8.3 The Reciprocal Functions

Three More Functions

The three basic trigonometric functions occur so often as the denominator of a fraction that it is convenient to give names to their reciprocals. We define three new trigonometric functions as follows.

Definition

Three More Functions

If $θ$ is an angle in standard position, and $P (x, y)$ is a point on the terminal side, then we define the following functions.

$\begin{aligned} The secant : & \sec (θ) = \frac{r}{x} \\ The cosecant : & \csc (θ) = \frac{r}{y} \\ The cotangent : & \cot (θ) = \frac{x}{y} \end{aligned}$

We can find exact values for all six trig functions at a given angle if we know the value of any one of them.

Example 1

If $sec ⁡ open parenthesis θ close parenthesis equals 3$ , and $- \frac{π}{2} \leq θ \leq 0$ , find exact values for the other five trig functions.

Because $- \frac{π}{2} \leq θ \leq 0$ , we draw a reference triangle in the fourth quadrant, as shown at right. Because $\sec (θ) = 3 = \frac{3}{1}$ , we label the horizontal leg with $x equals 1$ and the hypotenuse with $r equals 3$ .

From the Pythagorean theorem, we find $y = - \sqrt{8} = - 2 \sqrt{2}$ . We can now compute the values of the six trigonometric ratios.

$\begin{aligned} \cos (θ) & = \frac{x}{r} = \frac{1}{3} & \sec (θ) = \frac{r}{x} = \frac{3}{1} = 3 \\ \sin (θ) & = \frac{y}{r} = \frac{- 2 \sqrt{2}}{3} & \csc (θ) = \frac{r}{y} = \frac{3}{- 2 \sqrt{2}} = \frac{- 3 \sqrt{2}}{4} \\ \tan (θ) & = \frac{y}{x} = \frac{- 2 \sqrt{2}}{1} = - 2 \sqrt{2} & \cot (θ) = \frac{x}{y} = \frac{1}{- 2 \sqrt{2}} = \frac{- \sqrt{2}}{4} \end{aligned}$

If $csc ⁡ open parenthesis θ close parenthesis equals 4$ , and $90 degrees less than or equal to θ less than or equal to 180 degrees$ , find exact values for the other five trig functions.

$\cos (θ) = \frac{- \sqrt{15}}{4}$ , $\sin (θ) = \frac{1}{4}$ , $\tan (θ) = \frac{- 1}{\sqrt{15}}$ , $\sec (θ) = \frac{- 4}{\sqrt{15}}$ , $\cot (θ) = - \sqrt{15}$

By comparing the definitions of secant, cosecant, and cotangent to the three basic trigonometric functions, we find the following relationships.

Note

Reciprocal Trigonometric Functions

$\begin{matrix} The secant function : \sec (θ) = \frac{1}{\cos (θ)} \\ The cosecant function : \csc (θ) = \frac{1}{\sin (θ)} \\ The cotangent function : \cot (θ) = \frac{1}{\tan (θ)} \end{matrix}$

Calculators do not have keys for the secant, cosecant, and cotangent functions; instead, we calculate their values as reciprocals.

Use a calculator to approximate $csc ⁡ open parenthesis 132 degrees close parenthesis$ to three decimal places.

$1.346$

Of course, we can also evaluate the reciprocal trig functions for angles in radians, or for real numbers. Thus for example,

$\csc (3.5) = \frac{1}{\sin (3.5)} = - 2.8508 and \cot (- 4) = \frac{1}{\tan (- 4)} = - 0.8637$

In particular, the exact values for the reciprocal trig functions of the special angles are easily obtained.

Exact Values for the Special Angles
$θ$	$\sec (θ)$	$\sec (θ)$	$\cot (θ)$
$0$	$1$	undefined	undefined
$\frac{π}{6}$	$\frac{2 \sqrt{3}}{3}$	$2$	$the square root of 3$
$\frac{π}{4}$	$the square root of 2$	$the square root of 2$	$1$
$\frac{π}{3}$	$2$	$\frac{2 \sqrt{3}}{3}$	$\frac{1}{\sqrt{3}}$
$\frac{π}{2}$	undefined	$1$	$0$

Caution

The reciprocal functions are not the same as the inverse trig functions!

For example, $sec ⁡ open parenthesis 0.8 close parenthesis$ is not equal to $\cos^{- 1} (0.8)$ . Remember that $\cos^{- 1} (0.8)$ is an angle, namely the angle whose cosine is 0.8, while $sec ⁡ open parenthesis 0.8 close parenthesis$ is the reciprocal of the cosine of 0.8 radians, or $\frac{1}{\cos (0.8)}$ . You can check on your calculator that

$\cos^{- 1} (0.8) = 0.6435 radians, and \sec (0.8) = 1.4353$

Each of the reciprocal functions is undefined when its denominator is equal to zero. For example, the secant is undefined when $cos ⁡ open parenthesis θ close parenthesis equals 0$ , or when $θ$ is an odd multiple of $90 degrees$ .

For what angles is the cotangent undefined? Give your answers in degrees and in radians.

Multiples of $180 degrees$ , or multiples of $π$ .

Application to Right Triangles

In Chapter 2 we defined three trigonometric ratios for an acute angle; namely, sine, cosine, and tangent. When we take the reciprocals of those ratios, we obtain expressions for the secant, cosecant, and cotangent.

Note

Reciprocal Trigonometric Ratios

If $θ$ is one of the acute angles in a right triangle,

$\begin{aligned} \sec (θ) & = \frac{hypotenuse}{adjacent} \\ \csc (θ) & = \frac{hypotenuse}{opposite} \\ \cot (θ) & = \frac{adjacent}{opposite} \end{aligned}$

Although we can express any relationship between the sides of a right triangle using sine, cosine, and tangent, sometimes it is more convenient to use one of the reciprocal functions.

Example 4

The length, $L$ , of the shadow cast by a flagpole on a sunny day depends on the height, $h$ , of the flagpole and the angle, $θ$ , that the sun's rays make with ground.

Write an expression for the length, $L$ , of the shadow cast by a flagpole of height $h$ when the sun makes an angle of $θ$ from the ground.
Find the length (to the nearest 0.01 meter) of the shadow cast by a 3-meter flagpole when the sun makes an angle of $20 degrees$ from the ground.

From the figure, we see that $\frac{L}{h} = \cot (θ)$ , or $L = h \cot (θ)$ .
Substiting $3$ for $h$ and $20 degrees$ for $θ$ , we find
$L = 3 \cot (20^{\circ}) = 3 (2.7475) = 8.24$
The shadow is about 8.24 meters long.

The area $A$ of a regular polygon with $n$ sides having perimeter $L$ satisfies

$A = \frac{L^{2}}{4 n} \cot \frac{π}{n}$

Refer to the figure at right showing $n equals 6$ to prove this formula in the following steps.

Find an expression for the angle $θ$ in terms of $n$ .
Find an expression for the base of the triangle shown.
Find an expression for the height of the triangle.
Write an expression for the area of the triangle, and then for the area of the entire polygon.

$θ = \frac{π}{n}$
$b = \frac{L}{n}$
$h = \frac{L}{2 n} \cot (\frac{π}{n})$
$A_{T} = \frac{L^{2}}{4 n^{2}} \cot (\frac{π}{n}), A_{P} = \frac{L^{2}}{4 n} \cot (\frac{π}{n})$

Graphs of the Reciprocal Functions

We can obtain graphs of the reciprocal trig functions by plotting points, as we did for the sine, cosine and tangent functions. However, it is more enlightening to construct these graphs as the reciprocals of the three basic functions.

Example 5

Use the graph of $y = \cos (x)$ to construct a graph of $f (x) = \sec (x)$ .

Consider the graph of $y = \cos (x)$ shown at left below.

When $x = \frac{- π}{2}, \frac{π}{2}$ and $\frac{3 π}{2}, \cos (x) = 0$ , so $\sec (x)$ is undefined at these $x$ -values, and we insert vertical asymptotes at those $x$ -values to start our graph of $y = \sec (x)$ , as shown at right below.

To find some points on the graph, we look at points on the graph of $y = \cos (x)$ . At each $x$ -value, the $y$ -coordinate of the point on the graph of $y = \sec (x)$ is the reciprocal of $\cos (x)$ .

For example, at $x equals 0$ and $x equals 2 π$ , we have $cos ⁡ open parenthesis x close parenthesis equals 1$ , so

$\sec (x) = \frac{1}{1} = 1$

Thus, we plot the points $open parenthesis 0, 1 close parenthesis$ and $open parenthesis 2 π, 1 close parenthesis$ on the graph of $f (x) = \sec (x)$ .

Similarly, at $x = - π$ and $x = π$ , $\cos (x) = - 1$ , so the value of $\sec (x)$ is $\frac{1}{- 1} = - 1$ , and we plot the points $(- π, - 1)$ and $open parenthesis π, minus 1 close parenthesis$ on the graph of $f (x) = \sec (x)$ .

Finally, we notice that the values of $\cos (x)$ are decreasing toward $0$ as $x$ increases from $0$ to $\frac{π}{2}$ , so the graph of $f (x) = \sec (x)$ increases toward $∞$ on the same interval.

By similar arguments, we fill in the graph of $f (x) = \sec (x)$ between each of the vertical asymptotes, to produce the graph below.

Use the graph of $y = \tan (x)$ to sketch a graph of $g (x) = \cot (x)$ .

The graphs of the three new functions are shown below, with $x$ in radians. Note that the secant function is undefined at odd multiples of $\frac{π}{2}$ , the values at which $cos ⁡ open parenthesis x close parenthesis equals 0$ . The cosecant is undefined where $sin ⁡ open parenthesis x close parenthesis equals 0$ , namely at multiples of $π$ . The cotangent is also undefined at multiples of $π$ , because $tan ⁡ open parenthesis x close parenthesis equals 0$ at those values.

State the domain and range of the cosecant and cotangent functions.

Domain of cosecant: all real numbers except integer multiples of $π$ ; Range of cosecant: $(- \infty, - 1] \cup [1, \infty)$

Domain of cotangent: all real numbers except integer multiples of $π$ ; Range of cotangent: all real numbers

Solving Equations

From the graph of the secant function, we can see that the equation $\sec (θ) = k$ has two solutions between $0$ and $2 π$ if $k greater than or equal to 1$ or $k less than or equal to minus 1$ , but no solution for $minus 1 less than k less than 1$ . The same is true of the cosecant function: the equation $\csc (θ) = k$ has no solution for $minus 1 less than k less than 1$ .

Example 7

Solve $\csc (θ) = \frac{2 \sqrt{3}}{3}$ for $θ$ between $0$ and $2 π$ .

We take the reciprocal of each side of the equation to obtain

$\sin (θ) = \frac{3}{2 \sqrt{3}} = \frac{\sqrt{3}}{2}$

Because $\frac{\sqrt{3}}{2}$ is one of the special values, we recognize that one of the solutions is $θ = \frac{π}{3}$ . The sine and the cosecant are also positive in the second quadrant, so the second solution is $π - \frac{π}{3} = \frac{2 π}{3}$ .

Solve $\sec (θ) = - 1.6$ for $θ$ between $0$ and $2 π$ .

$θ equals 2.25, 4.04$

Using Identities

All six of the trigonometric ratios are related. If we know one of the ratios, we can use identities to find any of the others.

Example 8

If $sec ⁡ open parenthesis θ close parenthesis equals 3$ , and $- \frac{π}{2} \leq θ \leq 0$ , find an exact value for $\csc (θ)$ .

Because $\sec (θ) = \frac{1}{\cos (θ)}$ , we see that $\frac{1}{\cos (θ)} = 3$ , or $\cos (θ) = \frac{1}{3}$ . We use the Pythagorean identity to find the sine.

$\begin{aligned} \cos^{2} (θ) + \sin^{2} (θ) & = 1 & Substitute \frac{1}{3} for \cos (θ) . \\ {(\frac{1}{3})}^{2} + \sin^{2} (θ) & = 1 & Subtract {(\frac{1}{3})}^{2} = \frac{1}{9} . \\ \sin^{2} (θ) & = 1 - \frac{1}{9} = \frac{8}{9} \end{aligned}$

Because $θ$ lies in the fourth quadrant, where the sine function is negative, we choose the negative square root for $\sin (θ)$ . Once we know $\sin (θ)$ , we calculate its reciprocal to find $\csc (θ)$ .

$\sin (θ) = - \sqrt{\frac{8}{9}} = \frac{- 2 \sqrt{2}}{3}, and \csc (θ) = \frac{1}{\sin (θ)} = \frac{- 3}{2 \sqrt{2}} = \frac{- 3 \sqrt{2}}{4}$

If $\csc (θ) = \frac{- \sqrt{13}}{3}$ , and $π \leq θ \leq \frac{3 π}{2}$ , find an exact value for $\sec (θ)$ .

$\sec (θ) = \frac{- \sqrt{13}}{2}$

Identities are especially useful if the trig ratios are algebraic expressions, rather than numerical values. In the next example, we use the cotangent identity.

Example 9

If $\csc (x) = w$ and $0 < x < \frac{π}{2}$ , find an expression for $\cot (x)$ .

Because the sine is the reciprocal of the cosecant, we have

$\sin (x) = \frac{1}{\csc (x)} = \frac{1}{w}$

. We substitute $\frac{1}{w}$ for $\sin (x)$ in the Pythagorean identity to find

$\cos (x) = \pm \sqrt{1 - \sin^{2} (x)} = \pm \sqrt{1 - {(\frac{1}{w})}^{2}}$

We choose the positive root because cosine is positive in the first quadrant, and simplify to get

$\cos (x) = \sqrt{1 - \frac{1}{w^{2}}} = \sqrt{\frac{w^{2} - 1}{w^{2}}} = \frac{\sqrt{w^{2} - 1}}{| w |}$

We can replace $| w |$ by $w$ in this last expression because $w greater than 0$ . (Do you see why $w greater than 0$ ?) Finally, because the cotangent is the reciprocal of the tangent, we have

$\cot (x) = \frac{\cos (x)}{\sin (x)} = \frac{\frac{w^{2} - 1}{w}}{\frac{1}{w}} = \sqrt{w^{2} - 1}$

If $\sec (t) = \frac{2}{a}$ and $\frac{3 π}{2} < t < 2 π$ , find expressions for $\csc (t)$ and $\cot (t)$ .

$\csc (t) = \frac{- 2}{\sqrt{4 - a^{2}}}$ , $\cot (t) = \frac{- a}{\sqrt{4 - a^{2}}}$

We can often simplify trigonometric expressions by first converting all the trig ratios to sines and cosines.

Example 10

Simplify $\sec (θ) - \tan (θ) \sin (θ)$ .

We replace $\sec (θ)$ by $\frac{1}{\cos (θ)}$ and $\tan (θ)$ by $\frac{\sin (θ)}{\cos (θ)}$ to get

$\begin{aligned} \frac{1}{\cos (θ)} - \frac{\sin (θ)}{\cos (θ)} \cdot \sin (θ) & = \frac{1}{\cos (θ)} - \frac{\sin^{2} (θ)}{\cos (θ)} \\ = \frac{1 - \sin^{2} (θ)}{\cos (θ)} = \frac{\cos^{2} (θ)}{\cos (θ)} = \cos (θ) \end{aligned}$

In the previous example, you can verify that

$\sec (θ) - \tan (θ) \sin (θ) = \cos (θ)$

by graphing the functions $Y_{1} = \sec (θ) - \tan (θ) \sin (θ)$ and $Y sub 2 equals cos ⁡ open parenthesis θ close parenthesis$ to see that they are the same.

Show that $\sin^{2} (x) (1 + \cot^{2} (x)) = 1$ .

$\begin{aligned} \sin^{2} (x) (1 + \cot^{2} (x)) & = \sin^{2} (x) (1 + \frac{\cos^{2} (x)}{\sin^{2} (x)}) \\ = \sin^{2} (x) + \cos^{2} (x) \\ = 1 \end{aligned}$

There are two alternate versions of the Pythagorean identity which involve the reciprocal trig functions. These identities are useful when we know the value of $\tan (θ)$ or $\cot (θ)$ and want to find the other trig values.

You can memorize these identities, but they are easy to derive from the original Pythagorean identity, $\sin^{2} (θ) + \cos^{2} (θ) = 1$ . We will prove them in the Homework problems.

Example 11

If $\tan (α) = \frac{3}{5}$ and $α$ lies in the third quadrant, find exact values for $\sec (α)$ and $\cos (α)$ .

We cannot find the sine and cosine of an angle directly from the value of the tangent; in particular, it is not true that $sin ⁡ open parenthesis α close parenthesis equals 3$ and $cos ⁡ open parenthesis α close parenthesis equals 5$ ! (Do you see why?) Instead, we begin with the Pythagorean identity for the tangent.

$\begin{aligned} \sec^{2} (α) & = 1 + \tan^{2} (α) = 1 + (\frac{3}{5})^{2} \\ = \frac{25}{25} + \frac{9}{25} = \frac{34}{25} \\ \sec (α) & = \pm \sqrt{\frac{34}{25}} = \frac{\pm \sqrt{34}}{5} \end{aligned}$

Because $α$ is in the third quadrant, both its sine and cosine are negative. Therefore the reciprocals of cosine and sine, namely secant and cosecant, must also be negative, and hence $\sec (α) = \frac{- \sqrt{34}}{5}$ . The cosine of $α$ is the reciprocal of the secant, so $\cos (α) = \frac{- 5}{\sqrt{34}}$ .

If $\cot (ϕ) = \frac{- 3}{\sqrt{2}}$ and $ϕ$ lies in the second quadrant, find exact values for $\csc (ϕ)$ and $\sin (ϕ)$ .

$\csc (ϕ) = \sqrt{\frac{11}{2}}$ , $\sin (ϕ) = \sqrt{\frac{2}{11}}$

Activity

Reciprocal Ratios

Part 1

The figure shows a unit circle centered at $O$ . The line through $A C$ is tangent to the circle at $A$ . (Recall that a tangent to a circle is perpendicular to the radius that meets it.)

What is the radius of the circle?
Find line segments whose lengths are $\cos (θ)$ and $\sin (θ)$ .
Explain why $△ O A C$ is similar to $△ O B D$ .
Find a line segment in $△ O A C$ whose length is $\tan (θ)$ . Explain why your choice is correct.
Find a line segment in $△ O A C$ whose length is $\sec (θ)$ . Explain why your choice is correct.
Find a line segment in $△ O A C$ whose length is $\csc (θ)$ . Explain why your choice is correct.

Part 2

Here is another unit circle, with tangent $B D$ .

What is the measure of the angles $∠ O C D$ and $∠ O C B$ ?
Explain why $△ O C B$ is similar to $△ O A C$ .
Find a line segment in $△ O C B$ whose length is $\tan (θ)$ . Explain why your choice is correct.
Find a line segment in $△ O C B$ whose length is $\sec (θ)$ . Explain why your choice is correct.
Explain why $△ O C B$ is similar to $△ D C O$ .
Find a line segment in $△ D C O$ whose length is $\csc (θ)$ . Explain why your choice is correct.

Review the following skills you will need for this section.

Algebra Refresher

Solve $| 2 x - 6 | = 4$
Solve $| 1 - 3 x | = 10$
Simplify $\sqrt{(x - 4)^{2}}$
Simplify $\sqrt{(1 - x)^{2}}$
For what values of $x$ is $| x + 2 | = x + 2$ ?
For what values of $x$ is $| x - 3 | = 3 - x$ ?
Graph the function $f (x) = \frac{x}{| x |}$ .
Explain the difference between the graphs of $f (x) = (\sqrt{x})^{2}$ and $g (x) = \sqrt{x^{2}}$ .

$\underset{―}{}$

Algebra Refresher Answers

$x equals 1, 5$
$x = - 3, \frac{11}{3}$
$| x minus 4 |$
$| 1 minus x |$
$x greater than or equal to minus 2$
$x less than or equal to 3$
The domain of $f$ is $[ 0, ∞ close parenthesis$ , and the domain of $g$ is $open parenthesis minus ∞, ∞ close parenthesis$ .

Section 8.3 Summary

Vocabulary

Reciprocal
Secant
Cosecant
Cotangent

Concepts

Note
Three More Functions
If $θ$ is an angle in standard position, and $P (x, y)$ is a point on the terminal side, then we define the following functions.
$The secant : \sec (θ) = \frac{r}{x}$
$The cosecant : \csc (θ) = \frac{r}{y}$
$The cotangent : \cot (θ) = \frac{x}{y}$
Note
Reciprocal Trigonometric Ratios
If $θ$ is one of the acute angles in a right triangle,
$\begin{aligned} \sec (θ) & = \frac{hypotenuse}{adjacent} \\ \csc (θ) & = \frac{hypotenuse}{opposite} \\ \cot (θ) & = \frac{adjacent}{opposite} \end{aligned}$
Note
Reciprocal Trigonometric Functions
$The secant function : \sec (θ) = \frac{1}{\cos (θ)}$
$The cosecant function : \csc (θ) = \frac{1}{\sin (θ)}$
$The cotangent function : \cot (θ) = \frac{1}{\tan (θ)}$
We can obtain graphs of the secant, cosecant, and cotangent functions as the reciprocals of the three basic functions.
We can solve equations of the form $\sec (θ) = k$ , $\csc (θ) = k$ , and $\cot (θ) = k$ by taking the reciprocal of both sides.
If we know one of the trigonometric ratios for an angle, we can use identities to find any of the others.
Note
Cotangent Identity
$\cot (θ) = \frac{1}{\tan (θ)} = \frac{\cos (θ)}{\sin (θ)}, \sin (θ) \neq 0$
Note
Two More Pythagorean Identities
$1 + \tan^{2} (θ) = \sec^{2} (θ) 1 + \cot^{2} (θ) = \csc^{2} (θ)$
We can often simplify trigonometric expressions by first converting all the trig ratios to sines and cosines.

Study Questions

Delbert says that $\sec (x)$ is just another way of writing $\cos^{- 1} (x)$ , because $\cos^{- 1} (x) = \frac{1}{\cos (x)}$ . Is he correct? Explain your reasoning.
Each of the following functions is related to the sine function in a different way. Explain how.
$\cos (x), \csc (x), and \sin^{- 1} (x)$
Using Study Question #2 as an example, name three functions related to the tangent function, and explain how they are related.
Why do the graphs of $y = \csc (x)$ and $y = \cot (x)$ have vertical asymptotes at the same $x$ -values?

Skills

Evaluate the reciprocal trig functions for angles in degrees or radians #1–20
Find values or expressions for the six trig ratios #21–28
Evaluate the reciprocal trig functions in applications #29–32
Given one trig ratio, find the others #33–46, 71–80
Evaluate expressions exactly #47–52
Graph the secant, cosecant, and cotangent functions #53–58
Identify graphs of the reciprocal trig functions #59–64
Solve equations in secant, cosecant, and cotangent #65–70
Use identities to simplify or evaluate expressions #81–94

Homework 8-3

For Problems 1–8, evaluate. Round answers to 3 decimal places.

$csc ⁡ open parenthesis 27 degrees close parenthesis$

$2.203$

$sec ⁡ open parenthesis 8 degrees close parenthesis$

$cot ⁡ open parenthesis 65 degrees close parenthesis$

$0.466$

$csc ⁡ open parenthesis 11 degrees close parenthesis$

$sec ⁡ open parenthesis 1.4 close parenthesis$

$5.883$

$cot ⁡ open parenthesis 4.3 close parenthesis$

$\csc (\frac{5 π}{16})$

$1.203$

$\sec (\frac{7 π}{20})$

For Problems 9–16, evaluate. Give exact values.

$csc ⁡ open parenthesis 30 degrees close parenthesis$

$2$

$sec ⁡ open parenthesis 0 degrees close parenthesis$

$cot ⁡ open parenthesis 45 degrees close parenthesis$

$1$

$csc ⁡ open parenthesis 60 degrees close parenthesis$

$sec ⁡ open parenthesis 150 degrees close parenthesis$

$\frac{- 2 \sqrt{3}}{3}$

$cot ⁡ open parenthesis 120 degrees close parenthesis$

$csc ⁡ open parenthesis 135 degrees close parenthesis$

$the square root of 2$

$sec ⁡ open parenthesis 270 degrees close parenthesis$

For Problems 17–18, complete the tables with exact values.

$θ$	$0$	$\frac{π}{6}$	$\frac{π}{4}$	$\frac{π}{3}$	$\frac{π}{2}$	$\frac{2 π}{3}$	$\frac{3 π}{4}$	$\frac{5 π}{6}$	$π$
$\sec (θ)$
$\csc (θ)$
$\cot (θ)$

$θ$	$0$	$\frac{π}{6}$	$\frac{π}{4}$	$\frac{π}{3}$	$\frac{π}{2}$	$\frac{2 π}{3}$	$\frac{3 π}{4}$	$\frac{5 π}{6}$	$π$
$\sec (θ)$	$1$	$\frac{2 \sqrt{3}}{3}$	$the square root of 2$	$2$	undefined	$minus 2$	$- \sqrt{2}$	$- \frac{2 \sqrt{3}}{3}$	$minus 1$
$\csc (θ)$	undefined	$2$	$the square root of 2$	$\frac{2 \sqrt{3}}{3}$	$1$	$\frac{2 \sqrt{3}}{3}$	$the square root of 2$	$2$	undefined
$\cot (θ)$	undefined	$the square root of 3$	$1$	$\frac{\sqrt{3}}{3}$	$0$	$\frac{- \sqrt{3}}{3}$	$minus 1$	$- \sqrt{3}$	undefined

$θ$	$π$	$\frac{7 π}{6}$	$\frac{5 π}{4}$	$\frac{4 π}{3}$	$\frac{3 π}{2}$	$\frac{5 π}{3}$	$\frac{7 π}{4}$	$\frac{11 π}{6}$	$2 π$
$\sec (θ)$
$\csc (θ)$
$\cot (θ)$

Evaluate. Round answers to three decimal places.

$cos ⁡ open parenthesis 0.2 close parenthesis$
$(\cos (0.2))^{- 1}$
$\cos^{- 1} (0.2)$
$\frac{1}{\cos (0.2)}$
$\cos (\frac{1}{0.2})$
$sec ⁡ open parenthesis 0.2 close parenthesis$

$0.980$
$1.020$
$1.369$
$1.020$
$0.284$
$1.020$

Evaluate. Round answers to three decimal places.

$tan ⁡ open parenthesis 3.2 close parenthesis$
$\tan^{- 1} (3.2)$
$cot ⁡ open parenthesis 3.2 close parenthesis$
$\frac{1}{\tan (3.2)}$
$\tan (\frac{1}{3.2})$
$(\tan (3.2))^{- 1}$

For Problems 21–28, find exact values for the six trigonometric ratios of the angle $θ$ .

$\sin (θ) = \frac{4}{5}$ , $\cos (θ) = \frac{3}{5}$ , $\tan (θ) = \frac{4}{3}$ , $\sec (θ) = \frac{5}{3}$ , $\csc (θ) = \frac{5}{4}$ , $\cot (θ) = \frac{3}{4}$

$\sin (θ) = \frac{4}{\sqrt{41}}$ , $\cos (θ) = \frac{5}{\sqrt{41}}$ , $\tan (θ) = \frac{4}{5}$ , $\sec (θ) = \frac{\sqrt{41}}{5}$ , $\csc (θ) = \frac{\sqrt{41}}{4}$ , $\cot (θ) = \frac{5}{4}$

$\sin (θ) = \frac{5}{\sqrt{74}}$ , $\cos (θ) = \frac{- 7}{\sqrt{74}}$ , $\tan (θ) = \frac{- 5}{7}$ , $\sec (θ) = \frac{- \sqrt{74}}{7}$ , $\csc (θ) = \frac{\sqrt{74}}{5}$ , $\cot (θ) = \frac{- 7}{5}$

$\sin (θ) = \frac{- 5}{8}$ , $\cos (θ) = \frac{\sqrt{39}}{8}$ , $\tan (θ) = \frac{5}{\sqrt{39}}$ , $\sec (θ) = \frac{- 8}{\sqrt{39}}$ , $\csc (θ) = \frac{- 8}{5}$ , $\cot (θ) = \frac{\sqrt{39}}{5}$

The distance that sunlight must travel to pass through a layer of Earth's atmosphere depends on both the thickness of the atmosphere and the angle of the sun.

Write an expression for the distance, $d$ , that sunlight travels through a layer of atmosphere of thickness $h$ .
Find the distance (to the nearest mile) that sunlight travels through a 100-mile layer of atmosphere when the sun is $40 degrees$ above the horizon.

$d = h \csc (θ)$
155.572 miles

In railroad design, the degree of curvature of a section of track is the angle subtended by a chord 100 feet long.

Use the figure to write an expression for the radius, $r$ , of a curve whose degree of curvature is $θ$ . (Hint: The bisector of the angle $θ$ is perpendicular to the chord.)
Find the radius of a curve whose degree of curvature is $43 degrees$ .

When a plane is tilted by an angle $θ$ from the horizontal, the time required for a ball starting from rest to roll a horizontal distance of $l$ feet on the plane is

$t = \sqrt{\frac{l}{8} \csc (2 θ)} seconds$

How long, to the nearest 0.01 second, will it take the ball to roll 2 feet horizontally on a plane tilted by $12 degrees$ ?
Solve the formula for $l$ in terms of $t$ and $θ$ .

0.78 sec
$l equals 8 t squared sin ⁡ open parenthesis 2 θ close parenthesis$

After a heavy rainfall, the depth, $D$ , of the runoff flow at a distance $x$ feet from the watershed down a slope at angle $α$ is given by

$D = (k x)^{0.6} (\cot (α))^{0.3} inches$

where $k$ is a constant determined by the surface roughness and the intensity of the runoff.

How deep, to the nearest 0.01 inch, is the runoff 100 feet down a slope of $10 degrees$ if $k equals 0.0006$ ?
Solve the formula for $x$ in terms of $D$ and $α$ .

For Problems 33–38, write algebraic expressions for the six trigonometric ratios of the angle $θ$ .

$\sin (θ) = \frac{7}{\sqrt{x^{2} + 49}}$ , $\cos (θ) = \frac{x}{\sqrt{x^{2} + 49}}$ , $\tan (θ) = \frac{7}{x}$ , $\sec (θ) = \frac{\sqrt{x^{2} + 49}}{x}$ , $\csc (θ) = \frac{\sqrt{x^{2} + 49}}{7}$ , $\cot (θ) = \frac{x}{7}$

$\sin (θ) = S$ , $\cos (θ) = \sqrt{1 - S^{2}}$ , $\tan (θ) = \frac{S}{\sqrt{1 - S^{2}}}$ , $\sec (θ) = \frac{1}{\sqrt{1 - S^{2}}}$ , $\csc (θ) = \frac{1}{S}$ , $\cot (θ) = \frac{\sqrt{1 - S^{2}}}{S}$

$\sin (θ) = \frac{- \sqrt{9 - a^{2}}}{3}$ , $\cos (θ) = \frac{a}{3}$ , $\tan (θ) = \frac{- \sqrt{9 - a^{2}}}{a}$ , $\sec (θ) = \frac{3}{a}$ , $\csc (θ) = \frac{- 3}{\sqrt{9 - a^{2}}}$ , $\cot (θ) = \frac{- a}{\sqrt{9 - a^{2}}}$

The diagram shows a unit circle. Find six line segments whose lengths are, respectively, $\sin (t), \cos (t), \tan (t), \sec (t), \csc (t),$ and $\cot (t)$ .

$A C, O A, B D, O D, O E, E F$

Use the figure in Problem 39 to find each area in terms of the angle $t$ .

$△ O A C$
$△ O B D$
sector $O B C$
$△ O F E$

For Problems 41–46, sketch the reference angle, and find exact values for all six trigonometric functions of the angle.

$sec ⁡ open parenthesis θ close parenthesis equals 2, θ$ in Quadrant IV

$\sin (θ) = \frac{- \sqrt{3}}{2}$ , $\cos (θ) = \frac{1}{2}$ , $\tan (θ) = - \sqrt{3}$ , $sec ⁡ open parenthesis θ close parenthesis equals 2$ , $\csc (θ) = \frac{- 2 \sqrt{3}}{3}$ , $\cot (θ) = \frac{- \sqrt{3}}{3}$

$csc ⁡ open parenthesis ϕ close parenthesis equals 4, ϕ$ in Quadrant II

$csc ⁡ open parenthesis α close parenthesis equals 3, α$ in Quadrant I

$\sin (α) = \frac{1}{3}$ , $\cos (α) = \frac{2 \sqrt{2}}{3}$ , $\tan (α) = \frac{\sqrt{2}}{4}$ , $\sec (α) = \frac{3 \sqrt{2}}{4}$ , $csc ⁡ open parenthesis α close parenthesis equals 3$ , $\cot (α) = 2 \sqrt{2}$

$sec ⁡ open parenthesis β close parenthesis equals 4, β$ in Quadrant IV

$\cot (γ) = \frac{1}{4}, γ$ in Quadrant III

$\sin (γ) = \frac{- 4}{\sqrt{17}}$ , $\cos (γ) = \frac{- 1}{\sqrt{17}}$ , $tan ⁡ open parenthesis γ close parenthesis equals 4$ , $\sec (γ) = - \sqrt{17}$ , $\csc (γ) = \frac{- \sqrt{17}}{4}$ , $\cot (γ) = \frac{1}{4}$

$tan ⁡ open parenthesis θ close parenthesis equals 6, θ$ in Quadrant I

For Problems 47–52, evaluate.

$4 \cot (\frac{π}{3}) + 2 \sec (\frac{π}{4})$

$\frac{4 \sqrt{3}}{3} + 2 \sqrt{2}$

$\frac{1}{2} \csc (\frac{π}{6}) - \frac{1}{4} \cot (\frac{π}{6})$

$\frac{1}{2} \csc (\frac{5 π}{3}) \cot (\frac{3 π}{4})$

$\frac{\sqrt{3}}{3}$

$6 \cot (\frac{7 π}{6}) \sec (\frac{5 π}{4})$

${[\csc (\frac{2 π}{3}) - \sec (\frac{3 π}{4})]}^{2}$

$\frac{4 \sqrt{6}}{3} + \frac{10}{3}$

$\sec^{2} (\frac{5 π}{6}) \csc^{2} (\frac{4 π}{3})$

Complete the table and sketch a graph of $y = \sec (x)$ .

$x$	$0$	$\frac{π}{4}$	$\frac{π}{2}$	$\frac{3 π}{4}$	$π$	$\frac{5 π}{4}$	$\frac{3 π}{2}$	$\frac{7 π}{4}$	$2 π$
$\sec (x)$

$x$	$0$	$\frac{π}{4}$	$\frac{π}{2}$	$\frac{3 π}{4}$	$π$	$\frac{5 π}{4}$	$\frac{3 π}{2}$	$\frac{7 π}{4}$	$2 π$
$\sec (x)$	$1$	$the square root of 2$	undefined	$- \sqrt{2}$	$minus 1$	$- \sqrt{2}$	undefined	$the square root of 2$	$1$

Complete the table and sketch a graph of $y = \csc (x)$ .

$x$	$0$	$\frac{π}{4}$	$\frac{π}{2}$	$\frac{3 π}{4}$	$π$	$\frac{5 π}{4}$	$\frac{3 π}{2}$	$\frac{7 π}{4}$	$2 π$
$\csc (x)$

Use the graph of $y = \sin (x)$ to sketch a graph of its reciprocal, $y = \csc (x)$ .

Use the graph of $y = \cos (x)$ to sketch a graph of its reciprocal, $y = \sec (x)$ .

Complete the table and sketch a graph of $y = \cot (x)$ .

$x$	$0$	$\frac{π}{4}$	$\frac{π}{2}$	$\frac{3 π}{4}$	$π$	$\frac{5 π}{4}$	$\frac{3 π}{2}$	$\frac{7 π}{4}$	$2 π$
$\cot (x)$

$x$	$0$	$\frac{π}{4}$	$\frac{π}{2}$	$\frac{3 π}{4}$	$π$	$\frac{5 π}{4}$	$\frac{3 π}{2}$	$\frac{7 π}{4}$	$2 π$
$\cot (x)$	undefined	$1$	$0$	$minus 1$	undefined	$1$	$0$	$minus 1$	undefined

Use the graphs of $y = \cos (x)$ and $y = \sin (x)$ to sketch a graph of $y = \cot (x) = \frac{\cos (x)}{\sin (x)}$ .

For Problems 59–64,

Graph each function for $0 less than or equal to x less than or equal to 2 π$ , and write a simpler expression for the function.
Show algebraically that your new expression is equivalent to the original one.

$y = \frac{\csc (x)}{\cot (x)}$

$\begin{aligned} \frac{\csc (x)}{\cot (x)} & = \frac{\frac{1}{\sin (x)}}{\frac{\cos (x)}{\sin (x)}} \\ = \frac{1}{\sin (x)} \div \frac{\cos (x)}{\sin (x)} \\ = \frac{1}{\sin (x)} \cdot \frac{\sin (x)}{\cos (x)} \\ = \frac{1}{\cos (x)} \\ = \sec (x) \end{aligned}$

$y = \frac{\sec (x)}{\tan (x)}$

$y = \frac{\sec (x) \cot (x)}{\csc (x)}$

$\frac{\sec (x) \cot (x)}{\csc (x)} = \frac{\frac{1}{\cos (x)} \cdot \frac{\cos (x)}{\sin (x)}}{\frac{1}{\sin (x)}} = \frac{\frac{1}{\sin (x)}}{\frac{1}{\sin (x)}} = 1$

$y = \frac{\csc (x) \tan x}{\sec (x)}$

$y = \tan (x) \csc (x)$

$\tan (x) \csc (x) = \frac{\sin (x)}{\cos (x)} \cdot \frac{1}{\sin (x)} = \frac{1}{\cos (x)} = \sec (x)$

$y = \sin (x) \sec (x)$

For Problems 65–70, find all solutions between $0$ and $2 π$ .

$3 \csc (θ) + 2 = 8$

$\frac{π}{6}, \frac{5 π}{6}$

$- 2 \sec (θ) + 7 = 3$

$\sqrt{2} \sec (θ) = - 2$

$\frac{3 π}{4}, \frac{5 π}{4}$

$8 + \csc (θ) = 6$

$2 \cot (θ) = - \sqrt{12}$

$\frac{5 π}{6}, \frac{11 π}{6}$

$\sqrt{3} \cot (θ) = 1$

For Problems 71–76, use identities to find exact values or to write algebraic expressions.

If $\tan (α) = - 2$ and $\frac{π}{2} < α < π$ , find $\cos (α)$ .

$\frac{- \sqrt{5}}{5}$

If $\cot (β) = \frac{5}{4}$ and $π < β < \frac{3 π}{2}$ , find $\sin (β)$ .

If $\sec (x) = \frac{a}{2}$ and $0 < x < \frac{π}{2}$ , find $\tan (x)$ .

$\frac{\sqrt{a^{2} - 4}}{2}$

If $\csc (y) = \frac{1}{b}$ and $\frac{π}{2} < y < π$ , find $\cot (y)$ .

If $\csc (ϕ) = w$ and $\frac{3 π}{2} < ϕ < 2 π$ , find $\cos ϕ$ .

$\frac{\sqrt{w^{2} - 1}}{- w}$

If $\sec (θ) = \frac{3}{z}$ and $π < θ < \frac{3 π}{2}$ , find $\sin (θ)$ .

For Problems 77–80, find exact values for $\sec (s), \csc (s),$ and $\cot (s)$ .

$\sec (s) = \frac{- 5}{4}$ , $\csc (s) = \frac{5}{3}$ , $\cot (s) = \frac{- 4}{3}$

$\sec (s) = \frac{1}{\sqrt{1 - w^{2}}}$ , $\csc (s) = \frac{1}{w}$ , $\cot (s) = \frac{\sqrt{1 - w^{2}}}{w}$

For Problems 81–88, write the expression in terms of sine and cosine, and simplify.

$\sec (θ) \tan (θ)$

$\frac{\sin (θ)}{\cos^{2} (θ)}$

$\csc (ϕ) \cot (ϕ)$

$\frac{\csc (t)}{\cot (t)}$

$\sec (t)$

$\frac{\tan (v)}{\sec (v)}$

$\sec (β) - \tan (β)$

$\frac{1 - \sin (β)}{\cos (β)}$

$\cot (α) + \csc (α)$

$\sin (x) \tan (x) - \sec (x)$

$- \cos (x)$

$\csc y - \cos y \cot y$

Prove the Pythagorean identity $1 + \tan^{2} (θ) = \sec^{2} (θ)$ . (Hint: Start with the identity $\cos^{2} (θ) + \sin^{2} (θ) = 1$ and divide both sides of the equation by $\cos^{2} (θ)$ .)

$\begin{aligned} \cos^{2} (θ) + \sin^{2} (θ) & = 1 \\ \frac{\cos^{2} (θ)}{\cos^{2} (θ)} + \frac{\sin^{2} (θ)}{\cos^{2} (θ)} & = \frac{1}{\cos^{2} (θ)} \\ 1 + \tan^{2} (θ) & = \sec^{2} (θ) \end{aligned}$

Prove the Pythagorean identity $1 + \cot^{2} (θ) = \csc^{2} (θ)$ . (Hint: Start with the identity $\cos^{2} (θ) + \sin^{2} (θ) = 1$ and divide both sides of the equation by $\sin^{2} (θ)$ .)

Suppose that $cot ⁡ open parenthesis θ close parenthesis equals 5$ and $θ$ lies in the third quadrant.

Use the Pythagorean identity to find the value of $\csc (θ)$ .
Use identities to find the values of the other four trig functions of $θ$ .

$\csc (θ) = - \sqrt{26}$
$\sin (θ) = \frac{- \sqrt{26}}{26}, \cos (θ) = \frac{- 5 \sqrt{26}}{26}, \tan (θ) = \frac{1}{5}, \sec (θ) = \frac{- \sqrt{26}}{5}$

Suppose that $\tan (θ) = - 2$ and $θ$ lies in the second quadrant.

Use the Pythagorean identity to find the value of $\sec (θ)$ .
Use identities to find the values of the other four trig functions of $θ$ .

Write each of the other five trig functions in terms of $\sin (t)$ only.

$\cos (t) = \pm \sqrt{1 - \sin^{2} (t)}$ , $\tan (t) = \frac{\pm \sin (t)}{\sqrt{1 - \sin^{2} (t)}}$ , $\sec (t) = \frac{\pm 1}{\sqrt{1 - \sin^{2} (t)}}$ , $\csc (t) = \frac{1}{\sin (t)}$ , $\cot (t) = \frac{\pm \sqrt{1 - \sin^{2} (t)}}{\sin (t)}$

Write each of the other five trig functions in terms of $\cos (t)$ only.

Show that if the angles of a triangle are $A, B,$ and $C$ and the opposite sides are respectively $a, b,$ and $c,$ then

$a \csc (A) = b \csc (B) = c \csc (C)$

$\begin{aligned} \frac{a}{\sin (A)} & = \frac{b}{\sin (B)} = \frac{c}{\sin (C)} \\ a \cdot \frac{1}{\sin (A)} & = b \cdot \frac{1}{\sin (B)} = c \cdot \frac{1}{\sin (C)} \\ a \csc (A) & = b \csc (B) = c \csc (C) \end{aligned}$

The figure shows a unit circle and an angle $θ$ in standard position. Each of the six trigonometric ratios for $θ$ is represented by the length of a line segment in the figure. Find the line segment for each ratio, and explain your choice.

Trigonometry by Katherine Yoshiwara (yoshiwarabooks.org), GNU Free Documentation License 1.2 or later. Adapted for the XYZ HTML edition with the authors' permission (recorded 2026-07-04). License: GFDL-1.2-or-later.