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}

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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}

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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}

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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.

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\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.

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\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.

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\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.

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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),

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g(x) = 3f\left(2x - \textcolor{#2192ff}{\frac{\pi}{4}}\right) + 1From 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,

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g(x) = 3f\left(2x - \frac{\pi}{4}\right) + 1Finally 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,

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Therefore, the graph of g(x),

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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) + 3y = 1Therefore, 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) - 1y = 2Therefore, 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) = 5Therefore, the maximum point of f is,
5To find the minimum point, replace \cos{x} with 1,
f(x) = 2 - 3(1)f(x) = -1Therefore, the minimum point of f is,
-1Therefore, 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),

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

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

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

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

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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) = 2Therefore, the maximum point of f is,
2To find the minimum point, replace \cos{2x} with -1,
f(x) = \frac{3}{2}(-1) + \frac{1}{2}f(x) = -1Therefore, the minimum point of f is,
-1Therefore, the range of f is,
-1 \le f(x) \le 2A 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 = 1The 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} + 2To find the greatest value of y, replace \cos{2x} with 1,
y = 3(1) + 2y = 5To find the least value of y, replace \cos{2x} with -1,
y = 3(-1) + 2y = -1Therefore,
\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),

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

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

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

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

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(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.