1.5.1 Trigonometric Functions and their Transformations

In this topic we will learn how to:

• sketch and use the graphs of sine, cosine and tangent functions
• understand and use the transformations of the graphs of $y = f(x)$ given by $y = f(x) + a$, $y = f(x + a)$, $y = af(x)$ and $y = f(ax)$ and simple combinations of these for trigonometric functions

$$\textbf{\textcolor{gray}{Sine Function}}$$The sine function is typically denoted as,
$$y = \sin{x}$$

It has a period of $2\pi$ radians i.e it repeats every $2\pi$ radians.
$$\textbf{\textcolor{gray}{Cosine Function}}$$The cosine function is typically denoted as,
$$y = \cos{x}$$

It has a period of $2\pi$ radians i.e it repeats every $2\pi$ radians.
$$\textbf{\textcolor{gray}{Tangent Function}}$$The tangent function is typically denoted as,
$$y = \tan{x}$$

It has a period of $\pi$ radians.

Note: The red lines are asymptotes. An asymptote is a line that the function approaches but never reaches.$$\textbf{\large\textcolor{gray}{Transformations}}$$

$$\textcolor{gray}{y = \sin{x} + a}$$This is a translation in the $y$ direction by $a$ units.
Let’s look at an example of this transformation.

Example 1
Given that $f(x) = \sin x\$ for $\ 0 < x < 2\pi$ sketch the graph of $g(x) = f(x) + 1\$ for $\ 0 < x < 2\pi$.

$$\textcolor{gray}{y = \sin(x + a)}$$This is a translation in the $x$ direction by $-a$ units.
Let’s look at an example of this transformation.

Example 2
Given that $f(x) = \sin x\$ for $\ -\pi < x < \pi$ sketch the graph of $g(x) = f\left(x + \frac{\pi}{2}\right)\$ for $\ -\pi < x < \pi$.

$$\textcolor{gray}{y = a\sin{x}}$$This is a stretch in the $y$ direction by a stretch factor of $a$.
Let’s look at an example of this transformation.

Example 3
Given that $f(x) = \sin x\$ for $\ 0 < x < 2\pi$ sketch the graph of $g(x) = 2f(x)\$ for $\ 0 < x < 2\pi$.

$$\textcolor{gray}{y = \sin(ax)}$$This is a stretch in the $x$ direction by a stretch factor of $\frac{1}{a}$. It represents the number of periods of the graph.
Let’s look at an example of this transformation.

Example 4
Given that $f(x) = \sin x\$ for $\ 0 < x < 2\pi$ sketch the graph of $g(x) = f(2x)\$ for $\ 0 < x < 2\pi$.

Note: Notice how there are two periods of $\ g(x)$ in the interval $\ 0 < x < 2\pi$.

$$\textcolor{gray}{y = a\sin(bx + c) + d}$$The above is known as a combined transformation. $a$ represents a stretch in the $y$ direction. $b$ represents a stretch in the $x$ direction. $c$ represents a translation in the $x$ direction. $d$ represents a translation in the $y$ direction.
Let’s look at an example of a combined transformation.

Example 5
Given that $f(x) = \sin x\$ for $\ 0 < x < \frac{5}{2}\pi$ sketch the graph of $g(x) = 3f\left(2x - \frac{\pi}{4}\right) + 1$ for $0 < x < \frac{9}{4}\pi$.

Start by sketching the graph of $f(x)$,

$$g(x) = 3f\left(2x - \textcolor{#2192ff}{\frac{\pi}{4}}\right) + 1$$From $f(x)$ to $g(x)$ there is a translation in the $x$ direction by $\textcolor{#2192ff}{\frac{\pi}{4}}$, then there is a translation in the $y$ direction by $1$ unit,

$$g(x) = 3f\left(2x - \frac{\pi}{4}\right) + 1$$Finally there is a stretch in the $x$ direction by a stretch factor of $\frac{1}{2}$, then a stretch in the $y$ direction by a stretch factor of $3$,

Therefore, the graph of $g(x)$,

Note: The transformations for $y = \cos{x}$ and $y = \tan{x}$ work exactly the same as those for $y = \sin{x}$.
$$\textbf{\textcolor{gray}{Finding the range of a trig function}}$$To find the range of a trig function, we have to find the minimum and maximum points. To find the minimum point of a trig function substitute the part in the equation containing the trig function with $-1$.
Let’s look at an example.

Example 6
Find the minimum value of the function $y = 2\sin{x} + 3$.

Replace $\sin{x}$ with $-1$,
$$y = 2(-1) + 3$$$$y = 1$$Therefore, the minimum point of $y = 2\sin{x} + 3$ is $1$

To find the maximum point, substitute the part in the equation containing the trig function with $1$.
Let’s look at an example.

Example 7
Find the maximum value of the function $y = 3\sin{2x} - 1$.

Replace $\sin{2x}$ with $1$,
$$y = 3(1) - 1$$$$y = 2$$Therefore, the maximum point of $y = 3\sin{2x} - 1$ is $2$

Note: If the trig function is negative (see past paper question number 1), then replacing the trig function with $-1$ will give you the maximum point and $1$ will give you the minimum point.

Let’s look at some past paper questions.

1. The function $f$ is defined by $f(x) = 2 - 3\cos{x}$ for $0 \le x \le 2\pi$. (9709/11/M/J/19 number 9)

(a) State the range of $f$.
$$f(x) = 2 - 3\cos{x}$$To get the range of $f$, we have to find both the maximum and minimum point of $f$. To find the maximum point, replace $\cos{x}$ with $-1$,
$$f(x) = 2 - 3(-1)$$$$f(x) = 5$$Therefore, the maximum point of $f$ is,
$$5$$To find the minimum point, replace $\cos{x}$ with $1$,
$$f(x) = 2 - 3(1)$$$$f(x) = -1$$Therefore, the minimum point of $f$ is,
$$-1$$Therefore, the range of $f$ is,
$$-1 \le f(x) \le 5$$(b) Sketch the graph of $y = f(x)$.

Start by sketching the graph of $y = \cos x\$ for $\ 0 \le x \le 2\pi$. Let’s call this function $g(x)$,

The negative sign on $3\cos x$ means the function has been reflected in the $x$-axis,

Then there is a translation by $2$ units in the $y$ direction,

Finally there is a stretch in the $y$ direction by a stretch factor of $\ 3$,

Therefore, the graph of $y = f(x)$ is,

Note: Alternatively, you can sketch the graph by first creating a table of values. It is advisable to get comfortable with transformations, and with time you should be able to sketch a combined transformation in one step.

2. The function $f$ is defined by $f(x) = \frac{3}{2}\cos{2x} + \frac{1}{2}$ for $0 \le x \le \pi$. (9709/11/M/J/20 number 4)

(a) State the range of $f$.
$$f(x) = \frac{3}{2}\cos{2x} + \frac{1}{2}$$To get the range of $f$, we have to find both the maximum and minimum point of $f$. To find the maximum point, replace $\cos{2x}$ with $1$,
$$f(x) = \frac{3}{2}(1) + \frac{1}{2}$$$$f(x) = 2$$Therefore, the maximum point of $f$ is,
$$2$$To find the minimum point, replace $\cos{2x}$ with $-1$,
$$f(x) = \frac{3}{2}(-1) + \frac{1}{2}$$$$f(x) = -1$$Therefore, the minimum point of $f$ is,
$$-1$$Therefore, the range of $f$ is,
$$-1 \le f(x) \le 2$$A function $g$ is such that $g(x) = f(x) + k$, where $k$ is a positive constant. The x-axis is a tangent to the curve $y = g(x)$.
(b) State the value of $k$ and hence describe fully the transformation that maps the curve $y = f(x)$ on to $y = g(x)$.

For the x-axis to be a tangent to the curve the minimum point has to be $0$. To make the minimum point $0$, we have to add $1$ to our current minimum point $-1$. Therefore,
$$k = 1$$The transformation is,
$$\textmd{Translation in the y-direction by }1\textmd{ unit}$$Therefore, the final answer is,
$$k = 1$$$$\textmd{Translation in the y-direction by }1\textmd{ unit}$$(c) State the equation of the curve which is the reflection of $y = f(x)$ in the x-axis. Give your answer in the form $y = a\cos{2x} + b$ where $a$ and $b$ are constants.
$$f(x) = \frac{3}{2}\cos{2x} + \frac{1}{2}$$If we reflect the function in the $x$-axis the function becomes negative i.e f(x) becomes -f(x), therefore, we multiply the function by $-1$,
$$f(x) = -\left(\frac{3}{2}\cos{2x} + \frac{1}{2}\right)$$$$f(x) = -\frac{3}{2}\cos{2x} - \frac{1}{2}$$Therefore, the final answer is,
$$f(x) = -\frac{3}{2}\cos{2x} - \frac{1}{2}$$3. A curve has equation $y = 3\cos{2x} + 2$ for $0 \le x \le \pi$. (9709/12/O/N/20 number 11)

(a) State the greatest and least values of $y$.
$$y = 3\cos{2x} + 2$$To find the greatest value of $y$, replace $\cos{2x}$ with $1$,
$$y = 3(1) + 2$$$$y = 5$$To find the least value of $y$, replace $\cos{2x}$ with $-1$,
$$y = 3(-1) + 2$$$$y = -1$$Therefore,
$$\textmd{The greatest and least values of }y\textmd{ are }5\textmd{ and }-1\textmd{, respectively.}$$(b) Sketch the graph of $y = 3\cos{2x} + 2$ for $0 \le x \le \pi$.
$$y = \textcolor{red}{3}\cos{\textcolor{#2192ff}{2}x} + \textcolor{#0f0}{2}$$Start by sketching the graph of $y = \cos x\$ for $\ 0 \le x \le \pi$. We will call this function $g(x)$,

From $g(x)$, there is a stretch in the $x$ direction by a stretch factor $\frac{1}{2}$,

Then there is a translation in the $y$ direction by $2$ units,

Finally, there is a stretch in the $y$ direction by a stretch factor of $3$,

Therefore, the graph of $y = 3\cos{2x} + 2$ is,

(c) By considering the straight line $y = kx$, where $k$ is a constant, state the number of solutions of the equation $3\cos{2x} + 2 = kx\$ for $\ 0 \le x \le \pi$ in each of the following cases.

(i) $k = -3$
The line $y = -3x$ does not intersect the graph of $y = 3\cos{2x} + 2$, therefore,
There are no solutions.

(ii) $k = 1$
The line $y = -3x$ intersects the graph of $y = 3\cos{2x} + 2$ twice, therefore,
There are two solutions.

(iii) $k = 3$
The line $y = -3x$ intersects the graph of $y = 3\cos{2x} + 2$ once, therefore,
There is one solution.

Note: If this is hard to visualize, sketch the graphs of the lines on the graph of the curve to see how many times they intersect.

Functions $f$, $g$ and $h$ are defined for $x \in \mathbb{R}$ by

$f(x) = 3\cos{2x} + 2$
$g(x) = f(2x) + 4$
$h(x) = 2f\left(x + \frac{1}{2}\pi\right)$

(d) Describe fully a sequence of transformations that maps the graph of $y = f(x)$ on to $y = g(x)$.
$$g(x) = f(\textcolor{#2192ff}{2}x) + \textcolor{#0f0}{4}$$
Stretch in the $x$ direction by a stretch factor of $\frac{1}{2}$. Followed by a translation in the $y$ direction by $4$ units.

(e) Describe fully a sequence of transformations that maps the graph of $y = f(x)$ on to $y = h(x)$.
$$h(x) = \textcolor{#0f0}{2}f\left(x + \textcolor{#2192ff}{\frac{1}{2}\pi}\right)$$
Translation in the $x$ direction by $-\frac{1}{2}\pi$ units. Followed by a stretch in the $y$ direction by stretch factor $2$.