5.02 Graphing sine and cosine functions

Vertical dilations - amplitude of sine and cosine

The graphs of trigonometric functions like $y=\sin x$y=sinx and $y=\cos x$y=cosx are periodic and we can understand a lot about the behavior of the whole function by looking only at what happens within a single cycle or period.

Typically, each trigonometric graph will have a minimum value, a maximum value, and and average value. The distance between the maximum (or minimum) value and the average value is called the amplitude.

The amplitude is given by $\frac{\text{max - min }}{2}$max - min 2​. Notice that the amplitude is always a positive number.

Dilations and the amplitude

An equation of the form $y=a\sin x$y=asinx has an amplitude of $a$a units. If we start with the equation $y=\sin x$y=sinx, where $a=1$a=1, we can transform this equation by changing the amplitude.

Graphically this transformation corresponds to stretching the graph of $\sin x$sinx in the vertical direction when $a>1$a>1, and compressing the graph of $\sin x$sinx in the vertical direction when $00<a<1. We can also reflect the graph across the $x$x-axis by multiplying the whole function by $-1$−1.

This action of stretching and compressing the graph of a function is known as dilation. For trigonometric graphs like $\sin x$sinx and $\cos x$cosx, a vertical dilation by a factor of $a$a is equivalent to increasing the amplitude by a factor of $a$a.

 

Worked examples

Question 1

State the amplitude of the function $f\left(x\right)=5\sin x$f(x)=5sinx.

Think: When we compare $f\left(x\right)$f(x) to the standard function $\sin x$sinx we can see that all the function values of $5\sin x$5sinx will be five times larger than all the function values of $\sin x$sinx. This means that the amplitude of $5\sin x$5sinx is also five times larger than the amplitude of $\sin x$sinx.

Do: The amplitude of $\sin x$sinx is $1$1, so the amplitude of $f\left(x\right)=5\sin x$f(x)=5sinx is $5\times1=5$5×1=5.

Reflect: We can obtain the graph of $f\left(x\right)=5\sin x$f(x)=5sinx by starting with the graph of $y=\sin x$y=sinx and applying a vertical dilation by a factor of $5$5.

 

Question 2

The graph of $y=\cos x$y=cosx is reflected across the $x$x-axis, then compressed in the vertical direction so that its minimum value is $-\frac{3}{4}$−34​. What is the equation of the resulting function. What is the amplitude?

Think: Recall that a reflection across the $x$x-axis corresponds to multiplying the function by $-1$−1. This will "switch" the location of the maximum and minimum values of the graph, but the amplitude will still be a positive value.

Do: Let's keep track of how the equation of the graph changes at each stage of the transformation.

$y=\cos x$y=cosx	$\rightarrow$→	$y=-\cos x$y=−cosx	(Reflection across $x$x-axis)
$y=-\cos x$y=−cosx	$\rightarrow$→	$y=-\frac{3}{4}\cos x$y=−34​cosx	(Vertical compression)

The final equation of the resulting graph is $y=-\frac{3}{4}\cos x$y=−34​cosx. The amplitude of this equation is $\frac{3}{4}$34​.

Reflect: Compare the resulting equation with the original equation. The only difference is the constant multiple of $-\frac{3}{4}$−34​. Notice that although this number is negative, the amplitude of the resulting equation is positive. In general, the amplitude of the equation $y=a\cos x$y=acosx is $\left|a\right|$|a|.

 

Practice questions

QUESTION 3

QUESTION 4

 

Horizontal dilations - period changes for sine and cosine

We define the $\cos$cos and $\sin$sin functions as the horizontal and vertical coordinates of a point that moves on the unit circle. In the diagrams below, this is shown for an angle $\alpha$α in the first and second quadrants.

If we imagine the point moving counterclockwise on the unit circle so that the radius from the point makes an ever-increasing angle with the positive horizontal axis, eventually the angle exceeds $360^\circ$360°, but the values of the $\cos$cos and $\sin$sin functions repeat the values of the coordinates from the angle $360^\circ$360° smaller. We say $\sin$sin and $\cos$cos are periodic functions with period $360^\circ$360° or $2\pi$2π radians.

Transformations involving period

Again, consider the angle $\alpha$α made by the point moving around the unit circle. If a new angle $\alpha'$α′ is defined by $\alpha'=k\alpha$α′=kα, We know that $\sin\alpha'$sinα′ has period $360^\circ$360° or $2\pi$2π radians, but we see that $\alpha'$α′ reaches $360^\circ$360°  when $\alpha=\frac{360^\circ}{k}$α=360°k​. So, $\sin k\alpha$sinkα and $\cos k\alpha$coskα must have period $\frac{360^\circ}{k}$360°k​ or $\frac{2\pi}{k}$2πk​ radians with respect to $\alpha$α.

Worked example

Question 5

Graph the function $\sin2x$sin2x on the same axes as $\sin x$sinx.

Think: $\sin2x$sin2x begins to repeat when $2x=360^\circ$2x=360°. That is, when $x=180^\circ$x=180°. So, $\sin2x$sin2x has period $180^\circ$180°.

Do:

Reflect: Thus, we see that for functions $\sin kx$sinkx and $\cos kx$coskx where $k$k is a constant, the period of the function with respect to $kx$kx is $360^\circ$360° but the period with respect to $x$x is $\frac{360^\circ}{k}$360°k​.

 

Question 5

Identify the equation of the graph below

 

Think: This graph looks like the graph of a cosine function since it has the value $1$1 at $0$0. However, the period is $288^\circ$288°.

Do: We know that $\cos k\alpha^\circ$coskα° has period $\frac{360^\circ}{k}$360°k​ and, in this case, $\frac{360^\circ}{k}=288^\circ$360°k​=288°. Therefore, $k=\frac{360^\circ}{288}=1.25$k=360°288​=1.25.

The graph has equation $y=\cos1.25x$y=cos1.25x or $y=\cos\frac{5}{4}x$y=cos54​x.

 

Practice questions

Question 6

Question 7

 

Horizontal translation - phase shifts for sine and cosine

Phase shift for trigonometric functions means moving the graph of the function to the right or to the left. This transformation occurs when a constant is added to (or subtracted from) the angle to which the function is applied.

For example, the following functions include a phase shift transformation.

$y=\sin\left(\theta+45^\circ\right)$y=sin(θ+45°)
$y=\cos(x-\frac{\pi}{2})$y=cos(x−π2​)
$y=\tan\left(\alpha+180^\circ\right)$y=tan(α+180°)

 

Worked example

question 8

Describe the transformation from $y=\cos\theta$y=cosθ to $y=\cos\left(\theta+23^\circ\right)$y=cos(θ+23°).

Think: The following graph shows the functions $\cos\theta$cosθ and $\cos(\theta+23^\circ)$cos(θ+23°) on the same axes.

Do: It can be seen that the graph of $\cos(\theta+23^\circ)$cos(θ+23°) is the graph of $\cos\theta$cosθ shifted to the left by the amount $23^\circ$23°. 

Reflect: The dotted lines drawn on the diagram are intended to show that the function $\cos(\theta+23^\circ)$cos(θ+23°) when $\theta=20^\circ$θ=20° attains the same value reached by $\cos\theta$cosθ when $\theta=43^\circ$θ=43°. Thus, the shift is to the left.

 

Question 9

Identify the equation of the graph below.

Think: The following graph looks like the graph of $\sin\theta$sinθ with a phase shift of $60^\circ$60° to the right. 

Do: The graph must belong to the function given by $y=\sin(\theta-60^\circ)$y=sin(θ−60°). The phase shift to the right has been brought about by adding $-60^\circ$−60° to $\theta$θ.

 

Practice Questions

Question 10

Question 11

 

Combined transformations of sine and cosine curves and equations

Just as we have seen with other functions, we can graph multiple transformations from the parent functions $y=\sin x$y=sinx and $y=\cos x$y=cosx including vertical and horizontal dilations, reflections and translations.

Consider the graphs of $y=\sin x$y=sinx and $y=-2\sin\left(3x+45^\circ\right)+2$y=−2sin(3x+45°)+2 which are drawn below.

The graphs of $y=\sin x$y=sinx and $y=-2\sin\left(3x+45^\circ\right)+2$y=−2sin(3x+45°)+2

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 $y=-2\sin\left(3x+45^\circ\right)+2$y=−2sin(3x+45°)+2.

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 $y=-2\sin\left(3x+45^\circ\right)$y=−2sin(3x+45°), and we can do so by applying a horizontal translation. In order to see what translation to apply, however, we first factor the function into the form $y=-2\sin\left(3\left(x+15^\circ\right)\right)$y=−2sin(3(x+15°)).

In this form, we can see that the $x$x-values are increased by $15^\circ$15° inside the function. This means that to get a particular $y$y-value, we can put in an $x$x-value that is $15^\circ$15° smaller than before. Graphically, this corresponds to shifting the entire function to the left by $15^\circ$15°.

The graph of $y=-2\sin\left(3x+45^\circ\right)$y=−2sin(3x+45°)

 

Lastly, we translate the graph of $y=-2\sin\left(3x+45^\circ\right)$y=−2sin(3x+45°) upwards by $2$2 units, to obtain the final graph of $y=-2\sin\left(3x+45^\circ\right)+2$y=−2sin(3x+45°)+2.

The graph of $y=-2\sin\left(3x+45^\circ\right)+2$y=−2sin(3x+45°)+2

 

 

The general case

The general form of the equation of a sine or cosine curve is 

$f\left(x\right)=a\sin\left(bx-c\right)+d$f(x)=asin(bx−c)+d

$f\left(x\right)=a\cos\left(bx-c\right)+d$f(x)=acos(bx−c)+d

From above, we can summarize as follows:

For a value of $a$a where $\left|a\right|>1$|a|>1, the graph vertically distributes or stretches. For a trigonometric function, we say that the amplitude increases. If $\left|a\right|<1$|a|<1, the graph of $y=f\left(x\right)$y=f(x) vertically compresses. For a trigonometric function, we say that the amplitude decreases.

For a value of $b$b where $\left|b\right|>1$|b|>1, the graph horizontally compresses. If $\left|b\right|<1$|b|<1, then the graph horizontally distributes or stretches. In the case that the graph describes a trigonometric function, a horizontal compression means the period decreases and a horizontal distribution means the period increases.

 

Graphing strategies

We can either graph one transformation at a time to end up with the final graph, as done with the example of $y=-2\sin\left(3x+45^\circ\right)+2$y=−2sin(3x+45°)+2 or we can use the summary above to identify the key features and graph from those. Let's look at that second strategy.

Worked example

question 12

Graph $y=2\sin\left(\frac{x}{2}\right)+3$y=2sin(x2​)+3

Think: Let's start by identifying all of the key features of the graph.

• amplitude: $a=2$a=2
• reflection:  no reflection as $a$a is positive
• period: $\frac{360^\circ}{b}=\frac{360^\circ}{\frac{1}{2}}=720^\circ$360°b​=360°12​​=720°
• phase shift: $\frac{c}{b}=0$cb​=0 as $c=0$c=0 in this case
• vertical translation: $d=3$d=3

Do: Now we need to draw the curve that matches the key features. We'll do it in steps.

First, draw the central line (indicated by the vertical translation)

Next, mark on the maximum and minimum by measuring the amplitude above and below the central line

Next, check for a phase shift (this would shift the initial starting position)

There is no phase shift for this function as $c=0$c=0

Next, mark out the distance of the full period.  At this stage mark out half way and quarter way marks, this will help us sketch the curve. 

Next, check for a reflection (this would change the initial starting direction

There is no reflection for this function as $a>0$a>0

Next, Create some dots on the starting and ending positions of the cycle (on the central line), also mark out the maximum and minimum points (on the quarter lines).

Next, sketch the curve lightly, joining the preparatory dots. Developing the skills for smooth curve drawing takes practice so don't get disheartened.  

 

Practice questions

question 13

question 14

Question 15

 

 

Domain and range 

The domain of a function is the set of all values that the independent variable (usually $x$x) can take and the range of a function is the set of all values that the dependent variable (usually $y$y) takes.  Each number in the range is the image of one or more domain values under the action of the function.

Since there are no $x$x-values to exclude, the domain for both sine and cosine functions is the whole set of real numbers.

We may, if we wish, restrict the domain in any way we choose. Often we may only be interested in the behavior of the function on an interval, rather than on the whole set of real numbers, for example, $0^\circ\le x\le360^\circ$0°≤x≤360°.

The range is the set of values that the function $y$y can take as $x$x varies over the domain.  We can see from the graphs that the $y$y-values vary continuously between the maximum and the minimum values. For both of these functions, the maximum is $1$1 and the minimum $-1$−1. Thus, the range is $-1\le y\le1$−1≤y≤1.

For other cosine and sine functions we would need to establish the maximum and minimum by either reading off the graph or by using the following rule:

The range of $y=a\sin\left(bx-c\right)+d$y=asin(bx−c)+d or $y=a\cos\left(bx-c\right)+d$y=acos(bx−c)+d is 

$d-a\le y\le d+a$d−a≤y≤d+a.

This formula uses facts about amplitude and vertical translation. The amplitude is the amount by which the function goes above the central value given by $y=d$y=d.

 

Worked example

Question 16

State the domain and range of $f(x)=2.5\cos x+0.5$f(x)=2.5cosx+0.5 in radians.

Think: The function makes sense for all real numbers $x$x. That is, the natural domain is the set of real numbers. The range will be determined by the minimum and maximum.

Do: Since there is no phase shift, the maximum occurs when $x=0$x=0 and we can see that $f(0)=3$f(0)=3. The minimum would occur when $x=\pi$x=π, and we can see that $f(\pi)=-2$f(π)=−2. The function varies by $2.5$2.5 above and below its mid-value of $0.5$0.5.

The domain is all real numbers and the range is $-2\le y\le3$−2≤y≤3.

 

Practice Questions

Question 17

 

Applications of sine and cosine functions

Now that we're familiar with the properties of sine and cosine functions, we can apply them to real-world situations. Many phenomena in the world around us change periodically such as pendulums, springs, rotors and wheels. In general, when solving problems involving applications like these, the first step is translating the information provided into either an equation or graph, and answering the relevant questions that we might be interested in.

Worked example

Question 18

Peter is riding forwards on a unicycle. When he gets on to start riding, the pedals are horizontally inline. The height of the back pedal in centimeters is given by the equation $y=20\sin\theta+40$y=20sinθ+40 where $\theta$θ is the measure of the angle that the back pedal has rotated.

a) What is the maximum and minimum height that the back pedal reaches?

Think: The maximum height of the back pedal occurs whenever $\sin\theta=1$sinθ=1 and the minimum height occurs whenever $\sin\theta=-1$sinθ=−1.

Do: Let's substitute $\sin\theta=1$sinθ=1 and $\sin\theta=-1$sinθ=−1 into the equation.

$y$y	$=$=	$20\sin\theta+40$20sinθ+40	(writing down the equation)
$y$y	$=$=	$20\times1+40$20×1+40	(making the substitution)
$y$y	$=$=	$60$60 cm	(simplifying the expression)

So the maximum height of the back pedal is $60$60 centimeters.

$y$y	$=$=	$20\sin\theta+40$20sinθ+40	(writing down the equation)
$y$y	$=$=	$20\times\left(-1\right)+40$20×(−1)+40	(making the substitution)
$y$y	$=$=	$20$20 cm	(simplifying the expression)

So the minimum height of the back pedal is $20$20 centimeters.

b) How high is the back pedal from the ground when Peter gets on?

Think: Peter gets on when the pedals are horizontally level and no angle has been formed. In other words, Peter gets on when $\theta=0^\circ$θ=0°.

Do: Let's substitute $\theta=0^\circ$θ=0° into the equation.

$y$y	$=$=	$20\sin\theta+40$20sinθ+40	(writing down the equation)
$y$y	$=$=	$20\sin0^\circ+40$20sin0°+40	(making the substitution)
$y$y	$=$=	$20\times0+40$20×0+40	(simplifying the $\sin$sin function)
$y$y	$=$=	$40$40 cm	(simplifying the expression)

So the back pedal is $40$40 centimeters above ground when Peter gets on.

c) Graph the function given by the equation $y=20\sin\theta+40$y=20sinθ+40.

Think: We can draw the function $y=20\sin\theta+40$y=20sinθ+40 by first considering $y=\sin\theta$y=sinθ.

Do: If we take the graph of $y=\sin\theta$y=sinθ, stretch the amplitude and vertically translate the curve up, we get our desired function.

The graph of $y=20\sin\theta+40$y=20sinθ+40

 

d) How high is the back pedal from the ground when it has rotated through $150^\circ$150°?

Think: We wish to find the value of $y$y when $\theta=150^\circ$θ=150°. We can see graphically that this looks like $50$50 centimeters, but we can show this algebraically as well.

Do: Let's substitute $\theta=150^\circ$θ=150° into the equation.

$y$y	$=$=	$20\sin\theta+40$20sinθ+40	(writing down the equation)
$y$y	$=$=	$20\sin150^\circ+40$20sin150°+40	(making the substitution)
$y$y	$=$=	$20\times\frac{1}{2}+40$20×12​+40	(simplifying the $\sin$sin function)
$y$y	$=$=	$50$50 cm	(simplifying the expression)

So we can confirm that the back pedal is $50$50 centimeters above the ground after the pedal has rotated $150^\circ$150°.

e) How many degrees has the back pedal first rotated through when it is $30$30 cm off the ground?

Think: We can see from the graph of $y=20\sin\theta+40$y=20sinθ+40 and $y=30$y=30 that the pedal has rotated somewhere between $180^\circ$180° and $240^\circ$240° - perhaps the midpoint of $210^\circ$210°. Again, we can confirm this algebraically.

The graph of $y=20\sin\theta+40$y=20sinθ+40 (green) and $y=30$y=30 (blue)

 

Do: Let's substitute $y=30$y=30 into the equation.

$y$y	$=$=	$20\sin\theta+40$20sinθ+40	(writing down the equation)
$30$30	$=$=	$20\sin\theta+40$20sinθ+40	(making the substitution)
$-10$−10	$=$=	$20\sin\theta$20sinθ	(subtracting $40$40 from both sides)
$-\frac{1}{2}$−12​	$=$=	$\sin\theta$sinθ	(dividing by $20$20 on both sides)
$\theta$θ	$=$=	$\sin^{-1}\left(-\frac{1}{2}\right)$sin−1(−12​)	(taking the inverse $\sin$sin of both sides)
$\theta$θ	$=$=	$210^\circ$210°	(writing down the answer in degrees)

So after the back pedal rotates $210^\circ$210°, it is $30$30 centimeters above the ground.

f) Graph the height of the front pedal as a function over $\theta$θ.

Think: Whenever the back pedal is closest to the ground, the front pedal is furthest from the ground and visa versa. From this, the peaks and troughs of the back and front pedal should occur alternately.

Do: We draw the two functions alongside one another below.

The height of the back pedal (green) and the front pedal (blue) as functions of $\theta$θ.

 

Practice questions

question 19

question 20