = a(1) + \frac{1}{3} \\ \frac{10}{3} - \frac{1}{3} & = a \\ 3 & = a \end{align*}

The complete equation is: \(y = 3 \tan{\theta} + \frac{1}{3}\).

Therefore \(a = 3 \text{ and } q = \frac{1}{3}\).

The graph below shows a tangent curve with an equation of the form \(y = a \tan{\theta} + q\). Two points are labelled on the curve: Point A is at \((0^{\circ} ; 0)\), and Point B is at \((45^{\circ} ; 1)\).

Find \(a\) and \(q\).

f0193e5d92c3b2c9fb510c1f44119fe3.png

To find \(q\) we note that \(q\) shifts the graph up or down. To determine \(q\) we can look at any point on the graph. For instance point \(A\) is at \((0^{\circ};0)\). For an unshifted tangent graph point \(A\) would be at \((0^{\circ};\text{0})\). For this graph we see that the graph has not been shifted. Therefore \(q = 0\).

To find \(a\) we can substitute point \(B\) into the equation for the tangent graph:

\begin{align*} y & = a \tan{\theta} \\ 1 & = a \tan{45^{\circ}} \\ 1 & = a(1) \\ 1 & = a \end{align*}

The complete equation is: \(y = \tan{\theta}\).

Therefore \(a = 1 \text{ and } q = 0\).

Given the following graph, identify a function that matches each of the following equations:

9058a885a54734c1d762929cfe5dc1ba.png

\(y = \sin{\theta}\)

\(h(x)\)

\(y = \frac{1}{2}\sin{\theta}\)

\(k(x)\)

\(y = 3\sin{\theta}\)

\(f(x)\)

\(y = 2\sin{\theta}\)

\(g(x)\)

The graph below shows functions \(f(x)\) and \(g(x)\)

ca6ff29c90e3b12ca43b8db1365e3f05.png

What is the equation for \(g(x)\)?

\(g(x) = 4\sin{\theta}\)

With the assistance of the table below sketch the three functions on the same set of axes.

\[\begin{array}{| l | l | l | l | l | l | l | l | l | l | } \hline \theta & 0^{\circ} & 45^{\circ} & 90^{\circ} & 135^{\circ} & 180^{\circ} & 225^{\circ} & 270^{\circ} & 315^{\circ} & 360^{\circ} \\ \hline \tan{\theta} & 0 & 1 & \text{undefined} & -1 & 0 & 1 & \text{undefined} & -1 & 0 \\ \hline 3\tan{\theta} & 0 & 3 & \text{undefined} & -3 & 0 & 3 & \text{undefined} & -3 & 0 \\ \hline \frac{1}{2}\tan{\theta} & 0 & \frac{1}{2} & \text{undefined} & -\frac{1}{2} & 0 & \frac{1}{2} & \text{undefined} & -\frac{1}{2} & 0 \\ \hline \end{array}\]

We are given a table with values and so we plot each of these points and join them with a smooth curve.

50a24e97b8f442206103b2a0afb19948.png

With the assistance of the table below sketch the three functions on the same set of axes.

\[\begin{array}{| l | l | l | l | l | l | } \hline \theta & 0^{\circ} & 90^{\circ} & 180^{\circ} & 270^{\circ} & 360^{\circ} \\ \hline \cos{\theta} - 2 & -1 & -2 & -3 &-2 & -1 \\ \hline \cos{\theta} + 4 & 5 & 4 & 2 & 4 & 5 \\ \hline \cos{\theta} + 2 & 3 & 2 & 1 & 2 & 3 \\ \hline \end{array}\]

We are given a table with values and so we plot each of these points and join them with a smooth curve.

18dbb323af62bc96caba7bf2dfb3f08f.png

State the coordinates at \(E\) and the range of the function.

c4149d04d46c1076130ed3da6dd2f3b6.png

To find the coordinates of \(E\) we read the value off the graph. To find the range we note that this is a cosine graph and so the maximum value occurs at \(0^\circ\) (and at \(360^\circ\)). The minimum value occurs at \(180^\circ\). So we read off the value of \(y\) at \(0^\circ\) and at \(180^\circ\).

Therefore \(E(180^{\circ};-2) \text{ and } -2 \le y \le 0\).

124a4140037c446904c396bd728c5571.png

To find the coordinates of \(E\) we read the value off the graph. To find the range we note that this is a sine graph and so the maximum value occurs at \(90^\circ\). The minimum value occurs at \(270^\circ\). So we read off the value of \(y\) at \(90^\circ\) and at \(270^\circ\).

Therefore \(E(360^{\circ};2) \text{ and } 0 \le y \le 4\)

973ef030b54adfd0cd13baf5d859f02f.png

To find the coordinates of \(E\) we read the value off the graph. To find the range we note that this is a sine graph and so the maximum value occurs at \(90^\circ\). The minimum value occurs at \(270^\circ\). So we read off the value of \(y\) at \(90^\circ\) and at \(270^\circ\).

Therefore \(E(90^{\circ};-\text{0,5}) \text{ and } -\text{2,5} \le y \le -\text{0,5}\)

251450c5f31d050bd8c7dfea167e6ad1.png

To find the coordinates of \(E\) we read the value off the graph. To find the range we note that this is a cosine graph and so the maximum value occurs at \(0^\circ\) (and at \(360^\circ\)). The minimum value occurs at \(180^\circ\). So we read off the value of \(y\) at \(0^\circ\) and at \(180^\circ\).

Therefore \(E(180^{\circ};\text{0,5}) \text{ and } \text{0,5} \le y \le \text{4,5}\)

State the coordinates at \(E\) and the domain and range of the function in the interval shown.

283d604ab4287d2759100a0e5f23fd80.png
\(E(180^{\circ};1) \text{ , range } y \in \mathbb{R} \text{ and domain } 0 \le \theta \le 360, x \neq 90, x\neq 270\)

Using your knowledge of the effects of \(a\) and \(q\), sketch each of the following graphs, without using a table of values, for \(\theta \in \left[0°;360°\right]\)

\(y=2\sin\theta\)

In this case \(q = 0\) and so the basic sine graph is not shifted up or downwards. We also note that \(a = 2\) and so the graph is stretched by 2 units. The maximum value will be 2 and the minimum value will be \(-\text{2}\).

62389f7511353bc68c36ebe67d92366e.png

\(y=-4\cos\theta\)

In this case \(q = 0\) and so the basic cosine graph is not shifted up or downwards. We also note that \(a = -4\) and so the graph is stretched by \(-\text{4}\) units. The maximum value will be 4 and the minimum value will be \(-\text{4}\).

9b3987c50b1b68db011a760bd1faa3d3.png

\(y=-2\cos\theta + 1\)

In this case \(q = 1\) and so the basic cosine graph is shifted upwards by 1 unit. We also note that \(a = -2\) and so the graph is stretched by \(-\text{2}\) units. The maximum value will be 3 and the minimum value will be \(-\text{1}\).

e63aa8891a31dd7c5ccddcb5875af775.png

\(y=\sin\theta - 3\)

In this case \(q = -3\) and so the basic sine graph is shifted downwards by 3 units. We also note that \(a = 1\) and so the graph is not stretched. The maximum value will be \(-\text{2}\) and the minimum value will be \(-\text{4}\).

7a7f680871f136f54b4aba9a35c81d5d.png

\(y=\tan\theta - 2\)

In this case \(q = -2\) and so the basic tangent graph is shifted downwards by 2 units. We also note that \(a = 1\) and so the graph is not stretched. When \(\theta = 0^\circ\), \(y = -2\). Similarly when \(\theta = 180^\circ\), \(y = -2\) and when \(\theta = 360^\circ\), \(y = -2\).

8597ef12d438c9a56c5592eeffe62b1d.png

\(y=2\cos\theta - 1\)

In this case \(q = -1\) and so the basic cosine graph is shifted downwards by 1 units. We also note that \(a = 2\) and so the graph is stretched by 2 units. The maximum value will be \(\text{1}\) and the minimum value will be \(-\text{3}\).

5bddbae542b75ad88a616b5252d3df93.png

Give the equations for each of the following graphs:

baebea8055587f531fc92764c4c8fa5b.png

The general form of a cosine graph is \(y = a \cos \theta + q\). We note that in this case the graph is not shifted. We also note the graph is stretched by \(-\text{2}\) units.

Therefore \(y=-2\cos\theta\).

a64735ef115e7b510a7b3db9517e765a.png

The general form of a sine graph is \(y = a \sin \theta + q\). We note that in this case the graph is shifted upwards by 1 unit. We also note the graph is not stretched.

Therefore \(y=\sin\theta + 1\).

For which values of \(\theta\) is the function increasing, in the interval shown?

023e00229365890a21fa5f4104fbf933.png
\(90^{\circ} < \theta < 270^{\circ}\)

For which values of \(\theta\) is the function negative, in the interval shown?

4855411dac53c1533265ff0b04ecefa1.png
\(0^{\circ} < \theta < 210^{\circ} \text { and } 330^{\circ} < \theta < 360^{\circ}\)

For which values of \(\theta\) is the function positive, in the interval shown?

4b4c1e1eb693406b36cdbd1f981c5cb9.png
\(60^{\circ} < \theta < 300^{\circ}\)

Given the following graph.

e2cb6653035e2d63c5cc795f26622d6e.png

State the coordinates at \(A,~B,~C\) and \(D\).

We can read the values off the graph:

\(A = (90^{\circ};4),~B = (90^{\circ};-2),~C = (180 ^{\circ}; 4) \text{ and } D = ( 180^{\circ}; -2)\)

How many times in this interval does \(f(x)\) intersect \(g(x)\).

\(2\)

What is the amplitude of \(f(x)\).

\(2\)

Evaluate: \(f(360^{\circ}) - g(360^{\circ})\) .

Read off the value of \(f(360^{\circ})\) and \(g(360^{\circ})\) from the graph. Then subtract \(g(360^{\circ})\) from \(f(360^{\circ})\).

\begin{align*} f(360^{\circ}) - g(360^{\circ}) &= 2 - 0 \\ &=2 \end{align*}

Given the following graph.

3c2e411abd0e5be8d3d64acbc87c606b.png

State the coordinates at \(A,~B,~C\) and \(D\).

We read the values off the graph:

\(A = ( 90^{\circ}; 1),~B = ( 180^{\circ}; -3),~C = (270 ^{\circ};-1 ) \text{ and } D = ( 360^{\circ}; 3)\)

How many times in this interval does \(f(x)\) intersect \(g(x)\).

\(2\)

What is the amplitude of \(g(x)\).

\(3\)

Evaluate: \(f(90^{\circ}) - g(90^{\circ})\) .

Read off the value of \(f(90^{\circ})\) and \(g(90^{\circ})\) from the graph. Then subtract \(g(90^{\circ})\) from \(f(90^{\circ})\).

\begin{align*} f(90^{\circ}) - g(90^{\circ}) &= 1 - 0 \\ &=1 \end{align*}

Given the following graph:

2bdd2f9beb5791e3f885a23cf372c28f.png

State the coordinates at \(A,~B,~C\) and \(D\).

We read the values off the graph:

\(A = (90 ^{\circ};3 ),~B = (90 ^{\circ};2 ),~C = (180 ^{\circ}; -4) \text{ and } D = ( 270^{\circ};2)\)

How many times in this interval does \(f(x)\) intersect \(g(x)\).

\(3\)

What is the amplitude of \(g(x)\).

\(2\)

Evaluate: \(f(270^{\circ}) - g(270^{\circ})\) .

Read off the value of \(f(270^{\circ})\) and \(g(270^{\circ})\) from the graph. Then subtract \(g(270^{\circ})\) from \(f(270^{\circ})\).

\begin{align*} f(270^{\circ}) - g(270^{\circ}) &= -3 - (-2)\\ &=-1 \end{align*}