# 4.2: The Tangent Function

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It turns out that we regularly need to evaluate the ratio of the sine and cosine functions at the same angle, so it is convenient to define a new function to be their ratio.

##### Definition $$\PageIndex{1}$$: The tangent function.

For any real number $$t$$ for which $$\cos(t) \ne 0\text{,}$$ we define the tangent of $$t$$, denoted $$\tan(t)\text{,}$$ by

$\tan(t) = \frac{\sin(t)}{\cos(t)}\text{.} \nonumber$

##### Preview Activity $$\PageIndex{1}$$

Through the following questions, we work to understand the special values and overall behavior of the tangent function.

1. Without using computational device, find the exact value of the $$\tan(t)$$ at the following values: $$t = \frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}, \frac{2\pi}{3}, \frac{3\pi}{4}, \frac{5\pi}{6}\text{.}$$
2. Why is $$\tan \left( \frac{\pi}{2} \right)$$ not defined? What are three other input values $$x$$ for which $$\tan(x)$$ is not defined?
3. Point your browser to http://gvsu.edu/s/0yO (“zero-y-Oh”) to find a Desmos worksheet with data from the tangent function already input. Click on several of the orange points to compare your exact values in (a) with the decimal values given by Desmos. Add one entry to the table: $$x = \frac{11\pi}{24}\text{,}$$ $$y = T(\frac{11\pi}{24})\text{.}$$ At about what coordinates does this point lie? What are the respective values of $$\sin(\frac{11\pi}{24})$$ and $$\cos(\frac{11\pi}{24})\text{?}$$ Why is the value of $$\tan(\frac{11\pi}{24})$$ so large?
4. At the top of the input lists on the left side of the Desmos worksheet, click the circle to highlight the function $$T(x) = \tan(x)$$ and thus show its plot along with the data points in orange. Use the plot and your work above to answer the following important questions about the tangent function:
• What is the domain of $$y = \tan(x)\text{?}$$
• What is the period of $$y = \tan(x)\text{?}$$
• What is the range of $$y = \tan(x)\text{?}$$

## Two perspectives on the tangent function Figure $$\PageIndex{3}$$ An angle $$t$$ in standard position in the unit circle that intercepts an arc from $$(1,0)$$ to $$(a,b).$$ Figure $$\PageIndex{4}$$ A right triangle with legs adjacent and opposite angle $$\theta\text{.}$$

Because the tangent function is defined in terms of the two fundamental circular functions by the rule $$\tan(t) = \frac{\sin(t)}{\cos(t)}\text{,}$$ we can use our understanding of the sine and cosine functions to make sense of the tangent function. In particular, we can think of the tangent of an angle from two different perspectives: as an angle in standard position in the unit circle, or as an angle in a right triangle.

From the viewpoint of Figure $$\PageIndex{3}$$, as the point corresponding to angle $$t$$ traverses the circle and generates the point $$(a,b)\text{,}$$ we know $$\cos(t) = a$$ and $$\sin(t) = b\text{,}$$ and therefore the tangent function tracks the ratio of these two quantities, and is given by

$\tan(t) = \frac{\sin(t)}{\cos(t)} = \frac{b}{a}\text{.} \nonumber$

From the perspective of any right triangle (not necessarily in the unit circle) with hypotenuse “hyp” and legs “adj” and “opp” that are respectively adjacent and opposite the known angle $$\theta\text{,}$$ as seen in Figure 4.2.4, we know that $$\sin(\theta) = \frac{\text{opp}}{\text{hyp}}$$ and $$\cos(\theta) = \frac{\text{adj}}{\text{hyp}}\text{.}$$ Substituting these expressions for $$\sin(\theta)$$ and $$\cos(\theta)$$ in the rule for the tangent function, we see that

$\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = \frac{\frac{\text{opp}}{\text{hyp}}}{\frac{\text{adj}}{\text{hyp}}} = \frac{\text{opp}}{\text{adj}}\text{.} \nonumber$

We typically use the first perspective of tracking the ratio of the $$y$$-coordinate to the $$x$$-coordinate of a point traversing the unit circle in order to think of the overall behavior and graph of the tangent function, and use the second perspective in a right triangle whenever we are working to determine missing values in a triangle.

## Properties of the tangent function

Because the tangent function is defined in terms of the sine and cosine functions, its values and behavior are completely determined by those two functions. To begin, we know the value of $$\tan(t)$$ for every special angle $$t$$ on the unit circle that we identified for the sine and cosine functions. For instance, we know that

$\tan \left( \frac{\pi}{6} \right) = \frac{ \sin \left( \frac{\pi}{6} \right) }{ \cos \left( \frac{\pi}{6} \right) } = \frac{ \frac{1}{2} }{ \frac{\sqrt{3}}{2} } = \frac{1}{\sqrt{3}}\text{.} \nonumber$

Executing similar computations for every familiar special angle on the unit circle, we find the results shown in Table 4.2.5 and Table 4.2.6. We also note that anywhere $$\cos(t) = 0\text{,}$$ the value of $$\tan(t)$$ is undefined. We record such instances in the table by writing “u”.

 $$t$$ $$0$$ $$\frac{\pi}{6}$$ $$\frac{\pi}{4}$$ $$\frac{\pi}{3}$$ $$\frac{\pi}{2}$$ $$\frac{2\pi}{3}$$ $$\frac{3\pi}{4}$$ $$\frac{5\pi}{6}$$ $$\pi$$ $$\sin(t)$$ $$0$$ $$\frac{1}{2}$$ $$\frac{\sqrt{2}}{2}$$ $$\frac{\sqrt{3}}{2}$$ $$1$$ $$\frac{\sqrt{3}}{2}$$ $$\frac{\sqrt{2}}{2}$$ $$\frac{1}{2}$$ $$\0$$ $$\cos(t)$$ $$1$$ $$\frac{\sqrt{3}}{2}$$ $$\frac{\sqrt{2}}{2}$$ $$\frac{1}{2}$$ $$0$$ $$-\frac{1}{2}$$ $$-\frac{\sqrt{2}}{2}$$ $$-\frac{\sqrt{3}}{2}$$ $$-1$$ $$\tan(t)$$ $$0$$ $$\frac{1}{\sqrt{3}}$$ $$1$$ $$\frac{3}{\sqrt{3}}$$ u $$-\frac{3}{\sqrt{3}}$$ $$-1$$ $$-\frac{1}{\sqrt{3}}$$ $$\0$$
 $$t$$ $$\frac{7\pi}{6}$$ $$\frac{5\pi}{4}$$ $$\frac{4\pi}{3}$$ $$\frac{3\pi}{2}$$ $$\frac{5\pi}{3}$$ $$\frac{7\pi}{4}$$ $$\frac{11\pi}{6}$$ $$2\pi$$ $$\sin(t)$$ $$-\frac{1}{2}$$ $$-\frac{\sqrt{2}}{2}$$ $$-\frac{\sqrt{3}}{2}$$ $$-1$$ $$-\frac{\sqrt{3}}{2}$$ $$-\frac{\sqrt{2}}{2}$$ $$-\frac{1}{2}$$ $$0$$ $$\cos(t)$$ $$-\frac{\sqrt{3}}{2}$$ $$-\frac{\sqrt{2}}{2}$$ $$-\frac{1}{2}$$ $$0$$ $$\frac{1}{2}$$ $$\frac{\sqrt{2}}{2}$$ $$\frac{\sqrt{3}}{2}$$ $$0$$ $$\tan(t)$$ $$\frac{1}{\sqrt{3}}$$ $$1$$ $$\frac{3}{\sqrt{3}}$$ u $$-\frac{3}{\sqrt{3}}$$ $$-1$$ $$-\frac{1}{\sqrt{3}}$$ $$0$$

Table 4.2.6. Additional values of the sine, cosine, and tangent functions at special points on the unit circle.

Table 4.2.5 and Table 4.2.6 helps us identify trends in the tangent function. For instance, we observe that the sign of $$\tan(t)$$ is positive in Quadrant I, negative in Quadrant II, positive in Quadrant III, and negative in Quadrant IV. This holds because the sine and cosine functions have the same sign in the first and third quadrants, and opposite signs in the other two quadrants.

In addition, we observe that as $$t$$-values in the first quadrant get closer to $$\frac{\pi}{2}\text{,}$$ $$\sin(t)$$ gets closer to $$1\text{,}$$ while $$\cos(t)$$ gets closer to $$0$$ (while being always positive). Noting that $$\frac{\pi}{2} \approx 1.57\text{,}$$ we observe that

$\tan(1.47) = \frac{\sin(1.47)}{\cos(1.47)} \approx \frac{0.995}{0.101} = 9.887 \nonumber$

and

$\tan(1.56) = \frac{\sin(1.56)}{\cos(1.56)} \approx \frac{0.9994}{0.0108} = 92.6205\text{.} \nonumber$

Because the ratio of numbers closer and closer to $$1$$ divided by numbers closer and closer to $$0$$ (but positive) increases without bound, this means that $$\tan(t)$$ increases without bound as $$t$$ approaches $$\frac{\pi}{2}$$ from the left side. Once $$t$$ is slightly greater than $$\frac{\pi}{2}$$ in Quadrant II, the value of $$\sin(t)$$ stays close to $$1\text{,}$$ but now the value of $$\cos(t)$$ is negative (and close to zero). For instance, $$\cos(1.58) \approx -0.0092\text{.}$$ This makes the value of $$\tan(t)$$ decrease without bound (negative and getting further away from $$0$$) for $$t$$ approaching $$\frac{\pi}{2}$$ from the right side, and results in $$h(t) = \tan(t)$$ having a vertical asymptote at $$t = \frac{\pi}{2}\text{.}$$ The periodicity and sign behaviors of $$\sin(t)$$ and $$\cos(t)$$ mean this asymptotic behavior of the tangent function will repeat.

Plotting the data in the table along with the expected asymptotes and connecting the points intuitively, we see the graph of the tangent function in Figure 4.2.7. Figure $$\PageIndex{7}$$ A plot of the tangent function together with special points that come from the unit circle.

We see from Table 4.2.5 and Table 4.2.6 as well as from Figure $$\PageIndex{7}$$ that the tangent function has period $$P = \pi$$ and that the function is increasing on any interval on which it is defined. We summarize our recent work as follows.

##### Properties of the tangent function.

For the function $$h(t) = \tan(t)\text{,}$$

• its domain is the set of all real numbers except $$t = \frac{\pi}{2} \pm k\pi$$ where $$k$$ is any whole number;
• its range is the set of all real numbers;
• its period is $$P = \pi\text{;}$$
• is increasing on any interval on which the function is defined at every point in the interval.

While the tangent function is an interesting mathematical function for its own sake, its most important applications arise in the setting of right triangles, and for the remainder of this section we will focus on that perspective.

## Using the tangent function in right triangles

The tangent function offers us an additional choice when working in right triangles with limited information. In the setting where we have a right triangle with one additional known angle, if we know the length of the hypotenuse, we can use either the sine or cosine of the angle to help us easily find the remaining side lengths. But in the setting where we know only the length of one leg, the tangent function now allows us to determine the value of the remaining leg in a similarly straightforward way, and from there the hypotenuse.

##### Example $$\PageIndex{8}$$

Use the tangent function to determine the width, $$w\text{,}$$ of the river in Figure $$\PageIndex{9}$$ (Note that here we are revisiting the problem in Activity 4.1.4, which we previously solved without using the tangent function.) What other information can we now easily determine? Figure $$\PageIndex{9}$$ A right triangle with one angle and one leg known.
Solution

Using the perspective that $$\tan(\theta) = \frac{\text{opp}}{\text{adj}}$$ in a right triangle, in this context we have

$\tan(56.4^\circ) = \frac{w}{50} \nonumber$

and thus $$w = 50\tan(56.4)$$ is the exact width of the river. Using a computational device, we find that $$w \approx 75.256\text{.}$$

Once we know the river's width, we can use the Pythagorean theorem or the sine function to determine the distance from $$P$$ to $$A\text{,}$$ at which point all $$6$$ parts of the triangle are known.

The tangent function finds a wide range of applications in finding missing information in right triangles where information about one or more legs of the triangle is known.

##### Activity $$\PageIndex{2}$$

The top of a $$225$$ foot tower is to be anchored by four cables that each make an angle of $$32.5^{\circ}$$ with the ground. How long do the cables have to be and how far from the base of the tower must they be anchored?

##### Activity $$\PageIndex{3}$$

Supertall 1  high rises have changed the Manhattan skyline. These skyscrapers are known for their small footprint in proportion to their height, with their ratio of width to height at most $$1:10\text{,}$$ and some as extreme as $$1:24\text{.}$$ Suppose that a relatively short supertall has been built to a height of $$635$$ feet, as pictured in Figure $$\PageIndex{10}$$, and that a second supertall is built nearby. Given the two angles that are computed from the new building, how tall, $$s\text{,}$$ is the new building, and how far apart, $$d\text{,}$$ are the two towers? Figure $$\PageIndex{10}$$ Two supertall skyscrapers.
##### Activity $$\PageIndex{4}$$

Surveyors are trying to determine the height of a hill relative to sea level. First, they choose a point to take an initial measurement with a sextant that shows the angle of elevation from the ground to the peak of the hill is $$19^\circ\text{.}$$ Next, they move $$1000$$ feet closer to the hill, staying at the same elevation relative to sea level, and find that the angle of elevation has increased to $$25^\circ\text{,}$$ as pictured in Figure $$\PageIndex{11}$$. We let $$h$$ represent the height of the hill relative to the two measurements, and $$x$$ represent the distance from the second measurement location to the “center” of the hill that lies directly under the peak. Figure $$\PageIndex{11}$$ The surveyors' initial measurements.
1. Using the right triangle with the $$25^\circ$$ angle, find an equation that relates $$x$$ and $$h\text{.}$$
2. Using the right triangle with the $$19^\circ$$ angle, find a second equation that relates $$x$$ and $$h\text{.}$$
3. Our work in (a) and (b) results in a system of two equations in the two unknowns $$x$$ and $$h\text{.}$$ Solve each of the two equations for $$h$$ and then substitute appropriately in order to find a single equation in the variable $$x\text{.}$$
4. Solve the equation from (c) to find the exact value of $$x$$ and determine an approximate value accurate to $$3$$ decimal places.
5. Use your preceding work to solve for $$h$$ exactly, plus determine an estimate accurate to $$3$$ decimal places.
6. If the surveyors' initial measurements were taken from an elevation of $$78$$ feet above sea level, how high above sea level is the peak of the hill?

## Summary

• The tangent function is defined defined to be the ratio of the sine and cosine functions according to the rule
$\tan(t) = \frac{\sin(t)}{\cos(t)} \nonumber$

for all values of $$t$$ for which $$\cos(t) \ne 0\text{.}$$

• The graph of the tangent function differs substantially from the graphs of the sine and cosine functions, primarily because near values where $$\cos(t) = 0\text{,}$$ the ratio of $$\frac{\sin(t)}{\cos(t)}$$ increases or decreases without bound, producing vertical asymptotes. In addition, while the period of the sine and cosine functions is $$P = 2\pi\text{,}$$ the period of the tangent function is $$P = \pi$$ due to how the sine and cosine functions repeat the same values (with different signs) as a point traverses the unit circle.
• The tangent function finds some of its most important applications in the setting of right triangles where one leg of the triangle is known and one of the non-right angles is known. Computing the tangent of the known angle, say $$\alpha\text{,}$$ and using the fact that
$\tan(\alpha) = \frac{\text{opp}}{\text{adj}} \nonumber$

we can then find the missing leg's length in terms of the other and the tangent of the angle.

## Exercises

##### 7.

A wheelchair ramp is to be built so that the angle it forms with level ground is $$4^{\circ}\text{.}$$ If the ramp is going to rise from a level sidewalk up to a front porch that is $$3$$ feet above the ground, how long does the ramp have to be? How far from the front porch will it meet the sidewalk? What is the slope of the ramp?

##### 8.

A person is flying a kite and at the end of a fixed length of string. Assume there is no slack in the string.

At a certain moment, the kite is $$170$$ feet off the ground, and the angle of elevation the string makes with the ground is $$40^\circ\text{.}$$

1. How far is it from the person flying the kite to another person who is standing directly beneath the kite?
2. How much string is out between the person flying the kite and the kite itself?
3. With the same amount of string out, the angle of elevation increases to $$50^\circ\text{.}$$ How high is the kite at this time?
##### 9.

An airplane is flying at a constant speed along a straight path above a straight road at a constant elevation of $$2400$$ feet. A person on the road observes the plane flying directly at them and uses a sextant to measure the angle of elevation from them to the plane. The first measurement they take records an angle of $$36^\circ\text{;}$$ a second measurement taken $$2$$ seconds later is $$41^\circ\text{.}$$

How far did the plane travel during the two seconds between the two angle measurements? How fast was the plane flying?