7.4 The Other Trigonometric Functions – fix background – do the todos

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7.4 The Other Trigonometric Functions – fix background – do the todos

Learning Objectives

After completing this section, you should be able to…

Find exact values of the trigonometric functions secant, cosecant, tangent, and cotangent of $\frac{\pi}{3}, \frac{\pi}{4},$ and $\frac{\pi}{6}.$
Use reference angles to evaluate the trigonometric functions secant, tangent, and cotangent.
Use properties of even and odd trigonometric functions.
Recognize and use fundamental identities.
Evaluate trigonometric functions with a calculator.

A wheelchair ramp that meets the standards of the Americans with Disabilities Act must make an angle with the ground whose tangent is $\frac{1}{12}$ or less, regardless of its length. A tangent represents a ratio, so this means that for every 1 inch of rise, the ramp must have 12 inches of run. Trigonometric functions allow us to specify the shapes and proportions of objects independent of exact dimensions. We have already defined the sine and cosine functions of an angle. Though sine and cosine are the trigonometric functions most often used, there are four others. Together they make up the set of six trigonometric functions. In this section, we will investigate the remaining functions.

Finding Exact Values of the Trigonometric Functions Secant, Cosecant, Tangent, and Cotangent

We can also define the remaining functions in terms of the unit circle with a point $(x,y)$ corresponding to an angle of $t,$ as shown in Figure 1. As with the sine and cosine, we can use the $(x,y)$ coordinates to find the other functions.

This image is a graph of circle with angle of t inscribed and a radius of 1. Point of (x, y) is at intersection of terminal side of angle and edge of circle. — Figure 1

The first function we will define is the tangent. The tangent of an angle is the ratio of the y-value to the x-value of the corresponding point on the unit circle. In Figure 1, the tangent of angle $t$ is equal to $\frac{y}{x}, x \neq 0.$ Because the y-value is equal to the sine of $t,$ and the x-value is equal to the cosine of $t,$ the tangent of angle $t$ can also be defined as $\frac{\sin t}{\cos t}, \cos t \neq 0.$ The tangent function is abbreviated as $\tan.$ The remaining three functions can all be expressed as reciprocals of functions we have already defined.

The secant function is the reciprocal of the cosine function. In Figure 1, the secant of angle $t$ is equal to $\frac{1}{\cos t} = \frac{1}{x}, x \neq 0.$ The secant function is abbreviated as $\sec.$
The cotangent function is the reciprocal of the tangent function. In Figure 1, the cotangent of angle $t$ is equal to $\frac{\cos t}{\sin t} = \frac{x}{y}, y \neq 0.$ The cotangent function is abbreviated as $\cot.$
The cosecant function is the reciprocal of the sine function. In Figure 1, the cosecant of angle $t$ is equal to $\frac{1}{\sin t} = \frac{1}{y}, y \neq 0.$ The cosecant function is abbreviated as $\csc.$

If $t$ is a real number and $(x,y)$ is a point where the terminal side of an angle of $t$ radians intercepts the unit circle, then $\begin{array}{rcl} \text{tan } t & =& \frac{y}{x}, x\neq 0 \\ \text{sec } t & = & \frac{1}{x}, x\neq 0 \\ \text{csc } t & = & \frac{1}{y}, y\neq 0 \\ \text{cot } t & = & \frac{x}{y}, y\neq 0 \end{array}$

Example 1

Finding Trigonometric Functions from a Point on the Unit Circle

The point $\left(-\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$ is on the unit circle, as shown in Figure 2. Find $\sin t, \cos t, \tan t, \sec t, \csc t,$ and $\cot t.$

This is an image of a graph of circle with angle of t inscribed and with radius 1. Point of (negative square root of 3 over 2, 1/2) is at intersection of terminal side of angle and edge of circle. — Figure 2

Solution

Because we know the $(x,y)$ coordinates of the point on the unit circle indicated by angle $t,$ we can use those coordinates to find the six functions:

$\begin{array}{rllcll} \sin t & = y & = \frac{1}{2} \\ \cos t & = x & = -\frac{\sqrt{3}}{2} \\ \tan t & = \frac{y}{x} & = \frac{\frac{1}{2}}{-\frac{\sqrt{3}}{2}} & = \frac{1}{2} \left(-\frac{2}{\sqrt{3}}\right) & = -\frac{1}{\sqrt{3}} & = -\frac{\sqrt{3}}{3} \\ \sec t & = \frac{1}{x} & = \frac{1}{-\frac{\sqrt{3}}{2}} & = -\frac{2}{\sqrt{3}} & = -\frac{2\sqrt{3}}{3} \\ \csc t & = \frac{1}{y} & = \frac{1}{\frac{1}{2}} & = 2 \\ \cot t & = \frac{x}{y} & = \frac{-\frac{\sqrt{3}}{2}}{\frac{1}{2}} & = -\frac{\sqrt{3}}{2}\left(\frac{2}{1}\right) & = -\sqrt{3} \end{array}$

Try It 1

The point $\left(\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\right)$ is on the unit circle, as shown in Figure 3. Find $\sin t, \cos t, \tan t, \sec t, \csc t,$ and $\cot t.$

This is an image of a graph of circle with angle of t inscribed with radius 1. Point of (square root of 2 over 2, negative square root of 2 over 2) is at intersection of terminal side of angle and edge of circle. — Figure 3

Show Solution

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

Example 2

Finding the Trigonometric Functions of an Angle

Find $\sin t, \cos t, \tan t, \sec t, \csc t,$ and $\cot t.$ when $t = \frac{\pi}{6}.$

Solution

We have previously used the properties of equilateral triangles to demonstrate that $\sin \frac{\pi}{6} = \frac{1}{2}$ and $\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}.$ We can use these values and the definitions of tangent, secant, cosecant, and cotangent as functions of sine and cosine to find the remaining function values.

$\begin{array}{rcll} \tan \frac{\pi}{6} & = & \frac{\sin \frac{\pi}{6}}{\cos \frac{\pi}{6}} \\ & = & \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}} & = \frac{\sqrt{3}}{3} \\ \sec \frac{\pi}{6} & = & \frac{1}{\cos \frac{\pi}{6}} \\ & = & \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}} & = \frac{2\sqrt{3}}{3} \\ \csc \frac{\pi}{6} & = & \frac{1}{\sin \frac{\pi}{6}} = \frac{1}{\frac{1}{2}} & = 2 \\ \cot \frac{\pi}{6} & = & \frac{\cos \frac{\pi}{6}}{\sin \frac{\pi}{6}} \\ & = & \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} & = \sqrt{3} \end{array}$

Try It 2

Find $\sin t, \cos t, \tan t, \sec t, \csc t,$ and $\cot t.$ when $t = \frac{\pi}{3}.$

Show Solution

$\sin \frac{\pi}{3}= \frac{\sqrt{3}}{2}, \cos \frac{\pi}{3}= \frac{1}{2}, \tan \frac{\pi}{3}= \sqrt{3}, \sec \frac{\pi}{3}= 2, \csc \frac{\pi}{3}= \frac{2\sqrt{3}}{3}, \cot \frac{\pi}{3}= \frac{\sqrt{3}}{3}$

Because we know the sine and cosine values for the common first-quadrant angles, we can find the other function values for those angles as well by setting $x$ equal to the cosine and $y$ equal to the sine and then using the definitions of tangent, secant, cosecant, and cotangent. The results are shown in Table 1.

Angle	$0$	$\frac{\pi}{6}, \text{ or } 30^{\circ}$	$\frac{\pi}{4}, \text{ or } 45^{\circ}$	$\frac{\pi}{3}, \text{ or } 60^{\circ}$	$\frac{\pi}{2}, \text{ or } 90^{\circ}$
Cosine	1	$\frac{\sqrt{3}}{2}$	$\frac{\sqrt{2}}{2}$	$\frac{1}{2}$	0
Sine	0	$\frac{1}{2}$	$\frac{\sqrt{2}}{2}$	$\frac{\sqrt{3}}{2}$	1
Tangent	0	$\frac{\sqrt{3}}{3}$	1	$\sqrt{3}$	Undefined
Secant	1	$\frac{2\sqrt{3}}{3}$	$\sqrt{2}$	2	Undefined
Cosecant	Undefined	2	$\sqrt{2}$	$\frac{2\sqrt{3}}{3}$	1
Cotangent	Undefined	$\sqrt{3}$	1	$\frac{\sqrt{3}}{3}$	0

Using Reference Angles to Evaluate Tangent, Secant, Cosecant, and Cotangent

We can evaluate trigonometric functions of angles outside the first quadrant using reference angles as we have already done with the sine and cosine functions. The procedure is the same: Find the reference angle formed by the terminal side of the given angle with the horizontal axis. The trigonometric function values for the original angle will be the same as those for the reference angle, except for the positive or negative sign, which is determined by x– and y-values in the original quadrant. Figure 4 shows which functions are positive in which quadrant.

To help remember which of the six trigonometric functions are positive in each quadrant, we can use the mnemonic phrase “A Smart Trig Class.” Each of the four words in the phrase corresponds to one of the four quadrants, starting with quadrant I and rotating counterclockwise. In quadrant I, which is “A,” all of the six trigonometric functions are positive. In quadrant II, “Smart,” only sine and its reciprocal function, cosecant, are positive. In quadrant III, “Trig,” only tangent and its reciprocal function, cotangent, are positive. Finally, in quadrant IV, “Class,” only cosine and its reciprocal function, secant, are positive.

This image is a graph of circle with each quadrant labeled. Under quadrant I, labels for sin t, cos t, tan t, sec t, csc t, and cot t. Under quadrant II, labels for sin t and csc t. Under quadrant III, labels for tan t and cot t. Under quadrant IV, labels for cos t, sec t. — Figure 4 The trigonometric functions are each listed in the quadrants in which they are positive.

Given an angle not in the first quadrant, use reference angles to find all six trigonometric functions.

Measure the angle formed by the terminal side of the given angle and the horizontal axis. This is the reference angle.
Evaluate the function at the reference angle.
Observe the quadrant where the terminal side of the original angle is located. Based on the quadrant, determine whether the output is positive or negative.

Example 3

Using Reference Angles to Find Trigonometric Functions

Use reference angles to find all six trigonometric functions of $-\frac{5\pi}{6}.$

Solution

The angle between this angle’s terminal side and the x-axis is $\frac{\pi}{6},$ so that is the reference angle. Since $-\frac{5\pi}{6}$ is in the third quadrant, where both $x$ and $y$ are negative, cosine, sine, secant, and cosecant will be negative, while tangent and cotangent will be positive.

$\begin{array}{l} \cos \left(-\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2}, \sin \left(-\frac{5\pi}{6}\right) = -\frac{1}{2}, \tan \left(-\frac{5\pi}{6}\right) = \frac{\sqrt{3}}{3}, \\ \sec \left(-\frac{5\pi}{6}\right) = -\frac{2\sqrt{3}}{3}, \csc \left(-\frac{5\pi}{6}\right) = -2, \cot \left(-\frac{5\pi}{6}\right) = \sqrt{3} \end{array}$

Try It 3

Use reference angles to find all six trigonometric functions of $-\frac{7\pi}{4}.$

Show Solution

$\sin(-\frac{7\pi}{4})= \frac{\sqrt{2}}{2}, \cos(-\frac{7\pi}{4})= \frac{\sqrt{2}}{2}, \tan(-\frac{7\pi}{4})= 1, \sec(-\frac{7\pi}{4})= \sqrt{2}, \csc(-\frac{7\pi}{4})= \sqrt{2}, \cot(-\frac{7\pi}{4})= 1$

Using Even and Odd Trigonometric Functions

To be able to use our six trigonometric functions freely with both positive and negative angle inputs, we should examine how each function treats a negative input. As it turns out, there is an important difference among the functions in this regard.

Consider the function $f(x)=x^{2},$ shown in Figure 5. The graph of the function is symmetrical about the y-axis. All along the curve, any two points with opposite x-values have the same function value. This matches the result of calculation: $(4)^{2}=(-4)^{2}, (-5)^{2}=(5)^{2},$ and so on. So $f(x)=x^{2}$ is an even function, a function such that two inputs that are opposites have the same output. That means $f(-x)=f(x).$

This is an image of a graph of and upward facing parabola with points (-2, 4) and (2, 4) labeled. — Figure 5 The function $f(x)=x^{2}$ is an even function.

Now consider the function $f(x)=x^{3},$ shown in Figure 6. The graph is not symmetrical about the y-axis. All along the graph, any two points with opposite x-values also have opposite y-values. So $f(x)=x^{3}$ is an odd function, one such that two inputs that are opposites have outputs that are also opposites. That means $f(-x)=-f(x).$

This is an image of a graph of the function f of x = x to the third power with labels for points (-1, -1) and (1, 1). — Figure 6 The function $f(x)=x^{3}$ is an odd function.

We can test whether a trigonometric function is even or odd by drawing a unit circle with a positive and a negative angle, as in Figure 7. The sine of the positive angle is $y.$ The sine of the negative angle is $-y.$ The sine function, then, is an odd function. We can test each of the six trigonometric functions in this fashion. The results are shown in Table 2.

Graph of circle with angle of t and -t inscribed. Point of (x, y) is at intersection of terminal side of angle t and edge of circle. Point of (x, -y) is at intersection of terminal side of angle -t and edge of circle. — Figure 7

$\begin{array}{rcl} \sin t & = & y \\ \sin (-t) & = & -y \\ \sin t & \neq & \sin (-t) \end{array}$	$\begin{array}{rcl} \cos t & = & x \\ \cos (-t) & = & x \\ \cos t & = & \cos (-t) \end{array}$	$\begin{array}{rcl} \tan (t) & = & \frac{y}{x} \\ \tan (-t) & = & -\frac{y}{x} \\ \tan t & \neq & \tan (-t) \end{array}$
$\begin{array}{rcl} \sec t & = & \frac{1}{x} \\ \sec (-t) & = & \frac{1}{x} \\ \sec t & = & \sec (-t) \end{array}$	$\begin{array}{rcl} \csc t & = & \frac{1}{y} \\ \csc (-t) & = & \frac{1}{-y} \\ \csc t & \neq & \csc (-t) \end{array}$	$\begin{array}{rcl} \cot t & = & \frac{x}{y} \\ \cot (-t) & = & \frac{x}{-y} \\ \cot t & \neq & \cot (-t) \end{array}$

An even function is one in which $f(-x)=f(x).$

An odd function is one in which $f(-x)=-f(x).$

Cosine and secant are even:

$\begin{array}{rcl} \cos (-t) & = & \cos t \\ \sec (-t) & = & \sec t \end{array}$

Sine, tangent, cosecant, and cotangent are odd:

$\begin{array}{rcl} \sin (-t) & = & -\sin t \\ \tan (-t) & = & -\tan t \\ \csc (-t) & = & -\csc t \\ \cot (-t) & = & -\cot t \end{array}$

Example 4

Using Even and Odd Properties of Trigonometric Functions

If the secant of angle $t$ is 2, what is the secant of $-t?$

Solution

Secant is an even function. The secant of an angle is the same as the secant of its opposite. So if the secant of angle $t$ is 2, the secant of $-t$ is also 2.

Try It 4

If the cotangent of angle $t$ is $\sqrt{3},$ what is the cotangent of $-t?$

Show Solution

$-\sqrt{3}$

Recognizing and Using Fundamental Identities

We have explored a number of properties of trigonometric functions. Now, we can take the relationships a step further, and derive some fundamental identities. Identities are statements that are true for all values of the input on which they are defined. Usually, identities can be derived from definitions and relationships we already know. For example, the Pythagorean Identity we learned earlier was derived from the Pythagorean Theorem and the definitions of sine and cosine.

We can derive some useful identities from the six trigonometric functions. The other four trigonometric functions can be related back to the sine and cosine functions using these basic relationships:

$\tan t = \frac{\sin t}{\cos t}$
$\sec t = \frac{1}{\cos t}$
$\csc t = \frac{1}{\sin t}$
$\cot t = \frac{1}{\tan t} = \frac{\cos t}{\sin t}$

Example 5

Using Identities to Evaluate Trigonometric Functions

ⓐ Given $\sin (45^{\circ}) = \frac{\sqrt{2}}{2}, \cos (45^{\circ}) = \frac{\sqrt{2}}{2},$ evaluate $\tan (45^{\circ}).$
ⓑ Given $\sin \left(\frac{5\pi}{6}\right) = \frac{1}{2}, \cos \left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2},$ evaluate $\sec \left(\frac{5\pi}{6}\right).$

Solution

Because we know the sine and cosine values for these angles, we can use identities to evaluate the other functions.

ⓐ $\begin{array}{rcl} \tan (45^{\circ}) & = & \frac{\sin (45^{\circ})}{\cos (45^{\circ})} \\ & = & \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} \\ & = & 1 \end{array}$
ⓑ $\begin{array}{rcl} \sec \left(\frac{5\pi}{6}\right) & = & \frac{1}{\cos \left(\frac{5\pi}{6}\right)} \\ & = & \frac{1}{-\frac{\sqrt{3}}{2}} \\ & = & \frac{-2\sqrt{3}}{1} \\ & = & \frac{-2}{\sqrt{3}} \\ & = & -\frac{2\sqrt{3}}{3} \end{array}$

Try It 5

Evaluate $\csc \left(\frac{7\pi}{6}\right).$

Show Solution

-2

Example 6

Using Identities to Simplify Trigonometric Expressions

Simplify $\frac{\sec t}{\tan t}.$

Solution

We can simplify this by rewriting both functions in terms of sine and cosine.

$\begin{array}{rcll} \frac{\sec t}{\tan t} & = & \frac{\frac{1}{\cos t}}{\frac{\sin t}{\cos t}} \\ & = & \frac{1}{\cos t} \cdot \frac{\cos t}{\sin t} & \text{Multiply by the reciprocal.} \\ & = & \frac{1}{\sin t} = \csc t & \text{Simplify and use the identity.} \end{array}$

By showing that $\frac{\sec t}{\tan t}$ can be simplified to $\csc t,$ we have, in fact, established a new identity.

$\frac{\sec t}{\tan t} = \csc t$

Try It 6

Simplify $(\tan t)(\cos t).$

Show Solution

$\sin t$

Alternate Forms of the Pythagorean Identity

We can use these fundamental identities to derive alternate forms of the Pythagorean Identity, $\cos^{2}t + \sin^{2}t = 1.$ One form is obtained by dividing both sides by $\cos^{2}t.$

$\begin{array}{rcl} \frac{\cos^{2}t}{\cos^{2}t} + \frac{\sin^{2}t}{\cos^{2}t} & = & \frac{1}{\cos^{2}t} \\ 1 + \tan^{2}t & = & \sec^{2}t \end{array}$

The other form is obtained by dividing both sides by $\sin^{2}t.$

$\begin{array}{rcl} \frac{\cos^{2}t}{\sin^{2}t} + \frac{\sin^{2}t}{\sin^{2}t} & = & \frac{1}{\sin^{2}t} \\ \cot^{2}t + 1 & = & \csc^{2}t \end{array}$

Alternate Forms of the Pythagorean Identity

$1 + \tan^{2}t = \sec^{2}t$
$\cot^{2}t + 1 = \csc^{2}t$

Example 7

Using Identities to Relate Trigonometric Functions

If $\cos (t) = \frac{12}{13}$ and $t$ is in quadrant IV, as shown in Figure 8, find the values of the other five trigonometric functions.

This is an image of graph of circle with angle of t inscribed. Point of (12/13, y) is at intersection of terminal side of angle and edge of circle. — Figure 8

Solution

We can find the sine using the Pythagorean Identity, $\cos^{2}t + \sin^{2}t = 1,$ and the remaining functions by relating them to sine and cosine.

$\begin{array}{rcl} \left(\frac{12}{13}\right)^{2} + \sin^{2}t & = & 1 \\ \sin^{2}t & = & 1 - \left(\frac{12}{13}\right)^{2} \\ \sin^{2}t & = & 1 - \frac{144}{169} \\ \sin^{2}t & = & \frac{25}{169} \\ \sin t & = & \pm \sqrt{\frac{25}{169}} \\ \sin t & = & \pm \frac{\sqrt{25}}{\sqrt{169}} \\ \sin t & = & \pm \frac{5}{13} \end{array}$

The sign of the sine depends on the y-values in the quadrant where the angle is located. Since the angle is in quadrant IV, where the y-values are negative, its sine is negative, $-\frac{5}{13}.$

The remaining functions can be calculated using identities relating them to sine and cosine.

$\begin{array}{rcll} \tan t & = \frac{\sin t}{\cos t} & = \frac{-\frac{5}{13}}{\frac{12}{13}} & = -\frac{5}{12} \\ \sec t & = \frac{1}{\cos t} & = \frac{1}{\frac{12}{13}} & = \frac{13}{12} \\ \csc t & = \frac{1}{\sin t} & = \frac{1}{-\frac{5}{13}} & = \frac{-13}{5} \\ \cot t & = \frac{1}{\tan t} & = \frac{1}{-\frac{5}{12}} & = -\frac{12}{5} \end{array}$

Try It 7

If $\sec (t) = -\frac{17}{8}$ and $0 < t < \pi,$ find the values of the other five functions.

Show Solution

$\cos t= -\frac{8}{17}, \sin t= \frac{15}{17}, \tan t= -\frac{15}{8}, \csc t= \frac{17}{15}, \cot t= -\frac{8}{15}$

As we discussed at the beginning of the chapter, a function that repeats its values in regular intervals is known as a periodic function. The trigonometric functions are periodic. For the four trigonometric functions, sine, cosine, cosecant and secant, a revolution of one circle, or $2\pi,$ will result in the same outputs for these functions. And for tangent and cotangent, only a half a revolution will result in the same outputs.

Other functions can also be periodic. For example, the lengths of months repeat every four years. If $x$ represents the length time, measured in years, and $f(x)$ represents the number of days in February, then $f(x+4)=f(x).$ This pattern repeats over and over through time. In other words, every four years, February is guaranteed to have the same number of days as it did 4 years earlier. The positive number 4 is the smallest positive number that satisfies this condition and is called the period. A period is the shortest interval over which a function completes one full cycle—in this example, the period is 4 and represents the time it takes for us to be certain February has the same number of days.

Period of a function

The period $P$ of a repeating function $f$ is the number representing the interval such that $f(x+P)=f(x)$ for any value of $x.$

The period of the cosine, sine, secant, and cosecant functions is $2\pi.$

The period of the tangent and cotangent functions is $\pi.$

Example 8

Finding the Values of Trigonometric Functions

Find the values of the six trigonometric functions of angle $t$ based on Figure 9.

This is an image of a graph of circle with angle of t inscribed. Point of (1/2, negative square root of 3 over 2) is at intersection of terminal side of angle and edge of circle. — Figure 9

Solution

$\begin{array}{rcll} \sin t & = y & = -\frac{\sqrt{3}}{2} \\ \cos t & = x & = -\frac{1}{2} \\ \tan t & = \frac{\sin t}{\cos t} & = \frac{-\frac{\sqrt{3}}{2}}{-\frac{1}{2}} & = \sqrt{3} \\ \sec t & = \frac{1}{\cos t} & = \frac{1}{-\frac{1}{2}} & = -2 \\ \csc t & = \frac{1}{\sin t} & = \frac{1}{-\frac{\sqrt{3}}{2}} & = -\frac{2\sqrt{3}}{3} \\ \cot t & = \frac{1}{\tan t} & = \frac{1}{\sqrt{3}} & = \frac{\sqrt{3}}{3} \end{array}$

Try It 8

Find the values of the six trigonometric functions of angle $t$ based on Figure 10.

This is an image of a graph of circle with angle of t inscribed. Point of (0, -1) is at intersection of terminal side of angle and edge of circle. — Figure 10

Show Solution

$\sin t= -1, \cos t= 0, \tan t= \text{Undefined}, \sec t= \text{Undefined}, \csc t= -1, \cot t= 0$

Example 9

Finding the Value of Trigonometric Functions

If $\sin (t) = -\frac{\sqrt{3}}{2} \text{ and } \cos (t) = \frac{1}{2}, \text{find } \sec (t), \csc (t), \tan (t), \cot (t).$

Solution

$\begin{array}{rcl} \sec t & = \frac{1}{\cos t} & = \frac{1}{\frac{1}{2}} = 2 \\ \csc t & = \frac{1}{\sin t} & = \frac{1}{-\frac{\sqrt{3}}{2}} - \frac{2\sqrt{3}}{3} \\ \tan t & = \frac{\sin t}{\cos t} & = \frac{-\frac{\sqrt{3}}{2}}{\frac{1}{2}} = -\sqrt{3} \\ \cot t & = \frac{1}{\tan t} & = \frac{1}{-\sqrt{3}} = -\frac{\sqrt{3}}{3} \end{array}$

Try It 9

$\sin (t) = \frac{\sqrt{2}}{2} \text{ and } \cos (t) = \frac{\sqrt{2}}{2}, \text{find } \sec (t), \csc (t), \tan (t), \text{ and } \cot (t)$

Show Solution

$\sec t= \sqrt{2}, \csc t= \sqrt{2}, \tan t= 1, \cot t= 1$

Evaluating Trigonometric Functions with a Calculator

We have learned how to evaluate the six trigonometric functions for the common first-quadrant angles and to use them as reference angles for angles in other quadrants. To evaluate trigonometric functions of other angles, we use a scientific or graphing calculator or computer software. If the calculator has a degree mode and a radian mode, confirm the correct mode is chosen before making a calculation.

Evaluating a tangent function with a scientific calculator as opposed to a graphing calculator or computer algebra system is like evaluating a sine or cosine: Enter the value and press the TAN key. For the reciprocal functions, there may not be any dedicated keys that say CSC, SEC, or COT. In that case, the function must be evaluated as the reciprocal of a sine, cosine, or tangent.

If we need to work with degrees and our calculator or software does not have a degree mode, we can enter the degrees multiplied by the conversion factor $\frac{\pi}{180}$ to convert the degrees to radians. To find the secant of $30^{\circ},$ we could press $\begin{array}{c} \text{(for a scientific calculator): } \frac{1}{30 \times \frac{\pi}{180}} \text{ COS} \\ \text{or} \\ \text{(for a graphing calculator): } \frac{1}{\cos \left(\frac{30\pi}{180}\right)} \end{array}$

#TODO: fix the latex formatting issues around here

Given an angle measure in radians, use a scientific calculator to find the cosecant.

If the calculator has degree mode and radian mode, set it to radian mode.
Enter: $1 /$
Enter the value of the angle inside parentheses.
Press the SIN key.
Press the = key.

Given an angle measure in radians, use a graphing utility/calculator to find the cosecant.

If the graphing utility has degree mode and radian mode, set it to radian mode.
Enter: $1 /$
Press the SIN key.
Enter the value of the angle inside parentheses.
Press the ENTER key.

Example 10

Evaluating the Cosecant Using Technology

Evaluate the cosecant of $\frac{5\pi}{7}.$

Solution

For a scientific calculator, enter information as follows:

$\begin{array}{rcl} 1 / (5 \times \pi / 7) \text{ SIN} & = & \\ \csc \left(\frac{5\pi}{7}\right) & \approx & 1.279 \end{array}$

Try It 10

Evaluate the cotangent of $-\frac{\pi}{8}.$

Show Solution

$\approx -2.414$

7.4 Exercises Set

On an interval of $[0, 2\pi),$ can the sine and cosine values of a radian measure ever be equal? If so, where?
What would you estimate the cosine of $\pi$ degrees to be? Explain your reasoning.
For any angle in quadrant II, if you knew the sine of the angle, how could you determine the cosine of the angle?
Describe the secant function.
Tangent and cotangent have a period of $\pi.$ What does this tell us about the output of these functions?
$\tan \frac{\pi}{6}$
$\sec \frac{\pi}{6}$
$\csc \frac{\pi}{6}$
$\cot \frac{\pi}{6}$
$\tan \frac{\pi}{4}$
$\sec \frac{\pi}{4}$
$\csc \frac{\pi}{4}$
$\cot \frac{\pi}{4}$
$\tan \frac{\pi}{3}$
$\sec \frac{\pi}{3}$
$\csc \frac{\pi}{3}$
$\cot \frac{\pi}{3}$
$\tan \frac{5\pi}{6}$
$\sec \frac{7\pi}{6}$
$\csc \frac{11\pi}{6}$
$\cot \frac{13\pi}{6}$
$\tan \frac{7\pi}{4}$
$\sec \frac{3\pi}{4}$
$\csc \frac{5\pi}{4}$
$\cot \frac{11\pi}{4}$
$\tan \frac{8\pi}{3}$
$\sec \frac{4\pi}{3}$
$\csc \frac{2\pi}{3}$
$\cot \frac{5\pi}{3}$
$\tan 225^{\circ}$
$\sec 300^{\circ}$
$\csc 150^{\circ}$
$\cot 240^{\circ}$
$\tan 330^{\circ}$
$\sec 120^{\circ}$
$\csc 210^{\circ}$
$\cot 315^{\circ}$
If $\sin t = \frac{3}{4},$ and $t$ is in quadrant II, find $\cos t, \sec t, \csc t, \tan t,$ and $\cot t.$
If $\cos t = -\frac{1}{3},$ and $\t$ is in quadrant III, find $\sin t, \sec t, \csc t, \tan t,$ and $\cot t.$
If $\tan t = \frac{12}{5}$ , and $0 \le t < \frac{\pi}{2}$ , find $\sin t, \cos t, \sec t, \csc t, \text{and } \cot t.$
If $\sin t = \frac{\sqrt{3}}{2}$ and $\cos t = \frac{1}{2},$ find $\sec t, \csc t, \tan t,$ and $\cot t.$
If $\sin 40^{\circ} \approx 0.643$ and $\cos 40^{\circ} \approx 0.766,$ find $\sec 40^{\circ}, \csc 40^{\circ}, \tan 40^{\circ},$ and $\cot 40^{\circ}.$
If $\sin t = \frac{\sqrt{2}}{2},$ what is the $\sin (-t)?$
If $\cos t = \frac{1}{2},$ what is the $\cos (-t)?$
If $\sec t = 3.1,$ what is the $\sec (-t)?$
If $\csc t = 0.34,$ what is the $\csc (-t)?$
If $\tan t = -1.4,$ what is the $\tan (-t)?$
If $\cot t = 9.23,$ what is the $\cot (-t)?$
$\csc \frac{5\pi}{9}$
$\cot \frac{4\pi}{7}$
$\sec \frac{\pi}{10}$
$\tan \frac{5\pi}{8}$
$\sec \frac{3\pi}{4}$
$\csc \frac{\pi}{4}$
$\tan 98^{\circ}$
$\cot 33^{\circ}$
$\cot 140^{\circ}$
$\sec 310^{\circ}$
If $\tan (t) \approx 2.7,$ and $\sin (t) \approx 0.94,$ find $\cos (t).$
If $\tan (t) \approx 1.3,$ and $\cos (t) \approx 0.61,$ find $\sin (t).$
If $\csc (t) \approx 3.2,$ and $\cos (t) \approx 0.95,$ find $\tan (t).$
If $\cot (t) \approx 0.58,$ and $\cos (t) \approx 0.5,$ find $\csc (t).$
Determine whether the function $f(x) = 2\sin x \cos x$ is even, odd, or neither.
Determine whether the function $f(x) = 3\sin^{2}x \cos x + \sec x$ is even, odd, or neither.
Determine whether the function $f(x) = \sin x - 2\cos^{2}x$ is even, odd, or neither.
Determine whether the function $f(x) = \csc^{2}x + \sec x$ is even, odd, or neither.
$\csc t \tan t$
$\frac{\sec t}{\csc t}$
The amount of sunlight in a certain city can be modeled by the function $h = 15\cos \left(\frac{1}{600}d\right),$ where $h$ represents the hours of sunlight, and $d$ is the day of the year. Use the equation to find how many hours of sunlight there are on February 11, the 42^nd day of the year. State the period of the function.
The amount of sunlight in a certain city can be modeled by the function $h = 16\cos \left(\frac{1}{500}d\right),$ where $h$ represents the hours of sunlight, and $d$ is the day of the year. Use the equation to find how many hours of sunlight there are on September 24, the 267th day of the year. State the period of the function.
The equation $P = 20\sin (2\pi t) + 100$ models the blood pressure, $P,$ where $t$ represents time in seconds. (a) Find the blood pressure after 15 seconds. (b) What are the maximum and minimum blood pressures?
The height of a piston, $h,$ in inches, can be modeled by the equation $y = 3\sin x + 1,$ where $x$ represents the crank angle. Find the height of the piston when the crank angle is $55^{\circ}.$
The height of a piston, $h,$ in inches, can be modeled by the equation $y = 2\cos x + 5,$ where $x$ represents the crank angle. Find the height of the piston when the crank angle is $55^{\circ}.$