6.02 Period and phase shifts of trigonometric graphs

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.

Period

The period of a trigonometric function is defined as the time taken to complete one full cycle. For $y=\sin x$y=sinx and $y=\cos x$y=cosx, 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=sinx and $y=\cos x$y=cosx reveals a period of $2\pi$2π.

 

The tangent function $y=\tan x$y=tanx 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.

 

Horizontal dilations

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

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

Let's consider the graph of $y=\sin2x$y=sin2x sketched together with $y=\sin x$y=sinx:

From the graph, we can see that the function of $y=\sin2x$y=sin2x has a period of $\pi$π which is half the period of the original graph of $y=\sin x$y=sinx. 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=cos(x2​) sketched together with $y=\cos x$y=cosx:

From the graph, we can see that the function of $y=\cos\left(\frac{x}{2}\right)$y=cos(x2​) has a period of $4\pi$4π which is double the period of the original graph of $y=\cos x$y=cosx. 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=tan(−x3​) to see how it is dilated:

From the graph, we can see that the function of $y=\tan\left(-\frac{x}{3}\right)$y=tan(−x3​) has a period of $3\pi$3π which is triple the period of the original graph of $y=\tan x$y=tanx. 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 $|a|>1$|a|>1 horizontal stretch $0<|a|<1$0<|a|<1 vertical reflection $a<0$a<0	$\frac{2\pi}{a}$2πa​
$y=\cos\left(ax\right)$y=cos(ax)	horizontal compression $|a|>1$|a|>1 horizontal stretch $0<|a|<1$0<|a|<1 vertical reflection $a<0$a<0	$\frac{2\pi}{a}$2πa​
$y=\tan\left(ax\right)$y=tan(ax)	horizontal compression $|a|>1$|a|>1 horizontal stretch $0<|a|<1$0<|a|<1 vertical reflection $a<0$a<0	$\frac{\pi}{a}$πa​

Practice questions

Question 1

Question 2

Phase shifts

A phase shift of a trigonometric function is analogous to a horizontal shift of other functions. Recall that a function $y=f\left(x+b\right)$y=f(x+b) represents the function $y=f\left(x\right)$y=f(x) translated horizontally $b$b units (to the left for $b>0$b>0). The same applies to the trigonometric functions, and in this context this type of translation is called a phase shift.

We will see that amplitude and period are unaffected by phase shifts, as the whole graph is effectively being shifted left or right by this type of translation. It is called a phase shift because the translated graph is "out of phase" with the original function.

Let's look at some examples that demonstrate this type of translation.

The following graph shows the functions $\cos x$cosx and $\cos\left(x+0.4\right)$cos(x+0.4) on the same axes.

It can be seen that the graph of $\cos\left(x+0.4\right)$cos(x+0.4) is the graph of $\cos x$cosx shifted to the left by $0.4$0.4 units.

The following graph looks like the graph of $\sin x$sinx with a phase shift of $1.15$1.15 to the right.

This graph belongs to the function $\sin\left(x-1.15\right)$sin(x−1.15). The phase shift to the right has been brought about by adding $-1.15$−1.15 to $x$x.

The graphs of $f(x)=\tan x$f(x)=tanx and $f(x)=\tan\left(x-\frac{\pi}{4}\right)$f(x)=tan(x−π4​) are shown below.

We see that the function $\tan x$tanx has been moved to the right a distance $\frac{\pi}{4}$π4​. Note that the asymptotes also shift in the same direction an equivalent distance as the translation.

Practice questions

Question 3

Question 4

Combined translations and dilations of trigonometric functions

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.

Try experimenting with the value of each of these variables in the applet below.

Created with Geogebra

Worked examples

Example 1

Sketch the graph of $y=-2\sin\left(3x+\frac{\pi}{4}\right)+2$y=−2sin(3x+π4​)+2.

Think: Starting with the graph of $y=\sin x$y=sinx, we can work through a series of transformations so that it coincides with the graph of $y=-2\sin\left(3x+\frac{\pi}{4}\right)+2$y=−2sin(3x+π4​)+2.

Do: We can first reflect the graph of $y=\sin x$y=sinx about the $x$x-axis. This is represented by applying a negative sign to the function (multiplying the function by $-1$−1).

The graph of $y=-\sin x$y=−sinx

Then we can 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=−sinx 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=−2sinx 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 $y=-2\sin\left(3x+\frac{\pi}{4}\right)$y=−2sin(3x+π4​), and we can do so by applying a horizontal translation. In order to see what translation to apply, however, we first factorise the function into the form $y=-2\sin\left(3\left(x+\frac{\pi}{12}\right)\right)$y=−2sin(3(x+π12​)).

In this form, we can see that the $x$x-values are increased by $\frac{\pi}{12}$π12​ inside the function. This means that to get a particular $y$y-value, we can put in an $x$x-value that is $\frac{\pi}{12}$π12​ smaller than before. Graphically, this corresponds to shifting the entire function to the left by $\frac{\pi}{12}$π12​ units.

The graph of $y=-2\sin\left(3x+\frac{\pi}{4}\right)$y=−2sin(3x+π4​)

 

Lastly, we translate the graph of $y=-2\sin\left(3x+\frac{\pi}{4}\right)$y=−2sin(3x+π4​) upwards by $2$2 units, to obtain the final graph of $y=-2\sin\left(3x+\frac{\pi}{4}\right)+2$y=−2sin(3x+π4​)+2.

The graph of $y=-2\sin\left(3x+\frac{\pi}{4}\right)+2$y=−2sin(3x+π4​)+2

Example 2

Sketch the function $y=3\tan\left(2\left(x-\frac{\pi}{4}\right)\right)+1$y=3tan(2(x−π4​))+1 for the interval $-\pi\le x\le\pi$−π≤x≤π

Think: What transformations would take $y=\tan x$y=tanx to $y=3\tan\left(2\left(x-\frac{\pi}{4}\right)\right)+1$y=3tan(2(x−π4​))+1?

• We would dilate the graph by a factor of $3$3 from the $x$x-axis
• We need to dilate the graph by a factor of $\frac{1}{2}$12​ horizontally. Hence, the period becomes $\frac{\pi}{2}$π2​.
• We need to translate the graph by $1$1 unit vertically
• We need to translate the graph by $\frac{\pi}{4}$π4​ units horizontally

Do: List the parameters $a=3$a=3, $b=2$b=2, $c=\frac{\pi}{4}$c=π4​ and $d=1$d=1. Sketch a dotted line for the central line $y=1$y=1 and plot the point $\left(c,d\right)=\left(\frac{\pi}{4},1\right)$(c,d)=(π4​,1)

Find the period $P$P: $P=\frac{\pi}{b}=\frac{\pi}{2}$P=πb​=π2​ and draw dotted lines for the asymptotes half a period in both directions from the point $\left(\frac{\pi}{4},1\right)$(π4​,1). Then repeat at multiples of the period from these lines.

From the point $\left(\frac{\pi}{4},1\right)$(π4​,1), plot a point by going forwards $\frac{P}{4}=\frac{\pi}{8}$P4​=π8​ and up $a$a units ($3$3 units). Mirror this by plotting a second point backwards $\frac{\pi}{8}$π8​ from $\left(\frac{\pi}{4},1\right)$(π4​,1) and down $3$3 units.

Join the points with a smooth curve which also approaches the asymptotes.

Lastly, repeat the pattern for each period.

Reflect: Does the graph match how it should look after transformations? Does its cycle repeat the correct number of times for the domain given?

Practice questions

Question 5

Question 6