7. Trigonometry

iGCSE (2021 Edition)

Lesson

In this lesson, we will look at an important characteristic of a trigonometric function called it's period. We will also look at horizontal dilations and translations of the basic trigonometric functions. And finally, we will consider trigonometric functions with a combination of different dilations and translations.

The period of a trigonometric function is defined as the time taken to complete one full cycle. For $y=\sin x$`y`=`s``i``n``x` and $y=\cos x$`y`=`c``o``s``x`, the period can be measured between successive maximums or minimums, or any other repeated point on the graph with the same function value and characteristics. Analysis of the trigonometric functions $y=\sin x$`y`=`s``i``n``x` and $y=\cos x$`y`=`c``o``s``x` reveals a period of $2\pi$2π.

The tangent function $y=\tan x$`y`=`t``a``n``x` is different to the sine and cosine functions as it doesn't have defined maximum and minimum values. However, if we look at the distance along the horizontal axis between repeated points or asymptotes we can see that it has a period of $\pi$π.

A function can be dilated horizontally by multiplying the input by a scale factor. That is, a function $f\left(ax\right)$`f`(`a``x`) is the function $f\left(x\right)$`f`(`x`) horizontally dilated by a factor $\frac{1}{a}$1`a`.

For trigonometric functions that are horizontally dilated, there will be a change in the period.

Let's consider the graph of $y=\sin2x$`y`=`s``i``n`2`x` sketched together with $y=\sin x$`y`=`s``i``n``x`:

From the graph, we can see that the function of $y=\sin2x$`y`=`s``i``n`2`x` has a period of $\pi$π which is half the period of the original graph of $y=\sin x$`y`=`s``i``n``x`. This dilation has compressed the graph horizontally. It is important to notice that by changing the period of the function the amplitude, domain and range don't change.

Now let's consider the graph of $y=\cos\left(\frac{x}{2}\right)$`y`=`c``o``s`(`x`2) sketched together with $y=\cos x$`y`=`c``o``s``x`:

From the graph, we can see that the function of $y=\cos\left(\frac{x}{2}\right)$`y`=`c``o``s`(`x`2) has a period of $4\pi$4π which is double the period of the original graph of $y=\cos x$`y`=`c``o``s``x`. This dilation has stretched the graph horizontally. It is important to notice that by changing the period of the function the amplitude, domain and range don't change.

The tangent function behaves in a similar way. Let's consider the graph of $y=\tan\left(-\frac{x}{3}\right)$`y`=`t``a``n`(−`x`3) to see how it is dilated:

From the graph, we can see that the function of $y=\tan\left(-\frac{x}{3}\right)$`y`=`t``a``n`(−`x`3) has a period of $3\pi$3π which is triple the period of the original graph of $y=\tan x$`y`=`t``a``n``x`. This dilation has stretched the graph horizontally. The negative sign has reflected the graph about the vertical axis. Note that the range has not changed, but the domain has due to the changes in asymptote positions.

We can summarise the effects of horizontal dilations on trigonometric graphs as follows:

Function | Dilation | Period |
---|---|---|

$y=\sin\left(ax\right)$y=sin(ax) |
horizontal compression $|a|>1$| horizontal stretch $0<|a|<1$0<| vertical reflection $a<0$ |
$\frac{2\pi}{a}$2πa |

$y=\cos\left(ax\right)$y=cos(ax) |
horizontal compression $|a|>1$| horizontal stretch $0<|a|<1$0<| vertical reflection $a<0$ |
$\frac{2\pi}{a}$2πa |

$y=\tan\left(ax\right)$y=tan(ax) |
horizontal compression $|a|>1$| horizontal stretch $0<|a|<1$0<| vertical reflection $a<0$ |
$\frac{\pi}{a}$πa |

Consider the functions $f\left(x\right)=\cos x$`f`(`x`)=`c``o``s``x` and $g\left(x\right)=\cos4x$`g`(`x`)=`c``o``s`4`x`.

State the period of $f\left(x\right)$

`f`(`x`) in radians.Complete the table of values for $g\left(x\right)$

`g`(`x`).$x$ `x`$0$0 $\frac{\pi}{8}$π8 $\frac{\pi}{4}$π4 $\frac{3\pi}{8}$3π8 $\frac{\pi}{2}$π2 $\frac{5\pi}{8}$5π8 $\frac{3\pi}{4}$3π4 $\frac{7\pi}{8}$7π8 $\pi$π $g\left(x\right)$ `g`(`x`)$\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ $\editable{}$ State the period of $g\left(x\right)$

`g`(`x`) in radians.What transformation of the graph of $f\left(x\right)$

`f`(`x`) results in the graph of $g\left(x\right)$`g`(`x`)?Vertical dilation by a factor of $\frac{1}{4}$14

AHorizontal dilation by a factor of $4$4

BHorizontal dilation by a factor of $\frac{1}{4}$14

CVertical dilation by a factor of $4$4

DVertical dilation by a factor of $\frac{1}{4}$14

AHorizontal dilation by a factor of $4$4

BHorizontal dilation by a factor of $\frac{1}{4}$14

CVertical dilation by a factor of $4$4

DThe graph of $f\left(x\right)$

`f`(`x`) has been provided below.By moving the points, graph $g\left(x\right)$

`g`(`x`).Loading Graph...

Consider the function $y=\tan2x$`y`=`t``a``n`2`x`.

Answer the following questions in radians, where appropriate.

Determine the $y$

`y`-intercept.Determine the period of the function.

How far apart are the asymptotes of the function?

State the first asymptote of the function for $x\ge0$

`x`≥0.State the first asymptote of the function for $x\le0$

`x`≤0.Graph the function.

Loading Graph...

As is the case for other functions, we will often want to interpret trigonometric functions that involve multiple translations and dilations. We can combine all the dilations and translations previously mentioned into one equation as follows. Note carefully the structure of the equation and specifically that the input expression to the function is factorised to reveal the values of $a$`a` and $b$`b`.

Summary

To obtain the graph of $y=af\left(b\left(x\right)\right)+c$`y`=`a``f`(`b`(`x`))+`c` from the graph of $y=f\left(x\right)$`y`=`f`(`x`):

- $a$
`a`vertically dilates the graph of $y=f\left(x\right)$`y`=`f`(`x`), and corresponds to a change in amplitude - $b$
`b`horizontally dilates the graph of $y=f\left(x\right)$`y`=`f`(`x`), and corresponds to a change in period - $c$
`c`vertically translates the graph of $y=f\left(x\right)$`y`=`f`(`x`)

In the case that $b$`b` is negative, it has the additional property of reflecting the graph of $y=f\left(x\right)$`y`=`f`(`x`) about the vertical axis.

Sketch the graph of $y=2\sin3x+2$`y`=2`s``i``n`3`x`+2.

**Think:** Starting with the graph of $y=\sin x$`y`=`s``i``n``x`, we can work through a series of transformations so that it coincides with the graph of $y=2\sin3x+2$`y`=2`s``i``n`3`x`+2.

**Do:** We can first increase the amplitude of the function to match. This is represented by multiplying the $y$`y`-value of every point on $y=\sin x$`y`=`s``i``n``x` by $2$2.

The graph of $y=2\sin x$y=2sinx |

Next, we can apply the period change that is the result of multiplying the $x$`x`-value inside the function by $3$3. This means that to get a particular $y$`y`-value, we can put in an $x$`x`-value that is $3$3 times smaller than before. Notice that the points on the graph of $y=2\sin x$`y`=2`s``i``n``x` move towards the vertical axis by a factor of $3$3 as a result.

The graph of $y=2\sin3x$y=2sin3x |

Our next step will be to obtain the graph of $y=2\sin3x+2$`y`=2`s``i``n`3`x`+2, and we can do so by applying a vertical translation by two units upwards.

The graph of $y=2\sin3x+2$y=2sin3x+2 |

Understand the relationship between y = f(x) and y = |f(x)|, where f(x) may be linear, quadratic or trigonometric.

Understand amplitude and periodicity and the relationship between graphs of related trigonometric functions, e.g. sin x and sin 2x.

Draw and use the graphs of y = asinbx + c, y = acos bx + c, y = atan bx + c where a is a positive integer, b is a simple fraction or integer (fractions will have a denominator of 2, 3, 4, 6 or 8 only), and c is an integer.