Chapter 8: More Functions and Identities
8.3 The Reciprocal Functions
Algebra Refresher
Algebra Refresher
Review the following skills you will need for this section.
- Solve [latex]|2x-6|=4[/latex]
- Solve [latex]|1-3x|=10[/latex]
- Simplify [latex]\sqrt{(x-4)^2}[/latex]
- Simplify [latex]\sqrt{(1-x)^2}[/latex]
- For what values of [latex]x[/latex] is [latex]|x+2|=x+2[/latex]?
- For what values of [latex]x[/latex] is [latex]|x-3|=x-3[/latex]?
- Graph the function [latex]f(x)=\dfrac{x}{|x|}{.}[/latex]
- Explain the difference between the graphs of [latex]f(x)=(\sqrt{x})^2[/latex] and [latex]g(x)=\sqrt{x^2}{.}[/latex]
Algebra Refresher Answers
- [latex]\displaystyle x = 1,~5[/latex]
- [latex]\displaystyle x = -3,~\dfrac{11}{3}[/latex]
- [latex]\displaystyle |x-4|[/latex]
- [latex]\displaystyle |1-x|[/latex]
- [latex]\displaystyle x\ge -2[/latex]
- [latex]\displaystyle x \lt 3[/latex]
- The domain of [latex]f[/latex] is [latex][0, \infty){,}[/latex] and the domain of [latex]g[/latex] is [latex](-\infty, \infty){.}[/latex]
Section 8.3 The Reciprocal Functions
Learning Objectives
- Evaluate the reciprocal trig functions for angles in degrees or radians
- Find values or expressions for the six trig ratios
- Evaluate the reciprocal trig functions in applications
- Given one trig ratio, find the others
- Evaluate expressions exactly
- Graph the secant, cosecant, and cotangent functions
- Identify graphs of the reciprocal trig functions
- Solve equations in secant, cosecant, and cotangent
- Use identities to simplify or evaluate expressions
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 8.47. Three More Functions
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 8.48.
If [latex]\sec \theta = 3{,}[/latex] and [latex]-\dfrac{\pi}{2} \le \theta \le 0{,}[/latex] find exact values for the other five trig functions.
Solution
Because [latex]-\dfrac{\pi}{2} \le \theta \le 0{,}[/latex] we draw a reference triangle in the fourth quadrant, as shown at right. Because [latex]\sec \theta = 3 = \dfrac{3}{1}{,}[/latex] we label the horizontal leg with [latex]x=1[/latex] and the hypotenuse with [latex]r=3{.}[/latex]
From the Pythagorean theorem, we find [latex]y=-\sqrt{8}=-2\sqrt{2}{.}[/latex] We can now compute the values of the six trigonometric ratios.
[latex]\cos \theta =\dfrac{x}{r}=\dfrac{1}{3} \sec \theta =\dfrac{r}{x}=\dfrac{3}{1}=3\\ \sin \theta =\dfrac{y}{r}=\dfrac{-2\sqrt{2}}{3} \csc \theta =\dfrac{r}{y}=\dfrac{3}{-2\sqrt{2}}=\dfrac{-3\sqrt{2}}{4}\\ \tan \theta =\dfrac{y}{x}=\dfrac{-2\sqrt{2}}{1}=-2\sqrt{2} \cot \theta =\dfrac{x}{y}=\dfrac{1}{-2\sqrt{2}}=\dfrac{-\sqrt{2}}{4}[/latex]
Checkpoint 8.49.
If [latex]\csc \theta =4{,}[/latex] and [latex]90° \le \theta \le 180°{,}[/latex] find exact values for the other five trig functions.
Solution
[latex]\cos \theta = \dfrac{-\sqrt{15}}{4}{,}[/latex] [latex]~\sin \theta = \dfrac{1}{4}{,}[/latex] [latex]~\tan \theta = \dfrac{-1}{\sqrt{15}}{,}[/latex] [latex]~\sec \theta = \dfrac{-4}{\sqrt{15}}{,}[/latex] [latex]~\cot \theta = -\sqrt{15}[/latex]
Reciprocal Trigonometric Functions.
[latex]\begin{gather*} {The}~~ {secant} ~{function}:~~~~~~ {\sec \theta = \dfrac{1}{\cos \theta}}\\ {The}~~ {cosecant}~{function}:~~~~ {\csc \theta = \dfrac{1}{\sin \theta}}\\ {The}~~ {cotangent}~{function}:~~~ {\cot \theta = \dfrac{1}{\tan \theta}} \end{gather*}
Calculators do not have keys for the secant, cosecant, and cotangent functions; instead, we calculate their values as reciprocals.
Example 8.50.
Solution
With the calculator in degree mode, enter
[latex]\qquad\qquad\qquad[/latex] 1 ÷ COS 47 ) ENTER
to obtain [latex]\sec 47° \approx 1.466{.}[/latex] Or we can calculate [latex]\cos 47°[/latex] first, and then use the reciprocal key:
[latex]\qquad\qquad\qquad[/latex]COS 47 ) ENTER [latex]\small\boxed{x^{-1}}[/latex] ENTER
Checkpoint 8.51.
Use a calculator to approximate [latex]\csc 132°[/latex] to three decimal places.
Solution
[latex]1.346[/latex]
| Exact Values for the Special Angles | |||
| [latex]\theta[/latex] | [latex]\sec \theta[/latex] | [latex]\sec \theta[/latex] | [latex]\cot \theta[/latex] |
| [latex]0[/latex] | [latex]1[/latex] | undefined | undefined |
| [latex]\dfrac{\pi}{6}[/latex] | [latex]\dfrac{2\sqrt{3}}{3}[/latex] | [latex]2[/latex] | [latex]\sqrt{3}[/latex] |
| [latex]\dfrac{\pi}{4}[/latex] | [latex]\sqrt{2}[/latex] | [latex]\sqrt{2}[/latex] | [latex]1[/latex] |
| [latex]\dfrac{\pi}{3}[/latex] | [latex]2[/latex] | [latex]\dfrac{2\sqrt{3}}{3}[/latex] | [latex]\dfrac{1}{\sqrt{3}}[/latex] |
| [latex]\dfrac{\pi}{2}[/latex] | undefined | [latex]1[/latex] | [latex]0[/latex] |
Caution 8.52.
The reciprocal functions are not the same as the inverse trig functions!
For example, [latex]\sec 0.8[/latex] is not equal to [latex]\cos^{-1}(0.8){.}[/latex] Remember that [latex]~{\cos^{-1}(0.8)}~[/latex] is an angle; namely, the angle whose cosine is 0.8, while [latex]~{\sec 0.8}~[/latex] is the reciprocal of the cosine of 0.8 radians, or [latex]\dfrac{1}{\cos 0.8}{.}[/latex] You can check on your calculator that
[latex]\cos^{-1}(0.8)=0.6435~{radians},~~~~{and}~~~~\sec 0.8=1.4353[/latex]
Each of the reciprocal functions is undefined when its denominator is equal to zero. For example, the secant is undefined when [latex]\cos \theta = 0{,}[/latex] or when [latex]\theta[/latex] is an odd multiple of [latex]90°{.}[/latex]
Example 8.53
For which angles is the cosecant undefined?
Solution
The cosecant is undefined when its denominator, [latex]\sin \theta{,}[/latex] equals zero, and [latex]\sin \theta = 0[/latex] when [latex]\theta[/latex] is a multiple of [latex]180°{.}[/latex] In radians, [latex]\csc \theta[/latex] is undefined if [latex]\theta[/latex] is a multiple of [latex]\pi{.}[/latex]
Checkpoint 8.54.
For what angles is the cotangent undefined? Give your answers in degrees and in radians.
Solution
Multiples of [latex]180°{,}[/latex] or multiples of [latex]\pi{.}[/latex]
Note 8.55.
Although [latex]\tan \dfrac{\pi}{2}[/latex] is undefined, [latex]\cot \dfrac{\pi}{2}=0{.}[/latex]
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.
Reciprocal Trigonometric Ratios.
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 8.56.
The length, [latex]L{,}[/latex] of the shadow cast by a flagpole on a sunny day depends on the height, [latex]h{,}[/latex] of the flagpole and the angle, [latex]\theta{,}[/latex] that the sun's rays make with ground.
-
- Write an expression for the length, [latex]L{,}[/latex] of the shadow cast by a flagpole of height [latex]h[/latex] when the sun makes an angle of [latex]\theta[/latex] 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 [latex]20°[/latex] from the ground.
Solution
-
- From the figure, we see that [latex]\dfrac{L}{h}=\cot \theta{,}[/latex] or [latex]L=h\cot \theta{.}[/latex]
Substituting [latex]{3}[/latex] for [latex]h[/latex] and [latex]{20°}[/latex] for [latex]\theta{,}[/latex] we find
[latex]L = {3} \cot {20°} = 3(2.7475)=8.24[/latex]
The shadow is about 8.24 meters long.
Checkpoint 8.57.
The area [latex]A[/latex] of a regular polygon with [latex]n[/latex] sides having perimeter [latex]L[/latex] satisfies
[latex]A = \dfrac{L^2}{4n}\cot\dfrac{\pi}{n}[/latex]
Refer to the figure at right showing [latex]n=6[/latex] to prove this formula in the following steps.
- Find an expression for the angle [latex]\theta[/latex] in terms of [latex]n{.}[/latex]
- 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.
Solution
- [latex]\displaystyle \theta = \dfrac{\pi}{n}[/latex]
- [latex]\displaystyle b = \dfrac{L}{n}[/latex]
- [latex]\displaystyle h=\dfrac{L}{2n}\cot \dfrac{\pi}{n}[/latex]
- [latex]\displaystyle A_T=\dfrac{L^2}{4n^2}\cot \dfrac{\pi}{n},~~A_P=\dfrac{L^2}{4n}\cot \dfrac{\pi}{n}[/latex]
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 8.58
Use the graph of [latex]y=\cos x[/latex] to construct a graph of [latex]f(x)=\sec x{.}[/latex]
Solution
Consider the graph of [latex]y=\cos x[/latex] shown at left below. When [latex]x=\dfrac{-\pi}{2},~\dfrac{\pi}{2}[/latex] and [latex]\dfrac{3\pi}{2},~ \cos x=0{,}[/latex] so [latex]\sec x[/latex] is undefined at these [latex]x[/latex]-values, and we insert vertical asymptotes at those [latex]x[/latex]-values to start our graph of [latex]y=\sec x{,}[/latex] as shown at right below.
To find some points on the graph, we look at points on the graph of [latex]y=\cos x{.}[/latex] At each [latex]x[/latex]-value, the [latex]y[/latex]-coordinate of the point on the graph of [latex]y=\sec x[/latex] is the reciprocal of [latex]\cos x{.}[/latex]
For example, at [latex]x=0[/latex] and [latex]x=2\pi{,}[/latex] we have [latex]\cos x = 1{,}[/latex] so [latex]\sec x = \frac{1}{1} = 1{.}[/latex] Thus, we plot the points [latex](0,1)[/latex] and [latex](2\pi,1)[/latex] on the graph of [latex]f(x)=\sec x{.}[/latex] Similarly, at [latex]x=-\pi[/latex] and [latex]x=\pi{,}[/latex] [latex]\cos x = -1{,}[/latex] so the value of [latex]\sec x[/latex] is [latex]\frac{1}{-1} = -1{,}[/latex] and we plot the points [latex](-\pi,-1)[/latex] and [latex](\pi,-1)[/latex] on the graph of [latex]f(x)=\sec x{.}[/latex]
Finally, we notice that the values of [latex]\cos x[/latex] are decreasing toward [latex]0[/latex] as [latex]x[/latex] increases from [latex]0[/latex] to [latex]\dfrac{\pi}{2}{,}[/latex] so the graph of [latex]f(x)=\sec x[/latex] increases toward [latex]\infty[/latex] on the same interval. By similar arguments, we fill in the graph of [latex]f(x)=\sec x[/latex] between each of the vertical asymptotes to produce the graph below.
Checkpoint 8.59.
Use the graph of [latex]y=\tan x[/latex] to sketch a graph of [latex]g(x)=\cot x{.}[/latex]
Solution
The graphs of the three new functions are shown below, with [latex]x[/latex] in radians. Note that the secant function is undefined at odd multiples of [latex]\dfrac{\pi}{2}{,}[/latex] the values at which [latex]\cos x=0{.}[/latex] The cosecant is undefined where [latex]\sin x=0{;}[/latex] namely, at multiples of [latex]\pi{.}[/latex] The cotangent is also undefined at multiples of [latex]\pi{,}[/latex] because [latex]\tan x=0[/latex] at those values.
Example 8.60.
State the domain and range of the secant function.
Solution
Because the cosine is defined for all real numbers, the domain of the secant includes all real numbers except for values where the cosine is zero. These values are the odd multiples of [latex]\dfrac{\pi}{2}{;}[/latex] that is, [latex]\dfrac{\pi}{2},~\dfrac{3\pi}{2},~\dfrac{5\pi}{2},~ \ldots{,}[/latex] and their opposites. Because the range of the cosine consists of all [latex]y[/latex]-values with [latex]-1 \le y \le 1{,}[/latex] the range of the secant includes the reciprocals of those values; namely, [latex]y \ge 1[/latex] and [latex]y \le -1{.}[/latex]
Checkpoint 8.61.
State the domain and range of the cosecant and cotangent functions.
Solution
Domain of cosecant: all real numbers except integer multiples of [latex]\pi{;}[/latex] Range of cosecant: [latex](-\infty,-1] \cup [1, \infty)[/latex]
Domain of cotangent: all real numbers except integer multiples of [latex]\pi{;}[/latex] Range of cotangent: all real numbers
Solving Equations
From the graph of the secant function, we can see that the equation [latex]\sec \theta =k[/latex] has two solutions between [latex]0[/latex] and [latex]2\pi[/latex] if [latex]k \ge 1[/latex] or [latex]k \le -1[/latex] but no solution for [latex]-1 \lt k \lt 1{.}[/latex] The same is true of the cosecant function: the equation [latex]\csc \theta = k[/latex] has no solution for [latex]-1 \lt k \lt 1{.}[/latex]
Example 8.62.
Solve [latex]~~\csc \theta = \dfrac{2\sqrt{3}}{3}~~[/latex] for [latex]\theta[/latex] between [latex]0[/latex] and [latex]2\pi{.}[/latex]
Solution
We take the reciprocal of each side of the equation to obtain
[latex]\sin \theta = \dfrac{3}{2\sqrt{3}} = \dfrac{\sqrt{3}}{2}[/latex]
Because [latex]\dfrac{\sqrt{3}}{2}[/latex] is one of the special values, we recognize that one of the solutions is [latex]\theta =\dfrac{\pi}{3}{.}[/latex] The sine and the cosecant are also positive in the second quadrant, so the second solution is [latex]\pi - \dfrac{\pi}{3} = \dfrac{2\pi}{3}{.}[/latex]
Checkpoint 8.63.
Solve [latex]~~\sec \theta = -1.6~~[/latex] for [latex]\theta[/latex] between [latex]0[/latex] and [latex]2\pi{.}[/latex]
Solution
[latex]\theta=2.25, \, 4.04[/latex]
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 8.64
If [latex]\sec \theta = 3{,}[/latex] and [latex]-\dfrac{\pi}{2} \le \theta \le 0{,}[/latex] find an exact value for [latex]\csc \theta{.}[/latex]
Solution
Because [latex]\sec \theta = \dfrac{1}{\cos \theta}{,}[/latex] we see that [latex]\dfrac{1}{\cos \theta}=3{,}[/latex] or [latex]\cos \theta = \dfrac{1}{3}{.}[/latex] We use the Pythagorean identity to find the sine.
[latex]\cos^2 \theta + \sin^2 \theta = 1 {{Substitute}~ \frac{1}{3} { for } \cos \theta.}\\ \left(\dfrac{1}{3}\right)^2 + \sin^2 \theta = 1 {{Subtract}~ \left(\frac{1}{3}\right)^2=\frac{1}{9}~ {from both sides.}}\\ \sin^2 \theta = 1 - \dfrac{1}{9}=\dfrac{8}{9}[/latex]
Because [latex]\theta[/latex] lies in the fourth quadrant, where the sine function is negative, we choose the negative square root for [latex]\sin \theta{.}[/latex] Once we know [latex]\sin \theta{,}[/latex] we calculate its reciprocal to find [latex]\csc\theta{.}[/latex]
[latex]\sin \theta = -\sqrt{\dfrac{8}{9}} = \dfrac{-2\sqrt{2}}{3},~{ and }~\csc\theta = \dfrac{1}{\sin \theta} = \dfrac{-3}{2\sqrt{2}} = \dfrac{-3\sqrt{2}}{4}[/latex]
Checkpoint 8.65.
If [latex]\csc \theta = \dfrac{-\sqrt{13}}{3}{,}[/latex] and [latex]\pi \le \theta \le \dfrac{3\pi}{2}{,}[/latex] find an exact value for [latex]\sec \theta{.}[/latex]
Solution
[latex]\sec\theta = \dfrac{-\sqrt{13}}{2}[/latex]
Identities are especially useful if the trig ratios are algebraic expressions rather than numerical values. In the next example, we use the cotangent identity.
Cotangent Identity.
[latex]{\cot \theta = \dfrac{1}{\tan \theta} = \dfrac{\cos \theta}{\sin \theta},~~~~\sin \theta \not=0}[/latex]
Example 8.66.
If [latex]\csc x=w[/latex] and [latex]0 \lt x \lt \dfrac{\pi}{2}{,}[/latex] find an expression for [latex]\cot x{.}[/latex]
Solution
Because the sine is the reciprocal of the cosecant, we have [latex]\sin x = \dfrac{1}{\csc x} = \dfrac{1}{w}{.}[/latex] We substitute [latex]{\dfrac{1}{w}}[/latex] for [latex]\sin x[/latex] in the Pythagorean identity to find
[latex]\cos x = \pm \sqrt{1 - \sin^2 x} = \pm \sqrt{1 - \left({\dfrac{1}{w}}\right)^2}[/latex]
We choose the positive root because cosine is positive in the first quadrant and simplify to get
[latex]\cos x = \sqrt{1 - \dfrac{1}{w^2}} = \sqrt{\dfrac{w^2-1}{w^2}} = \dfrac{\sqrt{w^2 - 1}}{\abs{w}}[/latex]
We can replace [latex]\abs{w}[/latex] by [latex]w[/latex] in this last expression because [latex]w \gt 0{.}[/latex] (Do you see why [latex]w \gt 0{?}[/latex]) Finally, because the cotangent is the reciprocal of the tangent, we have
[latex]\cot x = \dfrac{\cos x}{\sin x} = \dfrac{\dfrac{w^2-1}{w}}{\dfrac{1}{w}}=\sqrt{w^2 - 1}[/latex]
Checkpoint 8.67.
If [latex]\sec t = \dfrac{2}{a}[/latex] and [latex]\dfrac{3\pi}{2} \lt t \lt 2\pi{,}[/latex] find expressions for [latex]\csc t[/latex] and [latex]\cot t{.}[/latex]
Solution
[latex]\csc t = \dfrac{-2}{\sqrt{1-a^2}}{,}[/latex] [latex]~ \cot t = \dfrac{-a}{\sqrt{1-a^2}}[/latex]
We can often simplify trigonometric expressions by first converting all the trig ratios to sines and cosines.
Example 8.68.
Simplify [latex]\sec \theta - \tan \theta \sin \theta{.}[/latex]
Solution
We replace [latex]~\sec \theta~[/latex] by [latex]~\dfrac{1}{\cos \theta}~[/latex] and [latex]~\tan \theta~[/latex] by [latex]~\dfrac{\sin \theta}{\cos \theta}~[/latex] to get
[latex]\dfrac{1}{\cos \theta} - \dfrac{\sin \theta}{\cos \theta} \cdot \sin \theta = \dfrac{1}{\cos \theta} - \dfrac{\sin^2 \theta}{\cos \theta}\\ = \dfrac{1 - \sin^2 \theta}{\cos \theta}= \dfrac{\cos^2 \theta}{\cos \theta} = \cos \theta[/latex]
In the previous example, you can verify that
[latex]\sec \theta - \tan \theta \sin \theta = \cos \theta[/latex]
by graphing the functions [latex]Y_1=\sec \theta - \tan \theta \sin \theta[/latex] and [latex]Y_2=\cos \theta[/latex] in the ZTrig window to see that they are the same.
Checkpoint 8.69.
Show that [latex]\sin^2 x(1+\cot^2 x) = 1{.}[/latex]
Solution
[latex]\sin^2x (1+\cot^2 x) = \sin^2 x\left(1 + \dfrac{\cos^2 x}{\sin^2 x}\right)=\sin^2 x + \cos^2 x = 1[/latex]
There are two alternate versions of the Pythagorean identity that involve the reciprocal trig functions. These identities are useful when we know the value of [latex]\tan \theta[/latex] or [latex]\cot \theta[/latex] and want to find the other trig values.
Two More Pythagorean Identities.
[latex]{1 + \tan^2 \theta = \sec^2\theta~~~~~~~~~~1 + \cot^2\theta = \csc^2 \theta}[/latex]
You should memorize these identities, but they are easy to derive from the original Pythagorean identity, [latex]\sin^2 \theta + \cos^2 \theta = 1{.}[/latex] We will prove them in the Homework problems.
Example 8.70.
If [latex]\tan \alpha = \dfrac{3}{5}[/latex] and [latex]\alpha[/latex] lies in the third quadrant, find exact values for [latex]\sec \alpha[/latex] and [latex]\cos \alpha{.}[/latex]
Solution
We cannot find the sine and cosine of an angle directly from the value of the tangent; in particular, it is not true that [latex]\sin \alpha = 3[/latex] and [latex]\cos \alpha = 5{!}[/latex] (Do you see why?) Instead, we begin with the Pythagorean identity for the tangent.
[latex]\sec^2 \alpha = 1 + \tan^2 \alpha = 1 + (\dfrac{3}{5})^2\\ = \dfrac{25}{25} + \dfrac{9}{25} = \dfrac{34}{25}\\ \sec \alpha = \pm \sqrt{\dfrac{34}{25}} = \dfrac{\pm \sqrt{34}}{5}[/latex]
Because [latex]\alpha[/latex] 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 [latex]\sec \alpha = \dfrac{-\sqrt{34}}{5}{.}[/latex] The cosine of [latex]\alpha[/latex] is the reciprocal of the secant, so [latex]\cos \alpha = \dfrac{-5}{\sqrt{34}}{.}[/latex]
Checkpoint 8.71.
If [latex]\cot \phi = \dfrac{-3}{\sqrt{2}}[/latex] and [latex]\phi[/latex] lies in the second quadrant, find exact values for [latex]\csc \phi[/latex] and [latex]\sin \phi{.}[/latex]
Solution
[latex]\csc \phi = \sqrt{\dfrac{11}{2}}{,}[/latex] [latex]~ \sin \phi = \sqrt{\dfrac{2}{11}}[/latex]
Section 8.3 Summary
Vocabulary
- Reciprocal
- Secant
- Cosecant
- Cotangent
CONCEPTS
-
Three More Functions.
If [latex]\theta[/latex] is an angle in standard position, and [latex]P(x,y)[/latex] is a point on the terminal side, then we define the following functions.
[latex]{The}~~ {secant}:~~~~~~ \sec \theta = \dfrac{r}{x}[/latex]
[latex]{The}~~ {cosecant}:~~~~ \csc \theta = \dfrac{r}{y}[/latex]
[latex]{The}~~ {cotangent}:~~~ \cot \theta = \dfrac{x}{y}[/latex]
-
Reciprocal Trigonometric Ratios.
If [latex]\theta[/latex] is one of the acute angles in a right triangle,
[latex]\sec \theta = \dfrac{{hypotenuse}}{{adjacent}}\\ \csc \theta = \dfrac{{hypotenuse}}{{opposite}}\\ \cot \theta = \dfrac{{adjacent}}{{opposite}}[/latex]
-
Reciprocal Trigonometric Functions.
[latex]{The}~~ {secant} ~{function}:~~~~~~ \sec \theta = \dfrac{1}{\cos \theta}[/latex]
[latex]{The}~~ {cosecant}~{function}:~~~~ \csc \theta = \dfrac{1}{\sin \theta}[/latex]
[latex]{The}~~ {cotangent}~{function}:~~~ \cot \theta = \dfrac{1}{\tan \theta}[/latex]
- 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 [latex]\sec \theta = k{,}[/latex] [latex]\csc \theta = k{,}[/latex] and [latex]\cot \theta = k[/latex] 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.
-
Cotangent Identity.
[latex]\cot \theta = \dfrac{1}{\tan \theta} = \dfrac{\cos \theta}{\sin \theta},~~~~\sin \theta \not=0[/latex]
-
Two More Pythagorean Identities.
[latex]1 + \tan^2 \theta = \sec^2\theta~~~~~~~~~~1 + \cot^2\theta = \csc^2 \theta[/latex]
- We can often simplify trigonometric expressions by first converting all the trig ratios to sines and cosines.
STUDY QUESTIONS
- Delbert says that [latex]\sec x[/latex] is just another way of writing [latex]\cos^{-1} x{,}[/latex] because [latex]\cos^{-1} x = \dfrac{1}{\cos x}{.}[/latex] Is he correct? Explain your reasoning.
- Each of the following functions is related to the sine function in a different way. Explain how.
[latex]\cos x,~~\csc x,~~{and} ~~\sin^{-1} x[/latex] - 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 [latex]y=\csc x[/latex] and [latex]y = \cot x[/latex] have vertical asymptotes at the same [latex]x[/latex]-values?