Front Matter1 Triangles and Circles2 The Trigonometric Ratios3 Laws of Sines and Cosines4 Trigonometric Functions5 Equations and Identities6 Radians7 Circular Functions8 More Functions and Identities9 Vectors10 Polar Coordinates and Complex Numbers

Authored in PreTeXt

## Section 8.3 The Reciprocal Functions

### Subsection 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.

You are watching: What is the reciprocal of cosine

Definition 8.47. Three More Functions.

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

egin{align*} ext{The}~~ extbf{secant}:~~~~~~ amp lert{sec heta = dfrac{r}{x}}\ ext{The}~~ extbf{cosecant}:~~~~ amp lert{csc heta = dfrac{r}{y}}\ ext{The}~~ extbf{cotangent}:~~~ amp lert{cot heta = dfrac{x}{y}}end{align*}

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 (sec heta = 3 ext{,}) and (-dfrac{pi}{2} le heta le 0 ext{,}) find exact values for the other five trig functions.

Because (-dfrac{pi}{2} le heta le 0 ext{,}) we draw a reference triangle in the fourth quadrant, as shown at right. Because (sec heta = 3 = dfrac{3}{1} ext{,}) we label the horizontal leg with (x=1) and the hypotenuse with (r=3 ext{.})

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

egin{align*}cos heta amp =dfrac{x}{r}=dfrac{1}{3} amp amp sec heta =dfrac{r}{x}=dfrac{3}{1}=3\sin heta amp =dfrac{y}{r}=dfrac{-2sqrt{2}}{3} amp amp csc heta =dfrac{r}{y}=dfrac{3}{-2sqrt{2}}=dfrac{-3sqrt{2}}{4}\ an heta amp =dfrac{y}{x}=dfrac{-2sqrt{2}}{1}=-2sqrt{2} amp amp cot heta =dfrac{x}{y}=dfrac{1}{-2sqrt{2}}=dfrac{-sqrt{2}}{4}end{align*}

Checkpoint 8.49.

If (csc heta =4 ext{,}) and (90degree le heta le 180degree ext{,}) find exact values for the other five trig functions.

(cos heta = dfrac{-sqrt{15}}{4} ext{,}) (~sin heta = dfrac{1}{4} ext{,}) (~ an heta = dfrac{-1}{sqrt{15}} ext{,}) (~sec heta = dfrac{-4}{sqrt{15}} ext{,}) (~cot heta = -sqrt{15})

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

Reciprocal Trigonometric Functions.

egin{gather*} ext{The}~~ extbf{secant} ~ ext{function}:~~~~~~ lert{sec heta = dfrac{1}{cos heta}}\ ext{The}~~ extbf{cosecant}~ ext{function}:~~~~ lert{csc heta = dfrac{1}{sin heta}}\ ext{The}~~ extbf{cotangent}~ ext{function}:~~~ lert{cot heta = dfrac{1}{ an heta}}end{gather*}

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

Example 8.50.

Use a calculator to approximate (sec 47degree) to three decimal places.

With the calculator in degree mode, enter

(qquadqquadqquad) 1 ÷ COS 47 ) ENTER

to obtain (sec 47degree approx 1.466 ext{.}) Or we can calculate (cos 47degree) first, and then use the reciprocal key:

(qquadqquadqquad)COS 47 ) ENTER (smalloxed{x^{-1}} ) ENTER

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

egin{equation*}csc 3.5 = dfrac{1}{sin 3.5}= -2.8508~~~~~~ ext{and}~~~~~~cot(-4) = dfrac{1}{ an(-4)} = -0.8637end{equation*}

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

Exact Values for the Special Angles | |||

( heta) | (sec heta) | (sec heta) | (cot heta) |

(0) | (1) | undefined | undefined |

(dfrac{pi}{6}) | (dfrac{2sqrt{3}}{3}) | (2) | (sqrt{3}) |

(dfrac{pi}{4}) | (sqrt{2}) | (sqrt{2}) | (1) |

(dfrac{pi}{3}) | (2) | (dfrac{2sqrt{3}}{3}) | (dfrac{1}{sqrt{3}}) |

(dfrac{pi}{2}) | undefined | (1) | (0) |

Caution 8.52.

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

For example, (sec 0.8) is not equal to (cos^{-1}(0.8) ext{.}) Remember that (~alert{cos^{-1}(0.8)}~) is an angle, namely the angle whose cosine is 0.8, while (~alert{sec 0.8}~) is the reciprocal of the cosine of 0.8 radians, or (dfrac{1}{cos 0.8} ext{.}) You can check on your calculator that

egin{equation*}cos^{-1}(0.8)=0.6435~ ext{radians},~~~~ ext{and}~~~~sec 0.8=1.4353end{equation*}

Each of the reciprocal functions is undefined when its denominator is equal to zero. For example, the secant is undefined when (cos heta = 0 ext{,}) or when ( heta) is an odd multiple of (90degree ext{.})

Example 8.53.

For which angles is the cosecant undefined?

The cosecant is undefined when its denominator, (sin heta ext{,}) equals zero, and (sin heta = 0) when ( heta) is a multiple of (180degree ext{.}) In radians, (csc heta) is undefined if ( heta) is a multiple of (pi ext{.})

Checkpoint 8.54.

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

Note 8.55.

Although ( an dfrac{pi}{2}) is undefined, (cot dfrac{pi}{2}=0 ext{.})

### Subsection 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.

If ( heta) is one of the acute angles in a right triangle,

egin{align*}lert{sec heta} amp = lert{dfrac{ ext{hypotenuse}}{ ext{adjacent}}}\lert{csc heta} amp = lert{dfrac{ ext{hypotenuse}}{ ext{opposite}}}\lert{cot heta} amp = lert{dfrac{ ext{adjacent}}{ ext{opposite}}}end{align*}

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, (L ext{,}) of the shadow cast by a flagpole on a sunny day depends on the height, (h ext{,}) of the flagpole and the angle, ( heta ext{,}) that the sun”s rays make with ground.

Write an expression for the length, (L ext{,}) of the shadow cast by a flagpole of height (h) when the sun makes an angle of ( heta) 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 (20degree) from the ground.

From the figure, we see that (dfrac{L}{h}=cot heta ext{,}) or (L=hcot heta ext{.})

Substiting (alert{3}) for (h) and (alert{20degree}) for ( heta ext{,}) we find

Find an expression for the angle ( heta) in terms of (n ext{.})

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.

(displaystyle heta = dfrac{pi}{n})

(displaystyle b = dfrac{L}{n})

(displaystyle h=dfrac{L}{2n}cot dfrac{pi}{n})

(displaystyle A_T=dfrac{L^2}{4n^2}cot dfrac{pi}{n},~~A_P=dfrac{L^2}{4n}cot dfrac{pi}{n})

### Subsection 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 (y=cos x) to construct a graph of (f(x)=sec x ext{.})

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

When (x=dfrac{-pi}{2},~dfrac{pi}{2}) and (dfrac{3pi}{2},~ cos x=0 ext{,}) 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 ext{,}) as shown at right below.

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

For example, at (x=0) and (x=2pi ext{,}) we have (cos x = 1 ext{,}) so (sec x = frac{1}{1} = 1 ext{.}) Thus, we plot the points ((0,1)) and ((2pi,1)) on the graph of (f(x)=sec x ext{.}) Similarly, at (x=-pi) and (x=pi ext{,}) (cos x = -1 ext{,}) so the value of (sec x) is (frac{1}{-1} = -1 ext{,}) and we plot the points ((-pi,-1)) and ((pi,-1)) on the graph of (f(x)=sec x ext{.})

Finally, we notice that the values of (cos x) are decreasing toward (0) as (x) increases from (0) to (dfrac{pi}{2} ext{,}) so the graph of (f(x)=sec x) increases toward (infty) 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.

Checkpoint 8.59.

Use the graph of (y= an x) to sketch a graph of (g(x)=cot x ext{.})

Answer.

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 (dfrac{pi}{2} ext{,}) the values at which (cos x=0 ext{.}) The cosecant is undefined where (sin x=0 ext{,}) namely at multiples of (pi ext{.}) The cotangent is also undefined at multiples of (pi ext{,}) because ( an x=0) 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 (dfrac{pi}{2} ext{,}) that is, (dfrac{pi}{2},~dfrac{3pi}{2},~dfrac{5pi}{2},~ ldots ext{,}) and their opposites.

Because the range of the cosine consists of all (y)-values with (-1 le y le 1 ext{,}) the range of the secant includes the reciprocals of those values, namely (y ge 1) and (y le -1 ext{.})

Checkpoint 8.61.

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

Answer.

Domain of cosecant: all real numbers except integer multiples of (pi ext{;}) Range of cosecant: ((-infty,-1> cup <1, infty))

Domain of cotangent: all real numbers except integer multiples of (pi ext{;}) Range of cotangent: all real numbers

### Subsection Solving Equations

From the graph of the secant function, we can see that the equation (sec heta =k) has two solutions between (0) and (2pi) if (k ge 1) or (k le -1 ext{,}) but no solution for (-1 lt k lt 1 ext{.}) The same is true of the cosecant function: the equation (csc heta = k) has no solution for (-1 lt k lt 1 ext{.})

Example 8.62.

Solve (~~csc heta = dfrac{2sqrt{3}}{3}~~) for ( heta) between (0) and (2pi ext{.})

Solution.

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

egin{equation*}sin heta = dfrac{3}{2sqrt{3}} = dfrac{sqrt{3}}{2}end{equation*}

Because (dfrac{sqrt{3}}{2}) is one of the special values, we recognize that one of the solutions is ( heta =dfrac{pi}{3} ext{.}) The sine and the cosecant are also positive in the second quadrant, so the second solution is (pi – dfrac{pi}{3} = dfrac{2pi}{3} ext{.})

Checkpoint 8.63.

Solve (~~sec heta = -1.6~~) for ( heta) between (0) and (2pi ext{.})

Answer.

( heta=2.25, , 4.04)

### Subsection 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 (sec heta = 3 ext{,}) and (-dfrac{pi}{2} le heta le 0 ext{,}) find an exact value for (csc heta ext{.})

Solution.

Because (sec heta = dfrac{1}{cos heta} ext{,}) we see that (dfrac{1}{cos heta}=3 ext{,}) or (cos heta = dfrac{1}{3} ext{.}) We use the Pythagorean identity to find the sine.

egin{align*}cos^2 heta + sin^2 heta amp = 1 amp amp lert{ ext{Substitute}~ frac{1}{3} ext{ for } cos heta.}\left(dfrac{1}{3}

ight)^2 + sin^2 heta amp = 1 amp amp lert{ ext{Subtract}~ left(frac{1}{3}

ight)^2=frac{1}{9}~ ext{from both sides.}}\sin^2 heta amp = 1 – dfrac{1}{9}=dfrac{8}{9}end{align*}

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

egin{equation*}sin heta = -sqrt{dfrac{8}{9}} = dfrac{-2sqrt{2}}{3},~ ext{ and }~csc heta = dfrac{1}{sin heta} = dfrac{-3}{2sqrt{2}} = dfrac{-3sqrt{2}}{4}end{equation*}

Checkpoint 8.65.

If (csc heta = dfrac{-sqrt{13}}{3} ext{,}) and (pi le heta le dfrac{3pi}{2} ext{,}) find an exact value for (sec heta ext{.})

Answer.

(sec heta = dfrac{-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.

Cotangent Identity.

egin{equation*}lert{cot heta = dfrac{1}{ an heta} = dfrac{cos heta}{sin heta},~~~~sin heta

ot=0}end{equation*}

Example 8.66.

If (csc x=w) and (0 lt x lt dfrac{pi}{2} ext{,}) find an expression for (cot x ext{.})

Solution.

Because the sine is the reciprocal of the cosecant, we have (sin x = dfrac{1}{csc x} = dfrac{1}{w} ext{.}) We substitute (alert{dfrac{1}{w}}) for (sin x) in the Pythagorean identity to find

egin{equation*}cos x = pm sqrt{1 – sin^2 x} = pm sqrt{1 – left(alert{dfrac{1}{w}}

ight)^2}end{equation*}

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

egin{equation*}cos x = sqrt{1 – dfrac{1}{w^2}} = sqrt{dfrac{w^2-1}{w^2}} = dfrac{sqrt{w^2 – 1}}{abs{w}}end{equation*}

We can replace (abs{w}) by (w) in this last expression because (w gt 0 ext{.}) (Do you see why (w gt 0 ext{?})) Finally, because the cotangent is the reciprocal of the tangent, we have

egin{equation*}cot x = dfrac{cos x}{sin x} = dfrac{dfrac{w^2-1}{w}}{dfrac{1}{w}}=sqrt{w^2 – 1}end{equation*}

Checkpoint 8.67.

If (sec t = dfrac{2}{a}) and (dfrac{3pi}{2} lt t lt 2pi ext{,}) find expressions for (csc t) and (cot t ext{.})

Answer.

(csc t = dfrac{-2}{sqrt{1-a^2}} ext{,}) (~ cot t = dfrac{-a}{sqrt{1-a^2}})

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

Example 8.68.

Simplify (sec heta – an heta sin heta ext{.})

Solution.

We replace (~sec heta~) by (~dfrac{1}{cos heta}~) and (~ an heta~) by (~dfrac{sin heta}{cos heta}~) to get

egin{align*}dfrac{1}{cos heta} – dfrac{sin heta}{cos heta} cdot sin heta amp = dfrac{1}{cos heta} – dfrac{sin^2 heta}{cos heta}\amp = dfrac{1 – sin^2 heta}{cos heta}= dfrac{cos^2 heta}{cos heta} = cos hetaend{align*}

In the previous example, you can verify that

egin{equation*}sec heta – an heta sin heta = cos hetaend{equation*}

by graphing the functions (Y_1=sec heta – an heta sin heta) and (Y_2=cos heta) in the ZTrig window to see that they are the same.

Checkpoint 8.69.

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

Answer.

(sin^2x (1+cot^2 x) = sin^2 xleft(1 + dfrac{cos^2 x}{sin^2 x}

ight)=sin^2 x + cos^2 x = 1)

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 ( an heta) or (cot heta) and want to find the other trig values.

Two More Pythagorean Identities.

egin{equation*}lert{1 + an^2 heta = sec^2 heta~~~~~~~~~~1 + cot^2 heta = csc^2 heta}end{equation*}

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

Example 8.70.

If ( an alpha = dfrac{3}{5}) and (alpha) lies in the third quadrant, find exact values for (sec alpha) and (cos alpha ext{.})

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 (sin alpha = 3) and (cos alpha = 5 ext{!}) (Do you see why?) Instead, we begin with the Pythagorean identity for the tangent.

egin{align*}sec^2 alpha amp = 1 + an^2 alpha = 1 + (dfrac{3}{5})^2\amp = dfrac{25}{25} + dfrac{9}{25} = dfrac{34}{25}\sec alpha amp = pm sqrt{dfrac{34}{25}} = dfrac{pm sqrt{34}}{5}end{align*}

Because (alpha) 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 alpha = dfrac{-sqrt{34}}{5} ext{.}) The cosine of (alpha) is the reciprocal of the secant, so (cos alpha = dfrac{-5}{sqrt{34}} ext{.})

Checkpoint 8.71.

If (cot phi = dfrac{-3}{sqrt{2}}) and (phi) lies in the second quadrant, find exact values for (csc phi) and (sin phi ext{.})

Answer.

(csc phi = sqrt{dfrac{11}{2}} ext{,}) (~ sin phi = sqrt{dfrac{2}{11}})

Activity 8.5. Reciprocal Ratios.

Part 1

The figure shows a unit circle centered at (O ext{.}) The line through (AC) is tangent to the circle at (A ext{.}) (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 heta) and (sin heta ext{.})

Explain why ( riangle OAC) is similar to ( riangle OBD ext{.})

Find a line segment in ( riangle OAC) whose length is ( an heta ext{.}) Explain why your choice is correct.

Find a line segment in ( riangle OAC) whose length is (sec heta ext{.}) Explain why your choice is correct.

Find a line segment in ( riangle OAC) whose length is (csc heta ext{.}) Explain why your choice is correct.

Part 2

Here is another unit circle, with tangent (BD ext{.})

What is the measure of the angles (angle OCD) and (angle OCB ext{?})

Explain why ( riangle OCB) is similar to ( riangle OAC ext{.})

Find a line segment in ( riangle OCB) whose length is ( an heta ext{.}) Explain why your choice is correct.

Find a line segment in ( riangle OCB) whose length is (sec heta ext{.}) Explain why your choice is correct.

Explain why ( riangle OCB) is similar to ( riangle DCO ext{.})

Find a line segment in ( riangle DCO) whose length is (csc heta ext{.}) Explain why your choice is correct.

Review the following skills you will need for this section.

Algebra Refresher 8.6.

Solve (abs{2x-6}=4)

Solve (abs{1-3x}=10)

Simplify (sqrt{(x-4)^2})

Simplify (sqrt{(1-x)^2})

For what values of (x) is (abs{x+2}=x+2) ?

For what values of (x) is (abs{x-3}=x-3) ?

Graph the function (f(x)=dfrac{x}{abs{x}} ext{.})

Explain the difference between the graphs of (f(x)=(sqrt{x})^2) and (g(x)=sqrt{x^2} ext{.})

(underline{qquadqquadqquadqquad})

Algebra Refresher Answers

(displaystyle x = 1,~5)

(displaystyle x = -3,~dfrac{11}{3})

(displaystyle abs{x-4})

(displaystyle abs{1-x})

(displaystyle xge -2)

(displaystyle x lt 3)

The domain of (f) is (<0, infty) ext{,}) and the domain of (g) is ((-infty, infty) ext{.})

### Subsection Section 8.3 Summary

Subsubsection Vocabulary

Reciprocal

Secant

Cosecant

Cotangent

Subsubsection Concepts

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

egin{equation*} ext{The}~~ extbf{secant}:~~~~~~ sec heta = dfrac{r}{x}end{equation*}

egin{equation*} ext{The}~~ extbf{cosecant}:~~~~ csc heta = dfrac{r}{y}end{equation*}

egin{equation*} ext{The}~~ extbf{cotangent}:~~~ cot heta = dfrac{x}{y}end{equation*}

Reciprocal Trigonometric Ratios.If ( heta) is one of the acute angles in a right triangle,

egin{align*}sec heta amp = dfrac{ ext{hypotenuse}}{ ext{adjacent}}\csc heta amp = dfrac{ ext{hypotenuse}}{ ext{opposite}}\cot heta amp = dfrac{ ext{adjacent}}{ ext{opposite}} end{align*}

Reciprocal Trigonometric Functions.

egin{equation*} ext{The}~~ extbf{secant} ~ ext{function}:~~~~~~ sec heta = dfrac{1}{cos heta}end{equation*}

egin{equation*} ext{The}~~ extbf{cosecant}~ ext{function}:~~~~ csc heta = dfrac{1}{sin heta}end{equation*}

egin{equation*} ext{The}~~ extbf{cotangent}~ ext{function}:~~~ cot heta = dfrac{1}{ an heta}end{equation*}

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 heta = k ext{,}) (csc heta = k ext{,}) and (cot heta = 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.

Cotangent Identity.

egin{equation*}cot heta = dfrac{1}{ an heta} = dfrac{cos heta}{sin heta},~~~~sin heta

ot=0end{equation*}

Two More Pythagorean Identities.

egin{equation*}1 + an^2 heta = sec^2 heta~~~~~~~~~~1 + cot^2 heta = csc^2 hetaend{equation*}

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

Subsubsection Study Questions

Delbert says that (sec x) is just another way of writing (cos^{-1} x ext{,}) because (cos^{-1} x = dfrac{1}{cos x} ext{.}) Is he correct? Explain your reasoning.

Each of the following functions is related to the sine function in a different way. Explain how.

egin{equation*}cos x,~~csc x,~~ ext{and} ~~sin^{-1} xend{equation*}

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?

Subsubsection 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

### Exercises Homework 8-3

Exercise Group.

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

1.

(csc 27degree)

2.

(sec 8degree)

3.

(cot 65degree)

4.

(csc 11degree)

5.

(sec 1.4)

6.

(cot 4.3)

7.

(csc dfrac{5pi}{16})

8.

(sec dfrac{7pi}{20})

Exercise Group.

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

9.

(csc 30degree)

10.

(sec 0degree)

11.

(cot 45)

12.

(csc 60degree)

13.

(sec 150degree)

14.

(cot 120degree)

15.

(csc 135degree)

16.

(sec 270degree)

Exercise Group.

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

17.

See more: Where Is Misty In Pokemon Crystal, I Have All The

( heta) | (0) | (dfrac{pi}{6}) | (dfrac{pi}{4}) | (dfrac{pi}{3}) | (dfrac{pi}{2}) | (dfrac{2pi}{3}) | (dfrac{3pi}{4}) | (dfrac{5pi}{6}) | (pi) |

(sec heta) | (hphantom{0000}) | (hphantom{0000}) | (hphantom{0000}) | (hphantom{0000}) | (hphantom{0000}) | (hphantom{0000}) | (hphantom{0000}) | (hphantom{0000}) | (hphantom{0000}) |

(csc heta) | (hphantom{0000}) | (hphantom{0000}) | (hphantom{0000}) | (hphantom{0000}) | (hphantom{0000}) | (hphantom{0000}) | (hphantom{0000}) | (hphantom{0000}) | (hphantom{0000}) |

(cot heta) | (hphantom{0000}) | (hphantom{0000}) | (hphantom{0000}) | (hphantom{0000}) | (hphantom{0000}) | (hphantom{0000}) | (hphantom{0000}) | (hphantom{0000}) | (hphantom{0000}) |

18.

( heta) | (pi) | (dfrac{7pi}{6}) | (dfrac{5pi}{4}) | (dfrac{4pi}{3}) | (dfrac{3pi}{2}) | (dfrac{5pi}{3}) | (dfrac{7pi}{4}) | (dfrac{11pi}{6}) | (2pi) |

(sec heta) | (hphantom{0000}) | (hphantom{0000}) | (hphantom{0000}) | (hphantom{0000}) | (hphantom{0000}) | (hphantom{0000}) | (hphantom{0000}) | (hphantom{0000}) | (hphantom{0000}) |

(csc heta) | (hphantom{0000}) | (hphantom{0000}) | (hphantom{0000}) | (hphantom{0000}) | (hphantom{0000}) | (hphantom{0000}) | (hphantom{0000}) | (hphantom{0000}) | (hphantom{0000}) |

(cot heta) | (hphantom{0000}) | (hphantom{0000}) | (hphantom{0000}) | (hphantom{0000}) | (hphantom{0000}) | (hphantom{0000}) | (hphantom{0000}) | (hphantom{0000}) | (hphantom{0000}) |

19.

Evaluate. Round answers to three decimal places.

(displaystyle cos 0.2)

(displaystyle (cos 0.2)^{-1})

(displaystyle cos^{-1}0.2)

(displaystyle dfrac{1}{cos 0.2})

(displaystyle cosdfrac{1}{0.2})

(displaystyle sec 0.2)

20.

Evaluate. Round answers to three decimal places.

(displaystyle an 3.2)

(displaystyle an^{-1} 3.2)

(displaystyle cot 3.2)

(displaystyle dfrac{1}{ an 3.2})

(displaystyle an dfrac{1}{3.2})

(displaystyle ( an 3.2)^{-1})

Exercise Group.

For Problems 21–28, find exact values for the six trigonometric ratios of the angle ( heta ext{.})

21.

22.

23.

24.

25.

26.

27.

28.

29.

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 ext{,}) that sunlight travels through a layer of atmosphere of thickness (h ext{.})

Find the distance (to the nearest mile) that sunlight travels through a 100-mile layer of atmosphere when the sun is (40degree) above the horizon.

30.

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 ext{,}) of a curve whose degree of curvature is ( heta ext{.}) (Hint: The bisector of the angle ( heta) is perpendicular to the chord.)

Find the radius of a curve whose degree of curvature is (43degree ext{.})

31.

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

egin{equation*}t=sqrt{dfrac{l}{8}csc(2 heta)}~~ ext{seconds}end{equation*}

How long, to the nearest 0.01 second, will it take the ball to roll 2 feet horizontally on a plane tilted by (12degree ext{?})

Solve the formula for (l) in terms of (t) and ( heta ext{.})

32.

After a heavy rainfall, the depth, (D ext{,}) of the runoff flow at a distance (x) feet from the watershed down a slope at angle (alpha) is given by

egin{equation*}D=(kx)^{0.6}(cot alpha)^{0.3}~~ ext{inches}end{equation*}

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 (10degree) if (k=0.0006 ext{?})

Solve the formula for (x) in terms of (D) and (alpha ext{.})

Exercise Group.

For Problems 33–38, write algebraic expressions for the six trigonometric ratios of the angle ( heta ext{.})

33.

34.

35.

36.

37.

38.

39.

The diagram shows a unit circle. Find six line segments whose lengths are, respectively, (sin t,~ cos t,~ an t,~ sec t,~ csc t,) and (cot t ext{.})

40.

Use the figure in Problem 39 to find each area in terms of the angle (t ext{.})

(displaystyle riangle OAC)

(displaystyle riangle OBD)

sector (OAC)

(displaystyle riangle OFE)

Exercise Group.

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

41.

(sec heta = 2,~~ heta) in Quadrant IV

42.

(csc phi = 4,~~phi) in Quadrant II

43.

(csc alpha = 3,~~alpha) in Quadrant I

44.

(sec eta = 4,~~eta) in Quadrant IV

45.

(cot gamma = dfrac{1}{4},~~gamma) in Quadrant III

46.

( an heta = 6,~~ heta) in Quadrant I

Exercise Group.

For Problems 47–52, evaluate.

47.

(4 cot dfrac{pi}{3} + 2sec dfrac{pi}{4})

48.

(dfrac{1}{2} csc dfrac{pi}{6} – dfrac{1}{4}cot dfrac{pi}{6})

49.

(dfrac{1}{2} csc dfrac{5pi}{3}cot dfrac{3pi}{4})

50.

(6 cot dfrac{7pi}{6} sec dfrac{5pi}{4})

51.

((csc dfrac{2pi}{3} – sec dfrac{3pi}{4})^2)

52.

(sec^2 dfrac{5pi}{6} csc^2 dfrac{4pi}{3})

53.

Complete the table and sketch a graph of (y=sec x ext{.})

(x) | (0) | (dfrac{pi}{4}) | (dfrac{pi}{2}) | (dfrac{3pi}{4}) | (pi) | (dfrac{5pi}{4}) | (dfrac{3pi}{2}) | (dfrac{7pi}{4}) | (2pi) |

(sec x) | (hphantom{0000}) | (hphantom{0000}) | (hphantom{0000}) | (hphantom{0000}) | (hphantom{0000}) | (hphantom{0000}) | (hphantom{0000}) | (hphantom{0000}) | (hphantom{0000}) |

54.

Complete the table and sketch a graph of (y=csc x ext{.})

(x) | (0) | (dfrac{pi}{4}) | (dfrac{pi}{2}) | (dfrac{3pi}{4}) | (pi) | (dfrac{5pi}{4}) | (dfrac{3pi}{2}) | (dfrac{7pi}{4}) | (2pi) |

(csc x) | (hphantom{0000}) | (hphantom{0000}) | (hphantom{0000}) | (hphantom{0000}) | (hphantom{0000}) | (hphantom{0000}) | (hphantom{0000}) | (hphantom{0000}) | (hphantom{0000}) |

55.

Use the graph of (y=sin x) to sketch a graph of its reciprocal, (y=csc x ext{.})

56.

Use the graph of (y=cos x) to sketch a graph of its reciprocal, (y=sec x ext{.})

57.

Complete the table and sketch a graph of (y=cot x ext{.})

(x) | (0) | (dfrac{pi}{4}) | (dfrac{pi}{2}) | (dfrac{3pi}{4}) | (pi) | (dfrac{5pi}{4}) | (dfrac{3pi}{2}) | (dfrac{7pi}{4}) | (2pi) |

(cot x) | (hphantom{0000}) | (hphantom{0000}) | (hphantom{0000}) | (hphantom{0000}) | (hphantom{0000}) | (hphantom{0000}) | (hphantom{0000}) | (hphantom{0000}) | (hphantom{0000}) |

58.

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

Exercise Group.

For Problems 59-64,

Graph each function in the ZTrig window, and write a simpler expression for the function.

Show algebraically that your new expression is equivalent to the original one.

59.

(y=dfrac{csc x}{cot x})

60.

(y=dfrac{sec x}{ an x})

61.

(y=dfrac{sec x cot x}{csc x})

62.

(y=dfrac{csc x an x}{sec x})

63.

(y= an x csc x)

64.

(y=sin x sec x)

Exercise Group.

For Problems 65–70, find all solutions between (0) and (2pi ext{.})

65.

(3csc heta + 2=8)

66.

(-2sec heta + 8=3)

67.

(sqrt{2}sec heta =-2)

68.

(8+csc heta =6)

69.

(2cot heta = -sqrt{12})

70.

(sqrt{3}cot heta =1)

Exercise Group.

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

71.

If ( an alpha = -2) and (dfrac{pi}{2} lt alpha lt pi ext{,}) find (cos alpha ext{.})

72.

If (cot eta = dfrac{5}{4}) and (pi lt eta lt dfrac{3pi}{2} ext{,}) find (sin eta ext{.})

73.

If (sec x = dfrac{a}{2}) and (0 lt alpha lt dfrac{pi}{2} ext{,}) find ( an x ext{.})

74.

If (csc y = dfrac{1}{b}) and (dfrac{pi}{2} lt y lt pi ext{,}) find (cot y ext{.})

75.

If (csc phi = w) and (dfrac{3pi}{2} lt alpha lt 2pi ext{,}) find (cos phi ext{.})

76.

If (sec heta = dfrac{3}{z}) and (pi lt alpha lt dfrac{pi}{2} ext{,}) find (sin heta ext{.})

Exercise Group.

For Problems 77–80, find exact values for (sec s,~ csc s,) and (cot s ext{.})

77.

78.

79.

80.

Exercise Group.

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

81.

(sec heta an heta)

82.

(csc phi cot phi)

83.

(dfrac{csc t}{cot t})

84.

(dfrac{ an v}{sec v})

85.

(sec eta – an eta)

86.

(cot alpha + csc alpha)

87.

(sin x an x – sec x)

88.

(csc y – cos y cot y)

89.

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

90.

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

91.

Suppose that (cot heta = 5) and ( heta) lies in the third quadrant.

Use the Pythagorean identity to find the value of (csc heta ext{.})

Use identities to find the values of the other four trig functions of ( heta ext{.})

92.

Suppose that ( an heta = -2) and ( heta) lies in the second quadrant.

Use the Pythagorean identity to find the value of (sec heta ext{.})

Use identities to find the values of the other four trig functions of ( heta ext{.})

93.

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

94.

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

95.

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

egin{equation*}a csc A = b csc B = c csc Cend{equation*}

96.

See more: Which Event Marked The Beginning Of The Civil War? Trigger Events Of The Civil War

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

## Discussion about this post