5.08 Circular functions: tangent

Recall from the introduction of the unit circle we can define the tangent function as follows:

• The tangent of the angle can be geometrically defined to be $y$y-coordinate of point $Q$Q, where $Q$Q is the intersection of the extension of the line $OP$OP and the tangent of the circle at $\left(1,0\right)$(1,0)
• Using similar triangles we can also define this algebraically as the ratio: $\tan\theta=\frac{\sin\theta}{\cos\theta}$tanθ=sinθcosθ​
• This also represents the gradient of the line that forms the angle $\theta$θ to the positive $x$x-axis

Let's revisit the applet from the introduction once more and focus our attention on the orange line. With the definitions above in mind, for what values of $\theta$θ is $\tan\theta$tanθ undefined? This would be the same as asking when is the slope of the line $OP$OP undefined, or when does $\cos\theta=0$cosθ=0. What is the value of $\tan0^\circ$tan0°? How often does the pattern repeat? That is what is the period of $\tan\theta$tanθ?

Change the angle in the applet below and try to imagine the shape of the graph of $\tan\theta$tanθ and key features.

Created with Geogebra

 

The graph of $y=\tan\theta$y=tanθ

 

The asymptotes of the function are where the angle $\theta$θ would cause the line $OP$OP to be vertical and hence the gradient is undefined. We can also see this through the definition $\tan\theta=\frac{\sin\theta}{\cos\theta}$tanθ=sinθcosθ​. The function is undefined where $\cos\theta=0$cosθ=0 and the graph approaches the vertical lines at these values forming asymptotes.

The fact that the tangent function repeats at intervals of $\pi$π can be verified by considering the unit circle diagram. Either by considering the gradient of the line $OP$OP and how the gradient will be the same for a point at an angle of $\theta$θ or at at angle of $\theta+\pi$θ+π. Or we can consider this algebraically by looking at the ratio $\frac{\sin\theta}{\cos\theta}$sinθcosθ​. If $\pi$π is added to an angle $\theta$θ, then the diagram  below shows that $\sin(\theta+\pi)$sin(θ+π) has the same magnitude as $\sin\theta$sinθ but opposite sign. The same relation holds between $\cos(\theta+\pi)$cos(θ+π) and $\cos\theta$cosθ.

We make use of the definition: $$

$\tan(\theta+\pi)$tan(θ+π)	$=$=	$\frac{\sin(\theta+\pi)}{\cos(\theta+\pi)}$sin(θ+π)cos(θ+π)​
	$=$=	$\frac{-\sin\theta}{-\cos\theta}$−sinθ−cosθ​
	$=$=	$\tan\theta$tanθ

 

Practice question

question 1

 

Transformations of $\tan x$tanx

Just as we transformed the trigonometric functions $y=\sin\theta$y=sinθ and $y=\cos\theta$y=cosθ we can apply parameters to the equation $y=\tan x$y=tanx to transform it to $y=a\tan\left(b\left(x-c\right)\right)+d$y=atan(b(x−c))+d.

Use the geogebra applet below to adjust the parameters in $y=a\tan\left(b\left(x-c\right)\right)+d$y=atan(b(x−c))+d and observe how it affects the graph. Try to answer the following questions.

• Which parameters affect the position of the vertical asymptotes? Which ones don't?
• Which parameters translate the graph, leaving the shape unchanged? Which ones affect the size?
• Which parameter changes the period of the graph? Does making this parameter larger make the period larger?

Created with Geogebra

The constants $a$a, $b$b, $c$c and $d$d transform the tangent graph. Let's summarise the impact of each:

 

 

Worked examples

Example 1

Illustrate change in dilation of the graph $y=\tan x$y=tanx by sketching the graphs of $y=\tan x$y=tanx together with $y=3\tan x$y=3tanx and $y=\frac{2}{7}\tan x$y=27​tanx.

Think: These are vertical dilations which means the graph of $y=\tan x$y=tanx is stretched vertically and the $y$y values are multiplied by the value of $a$a . All three graphs will share $x$x-intercepts and vertical asymptotes. We first sketch the base graph of $y=\tan x$y=tanx, shown in blue in the graph below. This graph will go through the points $\left(\frac{\pi}{4},1\right)$(π4​,1), $\left(0,0\right)$(0,0) and $\left(\frac{-\pi}{4},-1\right)$(−π4​,−1). The asymptotes will be located at $\frac{\pi}{2}$π2​, $\frac{-\pi}{2}$−π2​, $\frac{3\pi}{2}$3π2​, $\frac{-3\pi}{2}$−3π2​,....

The graph of $y=3\tan x$y=3tanx has a vertical dilation by a factor of $3$3. So we can plot the points $\left(\frac{\pi}{4},3\right)$(π4​,3) and $\left(\frac{-\pi}{4},-3\right)$(−π4​,−3) to show this stretch clearly. Similarly the graph $y=\frac{2}{7}\tan x$y=27​tanx has a vertical dilation by a factor of $\frac{2}{7}$27​. To show this we can plot the points $\left(\frac{\pi}{4},\frac{2}{7}\right)$(π4​,27​) and $\left(\frac{-\pi}{4},\frac{-2}{7}\right)$(−π4​,−27​). The graphs together are shown below.

 

 

 

Example 2

Sketch the graph of $f(x)=\tan\left(x-\frac{\pi}{4}\right)$f(x)=tan(x−π4​).

Think: This is a horizontal translation of the function $\tan x$tanx by $\frac{\pi}{4}$π4​ units to the right. We can sketch this by graphing the base graph of $y=\tan\theta$y=tanθ and shifting each point right by $\frac{\pi}{4}$π4​ (add $\frac{\pi}{4}$π4​ to the $x$x values).

The phase shift will also move the asymptotes and since $\tan x$tanx is undefined at $x=\frac{\pi}{2}+n\pi$x=π2​+nπ for all integers $n$n, the undefined points for $\tan\left(x-\frac{\pi}{4}\right)$tan(x−π4​) must be $x=\frac{3\pi}{4}+n\pi$x=3π4​+nπ.  The graph is shown below in purple.

 

Practice questions

question 2

question 3

question 4

Sketching tips

We can graph a tangent function by graphing the base function and then applying the transformations in stages to achieve the final sketch. Just as with sine and cosine functions we could also sketch the function using key features. The following steps may assist in sketching:

1. Ensure you formula is in the required format, $y=a\tan\left(b\left(x-c\right)\right)+d$y=atan(b(x−c))+d rearrange if necessary. Write down $a$a, $b$b, $c$c and $d$d.
2. Sketch in a dotted line for $y=d$y=d. This is the central line where the points of inflection will lie. If the graph has not been vertically translated this will be the $x$x-axis.
3. Plot the point $\left(c,d\right)$(c,d). This is where the point of inflection at the origin of $y=\tan x$y=tanx has been translated to.
4.  Find the period: $P=\frac{\pi}{b}$P=πb​ .
5. From the point $\left(c,d\right)$(c,d) draw dotted lines for the asymptotes half a period in both directions. That is at $x=c+\frac{P}{2}$x=c+P2​ and  $x=c-\frac{P}{2}$x=c−P2​. Then at integer multiples of the period, $P$P, in each direction from these asymptotes until you have the required domain.
6. To achieve a more accurate sketch and clearly show the dilation, plot the points $\left(c+\frac{P}{4},d+a\right)$(c+P4​,d+a)  and $\left(c-\frac{P}{4},d-a\right)$(c−P4​,d−a) 
7. Draw a smooth curve through your points and approaching the asymptotes.
8. Repeat this smooth curve for each period.

 

Worked example

Example 3

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? The order is "dilate then translate horizontally, dilate then translate vertically".

• We need to dilate the graph by a factor of $\frac{1}{2}$12​ horizontally. This means $x$x values are halved. For $\tan x$tanx the period is $\pi$π so this function has period of $\frac{\pi}{2}$π2​.

• We need to translate the graph by $\frac{\pi}{4}$π4​ units horizontally to the right (add $\frac{\pi}{4}$π4​ to $x$x values).
• We would dilate the graph by a factor of $3$3 vertically from the $x$x-axis which means every $y$y value is multiplied by $3$3.
• We need to translate the graph by $1$1 unit vertically which means every $y$y value is increased by $1$1. 

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: $period=\frac{\pi}{b}$period=πb​$=\frac{\pi}{2}$=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 $\frac{period}{4}$period4​$=\frac{\pi}{8}$=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.

Graph of $y=3\tan\left(2\left(x-\frac{\pi}{4}\right)\right)+1$y=3tan(2(x−π4​))+1

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

Axis intercepts

For our example above we did not have a $y$y-intercept as an asymptote coincided with the $y$y-axis. If this was not the case we could find the $y$y-intercept by evaluating the function at $x=0$x=0.

If we were required to label the $x$x-intercepts and there had been no vertical translation these would be found at the points of inflection at $x=c+Pn$x=c+Pn, where $P$P in the period and $n$n is any integer. However, if as in our example we have a vertical shift the $x$x-intercepts can be found with the assistance of technology or by solving the equation when $y=0$y=0. We will look further at solution to trigonometric equations later in this chapter. 

Practice questions

Question 5

Question 6

Question 7

Question 8