Basic Trigonometry - Montgomery Collegefaculty.montgomerycollege.edu/fkatira1/TrigTextFinal...

Basic Trigonometry

Stephen PerencevichDepartment of Mathematics

Montgomery College

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Revised July 22, 2016All firgures generated by Mathematica. R©

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Perencevich, StephenTrigonometry

Stephen PerencevichMontgomery CollegeRockville, MD [email protected]

c© 2016 George Canter Institute. No part of this publication may be re-produced, stored in a retreval system, or transmitted in any form or by anymeans, electronic, mechanical, photocopying, recording, or otherwise, withoutprior written permission of the author.

For Missy,

the flower of Lackawanna.

ii

Contents

1 Right-Triangle Trigonometry 21.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2 Right-Triangle Trigonometry . . . . . . . . . . . . . . . . . . . . 51.3 Trigonometric Functions of any Angle . . . . . . . . . . . . . . . 13

2 Trigonometric Identities 222.1 Basic Trigonometric Identities . . . . . . . . . . . . . . . . . . . . 22

Midcourse Review 26Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26Practice Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

3 Radian Measure 283.1 The Definition of Radian Measure . . . . . . . . . . . . . . . . . 28

4 Graphs of Trigonometric Functions 344.1 The Unit Circle . . . . . . . . . . . . . . . . . . . . . . . . . . . . 344.2 Graphs of Trigonometric Functions . . . . . . . . . . . . . . . . . 36

5 Inverse Trigonometric Functions 405.1 Inverse Trigonometric Fucntions . . . . . . . . . . . . . . . . . . 405.2 Trigonometric Equations . . . . . . . . . . . . . . . . . . . . . . . 44

Final Review 47Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47Practice Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

A The Greek Alphabet 49

B Midcourse Review Answers 50

C Final Review Answers 52

D Answers to Odd-Numbered Problems 54

Index 58

1

Chapter 1

Right-TriangleTrigonometry

1.1 IntroductionIn this section we will recall some theorems on triangles from geometry.

Theorem 1.1.1. The sum of the angles in any triangle is 180

A B

C

a

b

c

Figure 1.1.1

In terms of Figure 1.1.1: m∠A+m∠B +m∠C = 180.

Definition 1.1.2.

A B

C

D E

F

ab

c

de

f

Two triangles are similar, de-noted ∆ABC ∼ ∆DEF if themeasures of corresponding an-gles are equal and if the ratiosof the lengths of correspondingsides are equal:

a

d= b

e= c

f

2

CHAPTER 1. RIGHT-TRIANGLE TRIGONOMETRY 3

Theorem 1.1.3. (AA Similarity) If two pairs of angles in two triamgles arecongruent, then the triangles are similar.

A B

C

D E

F

ab

c

de

f

Figure 1.1.2: Similar Triangles

In terms of Figure 1.1.2: if m∠A = m∠D and m∠B = m∠E, then ∆ABC ∼∆DEF .

Recall that if two triangles are similar, then ratios of pairs of correspondingsides are equal. Again, in terms of Figure 1.1.2: if ∆ABC ∼ ∆DEF , then(among other ratios)

b

c= e

fand b

e= a

d

Conventions: Note that in Figures 1.1.1 and 1.1.2 the angles of the twotriangles are labeled by upper case letters and the sides opposite the angles arelabeled with the corresponding lower case letter. We will follow this conventionthroughout these notes. In geometry angles, which are geometrical objects, arerigously distinguished frm their measures, which are numbers. In these noteswe will allow not-technically-correct identification of an angle with its measure.Thus we will write A = 63 in place of m∠A = 63.

Theorem 1.1.4. (Pythagoras)In a right triangle with side lengths a, b, and c, where c is the length of the

hypotenuse,c2 = a2 + b2

a

b

c

Figure 1.1.3


Section 1.1 Problems

A B

C

D E

F

ab

c

de

f

Figure 1.1.4

For Problems 1–4 we are given that ∆ABC ∼ ∆DEF .

1. If a = 2 and b = 3, find de .

2. If a = 3, b = 4, and d = 5, finde.

3. If a = 2 and c = 5, find fd .

4. If b = 4, c = 7, and d = 5, findf .

For Problems 5–8 we are given that ∆ABC is a right triangle with C a rightangle.

A

B

C

a

b

c

Figure 1.1.5

5. If a = 2 and b = 3, find c.

6. If a = 3 and b = 5, find b.

7. If a =√

2 and b =√

3, find c.

8. If a = 1 and c = 2, find b.

9. In ∆ABC, if A = 23 andB = 55 find C.

10. In ∆DEF , if D = 33 andE = 47 find F .

11. In ∆XY Z, if X = Y and Z =100 find X.

12. In ∆PQR, if P = 2Q and R =120 find P .

13. If ∆ABC with C a right angleand A = 27, find B.

14. If ∆ABC with C a right angleand B = 62, find A.


1.2 Right-Triangle TrigonometryConsider a right triangle that contains an angle with mearure 27.

27°

Opp

Adj

Hyp

Figure 1.2.1

As usual, the side opposite the right angle is called the hypotenuse of thetriangle. The side that forms the 27 angle with the hypotenuse is called theadjecent side to the angle and the last side is called the opposite side of theangle. Any right triangle containing a 27 angle will be similar to the one inFigure 1.2.2.

In any right triangle containing an angle of measure 27, the ratios opphyp,

adjhyp, and opp

adj will be the same. These ratios are called the sine, cosine andtangent, respectively.

The same can be said of any acute angle. This leads to the following defini-tion.

Definition 1.2.1. Let θ ba an acute angle in a right triangle. We define thefollowing functions on θ.

Θ

Opp

Adj

Hyp

Figure 1.2.2

sin θ = opphyp

cos θ = adjhyp

tan θ = oppadj

In days gone by, the values of the trigonometric functions were tabulated andbooks on trigonometry contained tables such as Table 1.2.1. Today, of course,these values are easily computed on a scientific calculator.


θ sin θ cos θ tan θ

1 0.0175 0.9998 0.0175

2 0.0349 0.9994 0.0349

3 0.0523 0.9986 0.0524

4 0.0698 0.9976 0.06992

5 0.0872 0.9962 0.0875...

27 0.3907 0.9205 0.4245

Table 1.2.1: A Short Trigonometric Table

To solve a triangle means to find the measures of unknown angles andlengths of unknown sides of a given triangle.

Example 1.2.2.Solve the triangle given in the diagram.

c

3.5

a

34° A

B

C

Since the acute angles in a right triangleare complementary,

B = 90 − 34 = 56

tan 34 = a

3.5a = 3.5 tan 34

= 2.4

Using the Pythagoras theorem:

c2 = a2 + b2

c2 = 2.42 + 3.52

c =√

2.42 + 3.52

= 4.2

The values in Table 1.2.1 are decimal approximations of the values of thetrigonometric functions. If the exact value of one of the trigonometric functionshappens to be known, then the exact value of the others can be found.


Example 1.2.3.Given that for acute angle θ, sin θ = 3

7 , find cos θ and tan θ.Choose a right triangle containing θ with opposite side of length 3 and hy-

potenuse of length 7. Let x be the length of the remaining side.

7

x = 2 10

3

Θ

Using the Pythagoras theorem,

x2 + 32 = 72

x2 = 40x =√

40 = 2√

10

cos θ = 2√

107

tan θ = 32√

10

One of the earliest uses of trigonometry is in indirect measurement problems.The height of a tall building would be difficult to measure directly. It is easierto measure a horizontal distance along the graound and the angle of a sightlineand then use trigonometry to deduce the desired height.

Example 1.2.4.At a point 105 feet away from a building the angle between the ground the

top of the building is 55. Find the height of the building.

55°

105

h

Let h be the height of the building.Then,

tan 55 = h

105h = 105 tan 55

= 150.0

There are three remaining trigonometric functions: cosecant, secant, cotan-gent, which are simply the reciprocals of sine, cosine, and tangent, respectively.


Definition 1.2.5. For an angle θ we define,

csc θ = 1sin θ

sec θ = 1cos θ

cot θ = 1tan θ

Example 1.2.6.Given that for acute angle θ, cos θ = 4

5 , find the values of the remaining 5trigonometric functions on θ.

Choose a right triangle containing θ with adjacent side of length 4 andhypotenuse of length 5. Let y be the length of the remaining side.

5

4

y = 3

Θ


y2 + 42 = 52

y2 = 9y = 3

sin θ = 35 csc θ = 5

3

cos θ = 45 sec θ = 5

4

tan θ = 34 cot θ = 4

3

The Inverse Trigonometric Functions

Given an acute angle θ, the trigonometric functions assign a number to thatangle. The inverse trigonometric functions reverse that process. That is giventhe value of a trigonometric function on some angle, the inverse trigonometricfunction will give the measure of the angle. The inverse trigonometric functionsare denoted: sin−1, cos−1, and tan−1. Scientific calculators have buttons forthese functions.

For example, using a scientific calculator,

sin 52 = 0.788010754sin−1 0.788010754 = 52


Example 1.2.7.∆ABC is a right triangle with C a right angle. Given that a = 5.1 and

c = 8.4, solve ∆ABC.

8.4

b = 6.7

5.1

A

B

C

Using the Pythagoras theorem, b = 6.7.

sinA = 5.18.4

A = sin−1(

5.18.4

)A = 37.4

B = 90 − 37.4 = 52.6

Special Triangles

In most cases evaluating the trigonometric functions on a given angle θ is acalculator problem. There are three acute angles, 30, 45, and 60, on whichthe exact value of the trigonometric angles can be found.

Consider an eqilateral triangle with side length 2. An altitude drawn fromone of the angles bisects the angle and the opposite side.

60°

30°

h

1

2 2Using the Pythagoras theorem,

12 + h2 = 22

h2 = 3h =√

3

We now construct a triangle as follows: choose two perpendicular lines andconstruct a segment of length 1 on each line, each with the point of intersectionfor one endpoint.


45°

45°

1 r

1


12 + 12 = r2

r2 = 2r =√

2

We can now use the following triangles to evaluate the trigonometric func-tions on the angles, 30, 45, and 60.

30°

60°

12

345°

45°

12

1

Figure 1.2.3: The Special Trianges


For problems 1–9 ∆ABC isa right triangle with C aright angle. Solve the trian-gle with the given informa-tion. Give your answer cor-rect to one decimal place.

c

b

a

A

B

C

1. a = 8.7; A = 22.

2. b = 3.1; A = 39.

3. c = 8.1; B = 52.

4. a = 3.3; b = 4.5.

5. a = 4.3; c = 7.6

6. b = 5.8; c = 8.5.

7. b = 7.9; B = 47.

8. c = 9.1; A = 17.

9. c = 5.5; B = 55.


For problems 10–15 ∆ABCis a right triangle with C aright angle. Given the valueof one of the trigonomet-ric function, find the valuesof the remaining 5. Giveexact-value answers. Θ

10. sin θ = 37 .

11. cos θ = 513 .

12. tan θ = 52 .

13. sin θ = 12 .

14. cos θ =√

25

15. tan θ =√

3.

16. Using Figure 1.2.3, evaluate the six trigonometric functions on

(a) θ = 30

(b) θ = 45

(c) θ = 60

Angle ofDepression

Angle ofElevation

In application problems, the angle ofdepression is the angle between theline of sight of an observer lookingdownward and the horizontal. The an-gle of elevation is the angle betweenthe line of sight of an upward lookingobserver and the horizontal.

17. Suzy’s eyes are 5 feet above theground. She is standing 12 feetaway from the base of a flag poleand the angle of elevation fromher eyes to the top of the pole is23. What is the height of theflag pole?

18. An observer on the ground mea-sures the angle of elevation of anairplane to be 38. A secondoberver, standing 4.3 miles fromthe first observer, is directly be-low the plane. What is the alti-tude of the plane?

19. Find the angle of elevation ofthe top of a 63-foot-tall buildingfrom a point on the ground thatis 97 feet from the base of the

building.

20. In the previous problem find theangle of elevation of the top ofthe building from the eye level ofan observer whose eyes are at aheight of 5.4 feet. (The observerstands 97 feet from the base ofthe buikding.)

21. The angle of elevation to the topof an oak tree from a point on theground 59 feet from the base ofthe tree is 43. What is the dis-tance from the point to the topof the tree?


Evel Knievel was a daredevil whorode his motorcycle over rampsto jump various objects. He es-timates that, in order to be safe,the angle of the ramp with theground should be no more than17.

22. Evil Knievel is going to jumpover 5 school buses that have aheight of 9 feet. If his ramphas an angle of 17, what is thedistance along the ground fromthe base of the ramp to the firstschool bus?

23. In over to jump over a 12-foot-high pen containing mountain li-ons, Evel Knievel constructs aramp that measures 50 feet alongthe ground. What is the anglethat the ramp makes with theground?

24. Evel Knievel is going to con-struct a ramp th jump over somelocomotives that have a height 14feet. If the angle of the rampwith the ground is 17, what isthe shortest ramp (not the dis-tance along the ground) that ourdaredevil can construct?


1.3 Trigonometric Functions of any AngleThe trigonometric functions defined in Section 1.2 can only be evaluated onacute angles. In this section we expand the definitions so that the trigonometricfunctions can be evaluated on any angle.

Angles in Standard Position

Recall the following definition from geometry.

Definition 1.3.1.

Θ

An angle is the union of two rays witha common endpoint.

We will adpot a more dynamic view of angles. An angle in standard po-sition is angle in the coordinate plane with its vertex at the origin. The angleis formed by placing both rays along the positive x-axis, fixing one in placeand rotating the other. The fixed ray is called the initial ray and the rotatedray is called the terminal ray. In forming such an angle the terminal raymay be rotated either in the counterclockwise or clockwise directions. They aredistinguished by defining their measures to be positive or negative, respectively.

Θ > 0

Θ < 0

x

y

Angles formed by counterclock-wise rotation have positive mea-sures.

Angles formed by clockwise ro-tation have negative measures.

Figure 1.3.1: Angles in Standard Position


Recall that the four quadrants of the coordinate plane are defined as inFigure 1.3.2.

I

x > 0, y > 0

II

x < 0, y > 0

III

x < 0, y < 0

IV

x > 0, y < 0

Note the inequalities definingthe quadrants are strict, so thatthe coordinate axes from theboundries of the quadrants, butdo not belong to the quadrants.

Figure 1.3.2

Angles whose terminal rays lie on the coordinate axes are called boundaryangles

0°

360°

90°

180°

270°

III

III IV

-360°

0°

-270°

-180°

-90°

III

III IV

Figure 1.3.3: Boundary Angles

0This is a nonstandard usage. Many authors call these angles quandrantal angles. Wehave adopted the term boundary angle as more discriptive. The terminal ray of such an anglelies in the boundary of the quadrants, not in the quadrants.


Definition 1.3.2.

Θ'

Θ

x

y

For an angle, θ, in standard po-sition whose terminal ray is in-terior to one of the quadrants,the reference angle, θ′, of θ isthe acute angle formed by theterminal ray and the x-axis.

Example 1.3.3.Find the reference angle for the following angles: 150, −130, 300

0°

360°

90°

180°

270°

Θ'=30°

Θ=150°

-360°

0°

-270°

-180°

-90°

Θ'=50°

Θ=-130°

0°

360°

90°

180°

270°

Θ'=60°

Θ=300°

Angles in standard position are formed by rotation of the terminal ray. Thereis no reason to restrict the rotation to once-round. This can result in differentangles having the same terminal ray.


Definition 1.3.4.

0°

360°

90°

180°

270°

Θ

Θ+360°

Angles with the same terminalray are called coterminal an-gles.

Example 1.3.5.Find three angles coterminal with θ = 30.

(i) 30 + 360 = 390

(ii) 30 + 2 · 360 = 750

(iii) 30 − 360 = −330


The Definition of the Trigonometric Functions

We now turn to the definition of the trigonometric functions on angles instandard position. Consider first an angle in standard position whose terminalray lies in the first quadrant. Let (x, y) be a point on the terminal ray otherthan the origin. Form a triangle by droping a perpendicular from this point tothe x-axis. This is called the reference triangle if θ. Let r be the distancefrom the point to the origin.

0°

360°

90°

180°

270°

Hx, yL

x

yr

Θ

By the Pythagoras theorem

r2 = x2 + y2

Using the definitions of thetrigonometric functions of anacute angle,

sin θ = y

rcsc θ = r

y

cos θ = x

rsec θ = r

x

tan θ = y

xcot θ = x

y

We now make the observation that the values of the trigonometric functionsgiven above do not depend on the terminal ray of the angle lying in the firstquadrant, so we take them to be our general definition.

Definition 1.3.6. (The trigonometric functions.)Let θ be an angle in standard position, P (x, y) be a point on the terminal

ray of θ, and let r be the distance from P to the origin. We define the values ofthe trigonometric functions on θ as follows:

0°

360°

90°

180°

270°

Hx, yL

x

y r

Θ'

Θ

sin θ = y

rcsc θ = r

y

cos θ = x

rsec θ = r

x

tan θ = y

xcot θ = x

y


In Definition1.3.6, r is a distance, so r > 0. However, (x, y) are the coordi-nates of a point and so they may be positive, negative, or zero. Regardless ofthe values of x and y, r still satisfies the Pythagoras relation: r2 = x2 + y2.

Example 1.3.7.Let θ be an angle in standard position whose terminal ray passes through

the point (2,−5). Find the values of the six trigonometric functions on θ.

0°

360°

90°

180°

270°

H2, -5L

2

-5r

Θ'

Θ

r2 = x2 + y2

r2 = 22 + (−5)2

= 29r =√

29

sin θ = − 5√29

csc θ = −√

295

cos θ = 2√29

sec θ =√

292

tan θ = −52 cot θ = −2

5

0°

360°

90°

180°

270°

I: all > 0II: sin > 0

III: tan > 0 IV: cos > 0

As Example 1.3.11 shows, thevalues of the trigonometricfunctions can sometimes be neg-ative. The signs of the trigono-metric functions depend thesigns on the signs of the coor-dinates (x, y). The diagram tothe left is helpful in remember-ing which functions are positivein which quadrants.

Figure 1.3.4


Example 1.3.8.Given that sin θ = 3

4 and 90 < θ < 180 Find the values of the remainingfive trigonometric functions on θ.

Since sin θ = 34 > 0, the terminal ray of θ must lie in Quadrants I or II. The

fact that 90 < θ < 180 implies that the terminal ray of θ lies in Quadrant II.Form the reference triangle of θ using the point (x, 3).

0°

360°

90°

180°

270°

Hx, 3L

x

34

Θ'

Θ

r2 = x2 + y2

42 = x2 + 32

x2 = 7x = ±

√7

In this case, the solution we arelooking for is x = −

√7.


3cos−θ =

√7

4 sec−θ = 4√7

tan θ = − 3√7

cot θ = −√

73

Finally we observe that for any angle θ interior to one of the quadrants, θ′is an ordinary acute angle and,

0°

360°

90°

180°

270°

Hx, yL

x

y r

Θ'

Θ

sin θ′ = |y|r

csc θ′ = r

|y|

cos θ′ = |x|r

sec θ′ = r

|x|

tan θ′ = |y||x|

cot θ′ = |x||y|

Hence the value of a trigono-metric function on θ′ is the ab-solute value of the value of thefunction on θ.


Theorem 1.3.9. If θ is an angle in stard position whose terminal ray liesinterior to one of the quadrants, then

sin θ′ = | sin θ| csc θ′ = | csc θ|

cos θ′ = | cos θ| sec θ′ = | sec θ|

tan θ′ = | tan θ| cot θ′ = | cot θ|

Example 1.3.10.Find the values of sin 150, cos 150, and tan 150.

The reference angle of θ = 150 is θ′ = 30.

Π6

Π31

3

2

Π4

Π4

1

1

2

0°

360°

90°

180°

270°

Θ'=30°

Θ=150°

sin 30 = 12

cos 30 =√

32

tan 30 = 1√3

Using Theorem 1.3.9 and Fig-ure 1.3.4

sin 150 = +12

cos 150 = −√

32

tan 150 = − 1√3

Because it is easy to find points on the coordinate axes, we can directly usethe the definition of the trigonometric functions to evaluate on the boundryangles.


Example 1.3.11.Find the values of sin 90, cos 90, and tan 90.

The point P (0, 1) lies on the terminal ray of the 90 angle and has distancer = 1 to the origin.

0°

360°

90°

180°

270°

Hx,yL = H0,1L

r = 1 sin 90 = y

r= 1

1 = 1

cos 90 = x

r= 0

1 = 0

tan 90 = y

x= 1

0 (undefined)


For problems 1–9, sketch the given angle in standard position and find itsreference angle.

1. θ = 120.

2. θ = 200.

3. θ = 295.

4. θ = −220.

5. θ = 27

6. θ = −55.

7. θ = 135.

8. θ = 240.

9. θ = 310.

For problems 10–15, find three angles coterminal to the given angle.

10. θ = 120.

11. θ = 200.

12. θ = 295.

13. θ = −220.

14. θ = 27.

15. θ = 310.

For problems 16–21, evaluate the six trigonometric functions on the angle θwhose terminal ray passes through the given point. Give exact-value answers.

16. P (−2, 5).

17. P (−3,−4).

18. P (5,−3).

19. P (−1, 0).

20. P (2, 4)

21. (0,−1).

For problems 22–27, evaluate the six trigonometric functions on the givenangle Give exact-value answers.

22. θ = 120.

23. θ = 225.

24. θ = 330.

25. θ = 180.

26. θ = 240

27. θ = 270.

Chapter 2

Trigonometric Identities

2.1 Basic Trigonometric IdentitiesWhile there are six trigonometric functions, they are not independent. In thissection we will explore some relationships among them. We will use the right-triangle definition of the trigonometric functions for geometric motivation, butthe identities we find will hold for the generalized definitions as well.

The Reciprocal Identities

This first set of identities is simply a restatement of Definition 1.2.5:

Reciprocal Identities

sec θ = 1cos θ cos θ = 1

sec θcsc θ = 1

sin θ sin θ = 1csc θ

cot θ = 1tan θ tan θ = 1

cot θ

The trigonometric ratios depend on the fact that any two right trianglescontaining the acute angle θ must be similar. It is also true that any individ-ual right triangle that contains the angle acute θ must satisfy the Pythagorastheorem. The next two sets of identities establish relationships among the thetrigonometric ratios derived from similarity relations and the Pythagoras theo-rem.

The Quotient IdentitiesThis set of identities establishes relationships among ratios of trigonometric

functions.

22

CHAPTER 2. TRIGONOMETRIC IDENTITIES 23

a

b

c

Θ

Figure 2.1.1

Referring to Figure 2.1.1,

sin θcos θ = b/c

a/c

= b

c· ca

= b

a= tan θ

cos θsin θ = a/c

b/c

= a

c· cb

= a

b= cot θ

Quotient Identities

sin θcos θ = tan θ cos θ

sin θ = cot θ

The Pythagoras Identities

The right triangle in Figure 2.1.1 satisfies the Pythagoras theorem: c2 =a2 + b2. We can use this theroem to investigate relationships among the squaresof the quotients that define the trigonometric functions. First we will computethe sum of the squares of the sine and cosine ratios.1

cos2 θ + sin2 θ =(ac

)2+(b

c

)2= a2

c2 + b2

c2

= a2 + b2

c2 = c2

c2 = 1

Putting all this together, we have:

cos2 θ + sin2 θ = 1 (†)1Note that sin2 θ is the notation for (sin θ)2 and similarly for the other trigonometric

functions.


We can now use identity (†) together with the quotient and reciprocal iden-tities to generate two more identities.

Dividing identity (†) by cos θ:

cos2 θ

cos2 θ+ sin2 θ

cos2 θ= 1

cos2 θ

1 + tan2 θ = sec2 θ

Dividing identity (†) by sin θ:

cos2 θ

sin2 θ+ sin2 θ

sin2 θ= 1

sin2 θ

cot2 θ + 1 = csc2 θ

Pythagoras Identities

cos2 θ + sin2 θ = 11 + tan2 θ = sec2 θ

1 + cot2 θ = csc2θ

Using the basic identities established so far and some algebra, we can es-tablish other identities. The process of proving a trigonometric identity is oneof simplification. Pick one side of the identity and use algebra and the basicidentities given above to simplify it. The end of the process should yield theother side o fthe identity.

Example 2.1.1.Prove the identity: cos θ + sin θ tan θ = sec θ.

cos θ + sin θ tan θ = cos θ + sin θ sin θcos θ

= cos θcos θ ·

cos θ1 + sin2 θ

cos θ

= cos2 θ

cos θ + sin2 θ

cos θ

= cos2 θ + sin2 θ

cos θ= 1

cos θ= sec θ


Example 2.1.2.Prove the identity: cot θ

csc θ = cos θ.

cot θcsc θ = cos θ/ sin θ

1/ sin θ

= cos θsin θ ·

sin θ1

= cos θ

Example 2.1.3.Prove the identity: tan θ + cot θ = csc θ sec θ.

tan θ + cot θ = sin θcos θ + cos θ

sin θ

= sin θsin θ ·

sin θcos θ + cos θ

sin θ ·cos θcos θ

= sin2 θ + cos2 θ

sin θ cos θ= 1

sin θ cos θ= 1

sin θ ·1

cos θ= csc θ sec θ


For problems 1–10, prove the trigonometric identity.

1. tan θsin θ = sec θ.

2. tan θsec θ = sin θ.

3. sin θ + sin θ cot2 θ = csc θ.

4. sin θ cos θ tan θ = 1 = cos2 θ.

5. tan θ cos θ = sin θ

6. sin θ csc2 θ = csc θ.

7. 1 + tan2 θ

1− tan2 θ= 1

cos2 θ − sin2 θ.

8. cos2 θ − sin2 θ = 1− 2 sin2 θ.

9. sec θ − 1sec θ + 1 = 1− cos θ

1 + cos θ

10. tan2 θ − sin2 θ = tan2 θ sin2 θ.

Midcourse Review

Review1. Find the values of all six trigonometric

functions on angle A in the triangle below.

11

b

6

A

B

C

2. Find the reference angles for each of thefollowing angles.

(a) 120

(b) 212(c) 350

(d) −97(e) −330

(f) −45

3. Given that α is an acute angle and sinα =2/3, find the values of the remaining 5trigonometric functions on θ.

4. Given that 90 < θ < 180 and sin θ = 2/3,find the values of the remaining 5 trigono-metric functions on θ.

5. Solve the right triangle in the figure below.

c

3.5

a

33°A

B

C

6. Evaluate sin2 23 + cos2 23.

7. Given that sin 23 = 0.3907, find:

(a) sin 157

(b) sin 203

(c) sin 337

(d) cos 67

(e) cos 113

(f) cos(−113)

8. Use the special triangles to evaluate:

(a) cos 120

(b) tan 210

(c) sin 330

(d) cos 135

(e) tan(−60)(f) sin 150

9. Evaluate:

(a) cos 90

(b) tan 180

(c) sin 0

(d) sin 270

(e) tan 270

(f) cos 0

10. Given sin θ = −.225.

(a) Find θ′, the referance angle for θ.(b) Given that θ lies in Quad. IV, evalu-

ate:i. cos θ

ii. tan θ


c

4.7

3.1

A

B

C

12. Prove the following trigonometric identi-ties:

(a) cot θcsc θ = cos θ

(b) cotα+ tanα = secα cscα(c) tan β sin β + cosβ = secβ

(d) 11− sin θ −

11 + sin θ = 2 tan θ sec θ

26


Practice Test1. If θ is an acute angle and sin θ =

37, find tan θ. Give an exact-value answer.

2. In the triangle below, find thelength of side a. Give youranswer correct to one decimalplace.

c

4.7

a

33°A

B

C

Problem 2

3. Given that cos θ = 45 and 270 <

θ < 360, find the exact valueof sin θ.

4. Find the reference angle for θ =265.

5. Given that sin θ = −0.5731 andθ lies in Quad IV. For this prob-lem, give your answers correct tothree digits after the decimal.

(a) Find θ′, the reference anglefor θ.

(b) cos θ =(c) tan θ =

6. Given that tan θ = 5√2

, find the

exact value of cot(90 − θ).

7. Use your knowledge of the spe-cial triangles and the definitionof the trigonometric functions togive the exact value of the fol-lowing:

(a) sin 60

(b) cos 90

(c) tan 300

(d) sin 150

(e) cos 240

(f) tan 180

8. In the triangle below, find themeasure of angle A. Give youranswer correct to two decimalplaces.

c

6.2

4.3

A

B

C

Problem 8

9. Find the value of the six trigono-metric functions on the angleshown in the diagram below

Θ

x

y

-4 -3 -2 -1 1 2 3 4

-6

-5

-4

-3

-2

-1

1

2

3

4

5

6

Problem 9

10. Prove the following identity:csc θ(csc θ − sin θ) = cot2 θ.

Chapter 3

Radian Measure

3.1 The Definition of Radian MeasureUp to now we have been measuring angles in degrees without much comment.We now wish to step back and consider what is meant by degree mrasure. Inorder to define an angle measure, divide a circle into 2n equal parts by choosingn equally spaced points on the top half of the circle and drawing diameters fromthose points. The result is that 2n equal central angles of the circle are formed.We then take the measure of one of these central angles to be the standard anglemeasure.

n = 8

The measure of an angle is then the number of nonoverlapping standardangles that fit into the angle. Degree measure is the result of choosing to dividethe circle into 360 equal parts. The choice of 360 is abritrary. Another choicewill result in an equally valid angle measure, rather like switching from feetto inches in linear measure. In contrast to linear measure, it is possible toconstruct an intrinsic or natural measure for angles, that is a measure that doesnot involve an arbitrary choice. That is the content of this section.

28

CHAPTER 3. RADIAN MEASURE 29

Definition 3.1.1. (Radian Measure)Let θ be an central angle of a circle or radius r, and let s be the length of

the arc subtended by θ.

Θ

r

s The radian measure of θ is givenby:

θ = sr

d

c

It may seem that Definition 3.1.1 depends on the choice of a circle,but since all circles are similar, the defining ratio will remain the samefrom circle to circle. The definition is also reminiscent of the definitionof π. Recall that π is the ratio of the circumference to the diameter ofa circle (Again, this does not depend on the circle chosen.):

π = c

d

In fact, this observation will be the key to converting angle measuresbetween degrees and radians.

Radian-Degree Conversions

Θ = 180°

r=1

2d

s =1

2c

If θ = 180, then the radianmeasure of θ is given by

θ = s

r= c/2d/2 = c

d= π

Hence,180 = π


We note that radian measure is actually unitless. The radius and arc lengthare measured in linear units, say inches. Then

θ = sinrin

= s

r(no units)

Nevertheless, it is a useful mnemonic device to append rad to radian measure asif it were a unit. For the purposes of converting between the two angle measuresit is also useful to append deg to an angle rather than the usual . With theseconventions, we have two conversion factors:

180 deg = π rad

180 degπ rad = 1 π rad

180 deg = 1

Example 3.1.2.

(i) Convert 60 to radian measure;

(ii) Convert 5π6 to degree measure.

(i) 60deg · π rad180deg = π

3

(ii) 5π6 rad · 180 deg

πrad = 150

Radian measure is rarely used when actually measuring angles in applica-tions in engineering, phisics or even more every-day applications such as car-pentry. The reason is that it is somewhat clumbsy. An angle of mearure 1radian has a degree measure of 57.3. The real utility of radian measure isthat since it is unitless, it converts the trigonometric functions from functionsassigning numbers to geometric objects (angles) to ordinary functions on thereal numbers.

Figure 3.1.1 shows the boundry angles and the special triangles with anglesmeasured in radians. It would be worth while to become familiar with the radianvalues of the special angles.


0

2Π

Π

2

Π

3 Π

2

III

III IV

Π6

Π31

3

2

Π4

Π4

1

1

2

Figure 3.1.1: Special Angles in Radians

Example 3.1.3.Using diagrams in Figure3.1.1,

(i) sin π6 = 12

(ii) cos π4 = 1√2

(iii) tan π3 =√

3

It is just as easy to evaluate the trigonometric functions on the boundryangles.

Example 3.1.4.

0

2Π

Π

2

Π

3 Π

2

H1, 0L

H0, 1L

H-1, 0L

H0, -1L

r = 1

Using the diagram on theleft and the definition of thetrigonometric functions,

sin π2 = y

r= 1

1 = 1

cosπ = x

r= −1

1 = −1

tan 3π2 = y

x= −1

0 (undefined)


The procedure for evaluating the trigonometric functions is exactly thatsame as that for angles measured in degrees.

Example 3.1.5.Evaluate tan 2π

3 .2π3 lies in Quad II and has reference angle π − 2π

3 = π

3 .

0

2Π

Π2

Π

3Π2

Θ'=Π3Θ=2Π3 Using Figure 3.1.1,

tan 3π3 = − tan π3 = −

√3


For problems 1–9, convert the angle to radian measure.

1. θ = 120.

2. θ = 90.

3. θ = 45.

4. θ = 210.

5. θ = 300

6. θ = 270.

7. θ = 135.

8. θ = 240.

9. θ = 315.

For problems 10–15, find three angles coterminal to the given angle.

10. θ = π

6 .

11. θ = 3π4 .

12. θ = 2π.

13. θ = 5π4 .

14. θ = 11π6

15. θ = 5π3 .

For problems 16–24, sketch the angle in standard position and find the ref-erence angle for the given angle.


16. θ = 7π6 .

17. θ = 5π4 .

18. θ = 2π.

19. θ = −5π6 .

20. θ = −5π4 .

21. θ = 11π3

22. θ = 5π3 .

23. θ11π6

24. θ = 5π3 .

For problems 22–30, evaluate the trigonometric function.

22. sin π6 .

23. tan 3π4 .

24. θ = 2π.

25. cos 5π3 .

26. sin 7π6

27. cos 3π2 .

28. cosπ.

29. sin π

30. sin 7π4 .

Chapter 4

Graphs of TrigonometricFunctions

4.1 The Unit CircleLet us recall the definition of the sine and cosine functions. If θ is an angle instandard position and (x, y) is a point on the terminal ray of θ and r is thedistance from (x, y) to the origin, then,

sin θ = y

r(4.1)

cos θ = x

r(4.2)

0

2Π

Π2

Π

3Π2

Hx, yL

r = 1

H1, 0LH-1, 0L

H0, 1L

H0, -1L

Θ

If (x, y) is the point of intersec-tion of the terminal ray with theunit circle, then, since r = 1,

sin θ = y

r= y

cos θ = x

r= x

Figure 4.1.1: The Unit Circle

34

CHAPTER 4. GRAPHS OF TRIGONOMETRIC FUNCTIONS 35

This is often taken as the definition for the sine and cosine functions. Wewill state this observation as a theorem.

Theorem 4.1.1. If (x, y) is the point of intersection of the terminal ray of θwith the unit circle, then (x, y) = (cos θ, sin θ).

0

2Π

Π2

Π

3Π2

Hcos Θ, sin ΘL

r = 1

Θ

Example 4.1.2.For θ = 5π

6 :

(i) Draw the terminal ray of θ;

(ii) Find the reference angle of θ;

(iii) Find the coordnates of the point of intersection of θ with the unit circle.

Since π

2 <5π6 < π, θ lies in the second quadrant. Hence θ′ = π − 5π

6 = π

6 .Π6

Π31

3

2

0

2Π

Π2

Π

3Π2

Θ'=Π6

Θ=5Π6Hx, yL

cos 5π6 = −

√3

2sin 5π

6 = 12

(x, y) =(−√

3/2, 1/2)



For each angle given,

(i) Draw the terminal ray of θ;

(ii) Find the reference angle of θ;

(iii) Find the coordnates of the point of intersection of θ with the unit circle.

1. θ = π

4

2. θ = 4π3

3. θ = 11π6

4. θ = π

2

5. θ = π

6. θ = 3π3

4.2 Graphs of Trigonometric FunctionsWe now turn to the graphs of the sine and cosine functions. Let θ increase from0 to 2π as in Figure 4.2.1. The y-coordinate of the point of intersection of theterminal ray of θ and the unit circle begins at 0, increases to a maximum valueof 1 at θ = π/2 returns to 0 at θ = π, decreases to a minimum value of −1 atθ = 3π/2, and then returns back to 0 at θ = 2π. The graph of y = sin θ is thanas shown in Figure 4.2.1.

H1,0LH-1,0L

H0,1L

H0,-1L

Π

2

Π

3 Π

2

Π3

3Π4

7Π6

5Π3

0 Π

3

Π

2

3 Π

4Π

7 Π

6

3 Π

2

5 Π

32 Π

Figure 4.2.1: y = sin θ


Of course increasing θ above would result in the terminal ray of θ tracingover the same points on the unit circle several times, so the shape of the sinecurve in Figure 4.2.1 will simply repeat itself. Similarly for negative values ofθ. We say that the sine function is periodic with period 2π.

-2 Π -Π Π 2 Π 3 Π 4 ΠΘ

-1

1

y = sin Θ

Figure 4.2.2

A similar analysis of the cosine function yields the following cosine curve.

-2 Π -Π Π 2 Π 3 Π 4 ΠΘ

-1

1

y = cos Θ

Figure 4.2.3

In general, it suffices to sketch a single period of a periodic curve. Figure 5.1.1shows the sine and cosine curves for 0 ≤ θ ≤ 2π. These curves should bememorized.

Π

2Π

3 Π

2

Θ

-1

1

y = sin Θ

(a) y = sin θ

Π

2Π

3 Π

2

Θ

-1

1

y = cos Θ

(b) y = cos θ

Figure 4.2.4


Consider now the circle of radius A centered at the origin. If θ is an anglein standard position and (x, y) is a point on the terminal ray of θ and r is thedistance from (x, y) to the origin, then, by definition,

sin θ = y

r

cos θ = x

r

0

2Π

Π2

Π

3Π2

Hx, yL

r = A

HA, 0LH-A, 0L

H0, AL

H0, -AL

Θ

If (x, y) is the point of intersec-tion of the terminal ray with theunit circle, then, since r = A,

sin θ = y

A⇒ y = A sin θ

cos θ = x

A⇒ x = A cos θ

Figure 4.2.5

Whereas the minimum and maximum values of y = sin θ and y = cos θ are−1 and 1, the minimum and maximum values of y = A sin θ and y = A cos θare −A and A. The graphs retain the same basic shapes, but are stretchedvertically.

Example 4.2.1.On the same set of axes, plot:

(i) y = sin θ and y = 2 sin θ

(ii) y = cos θ and y = 2 cos θ

Π

2Π

3 Π

2

Θ

-2

-1

1

2

y

(a) y = 2 sin θ

Π

2Π

3 Π

2

Θ

-2

-1

1

2

y

(b) y = 2 cos θ

Figure 4.2.6

Like with any other function, the graphs of y = − sin θ and y = − cos θ arethe graphs of y = sin θ and y = cos θ reflected across the x-axis.

Example 4.2.2.On the same set of axes, plot:

(i) y = − sin θ and y = −2 sin θ

(ii) y = − cos θ and y = −2 cos θ

Π

2Π

3 Π

2

Θ

-2

-1

1

2

y

(a) y = 2 sin θ

Π

2Π

3 Π

2

Θ

-2

-1

1

2

y

(b) y = 2 cos θ

Figure 4.2.7

Definition 4.2.3. For the functions

y = A sin θ (4.3)y = A cos θ (4.4)

|A| is called the amplidtue of the function.


For problems 1-6 graph the given trigonometric function for 0 ≤ θ ≤ 2π.Label the θ-axis with multiples of π2 .

1. y = 3 sin θ

2. y = −5 sin θ

3. y = 4 cos θ

4. y = 1.5 sin θ

5. y = 12 cos θ

6. y = −3 cos θ

39


Chapter 5

Inverse TrigonometricFunctions

5.1 Inverse Trigonometric FucntionsLet us recall than, interms of their graphs, functions assign points on the x-axisto points on the y-axis and that the inverse of a function reverses the process.

x

f HxL

x

y

f -1HyL

y

x

y

Figure 5.1.1

Recall also that a function may not have multiple outputs for a single input.This presents a problem for the sine function: it’s inverse is not a function.

-2 Π -Π Π 2 Π 3 ΠΘ

-1

1

y = sin Θ

Figure 5.1.2


We do wish to define an inverse function for the sine function. To this endwe recall that a function has three components: a domain, a range, and a rulethat assigns elements of the domain to elements of the range. When defining afunction, we have complete control over the first two components.

We now define a new function whose rule agrees with that of the sine func-tion, but we take the domain to be the interval −π2 ≤ θ ≤ π

2 . We denote thisfunction Sin θ.1 This new (restricted) function has the same range as the sinefunction (the interval [−1, 1]), but has the advantage that each element of therange is the image of only one element of the domain.

-Π -Π

2

Π

2Π

Θ

-1

1

y = Sin Θ

Figure 5.1.3

We denote the inverse of Sin θ by Sin −1x.

Sin −1x = θ if and only if(i) sin θ = x

(ii) − π

2 ≤ θ ≤π

2

We make a similar restriction for the cosine function, except that we cannotuse the same interval. We define Cos θ = cos θ for 0 ≤ θ ≤ π.

1Note this function has a capital S to distinguish it from the ordinary (nonrestricted) sinefunction.


-Π -Π

2

Π

2Π

3 Π

2

Θ

-1

1

y = Cos Θ

Figure 5.1.4

Cos −1x = θ if and only if(i) cos θ = x

(ii) 0 ≤ θ ≤ π

For the tangent function, the restriction is similar to that of the sine function.Tan θ = tan θ for −π2 < θ <

π

2 .

-Π

2

Π

2

Θ

y = Tan Θ

Tan −1x = θ if and only if(i) tan θ = x

(ii) − π

2 < θ <π

2

Figure 5.1.5

Example 5.1.1.

Evaluate Cos −1(−1

2

).

By the definition of Cos −1, this problem is equivalent to: find θ such thatcos θ = −1

2 and 0 ≤ θ ≤ π.

Since cos θ = −12 < 0, the terminal ray of θ lies in either Quad. II or Quad.

III.


Π6

Π31

3

20

2Π

Π2

Π

3Π2

Π3

Π3

cos θ = −12 ⇒ cos θ′ = 1

2⇒ θ′ = π

3

Since we require that 0 ≤ θ ≤ π, θ mustlie in Quad. II and

θ = π − π

3 = 2π3

Figure 5.1.6

Example 5.1.2.

Evaluate Sin −1(− 1√

2

).

By the definition of Sin −1, this problem is equivalent to: find θ such thatsin θ = − 1√

2and −π2 ≤ θ ≤

π

2 .

Since sin θ = − 1√2< 0, the terminal ray of θ lies in either Quad. III or

Quad. IV.

Π4

Π4

1

1

2

0

2Π

Π2

Π

3Π2

Π4Π4

sin θ = − 1√2⇒ sin θ′ = 1√

2⇒ θ′ = π

4

Since we require that −π2 ≤ θ ≤π

2 must lie in Quad. IV and

θ = −π4

Figure 5.1.7

Example 5.1.3.


Evaluate Tan −1(−5

4

). Since tan θ = −5

4 , θ is not a special angle, so we

use a scientific calculator: Be sure thatyour calculatoris in radianmode.

Tan −1(−5

4

)= −0.8961


For problems 1-9 evaluate the inverse trigonometric function. Give exact-value answers in radians.

Π6

Π31

3

2

Π4

Π4

1

1

2

1. Sin −1 1√2

2. Cos −1(−√

32

)3. Tan −11

4. Sin −1(−1

2

)5. Cos −11

6. Cos −1(−1

2

)7. Tan −1 (−√3

)8. sin−1 1

9. cos−1(

12

)

5.2 Trigonometric EquationsThe simplest trigonometric is one of the form:

sin θ = −12 (5.1)

To solve Equation 5.2 means to find all angles θ that satisfy the equation.

Since −12 < 0, we should expect to find solutions only in Quadrants III and

IV, where the sine function is negative. Recall that sin θ′ = | sin θ|, where θ′ isthe reference angle of θ.

Π6

Π31

3

2

0

2Π

Π2

Π

3Π2

Π6Π6

sin θ = −12 ⇒ sin θ′ = 1

2

θ′ = Sin −1(

12

)= π

6

Two solutions of Equa-tion 5.2 are:

θ = 2π − π

6 = 11π6

θ = π + π

6 = 7π6

Figure 5.2.1


Any other solution of Equation 5.2 are coterminal with these two, so thegeneral solution is

θ = 7π6 + 2kπ

θ = 11π6 + 2kπ

k = 0,±1,±2,±3, · · ·

In general, it suffices to find all solutions θ of a trigonometric equation thatsatisfy 0 ≤ θ < 2π. Any other solution will be coterminal with one of these.

In most cases, some algebra is involved to put a trigonometric equation inthe form of Equation 5.2.

Example 5.2.1.Find all solutions to √

2 cos θ − 1 = 0

that satisfy 0 ≤ θ < 2π.We first solve algebraically for cos θ:

√2 cos θ − 1 = 0√

2 cos θ = 1

cos θ = 1√2

Since cos θ = 1√2> 0, we will find solutions only in Quadrants I and IV,

where the cosine function is positive.

Π4

Π4

1

1

2

0

2Π

Π2

Π

3Π2

Π4

Π4

cos θ = 1√2

cos θ′ = 1√2

θ′ = Cos −1(

1√2

)= π

4

The two solutions of the equation are:

θ = π

4θ = 2π − π

4 = 7π4

Figure 5.2.2


Of course, the solutions to trigonometric equations are not always specialangles.

Example 5.2.2.Find all solutions to

2 tan θ + 3 = 0

that satisfy 0 ≤ θ < 2π.We first solve algebraically for cos θ:

2 tan θ + 3 = 02 tan θ = −3

tan θ = −32

Since tan θ = −32 < 0, we will find solutions only in Quadrants II and IV,

where the tangent function is negative.

Be sure youcalculator is inradian mode.

0

2Π

Π2

Π

3Π2

Θ'

Θ'

tan θ = −32

tan θ′ = 32

θ′ = Tan −1(

32

)= 0.9828

The two solutions of the equa-tion are:

θ = 2π − θ′ = 5.3004θ = π − θ′ = 2.1588

Figure 5.2.3


For problems 1-9 find all solutions for the trigonometric equation satisfying0 ≤ θ < 2π. Where possible, give exact-value answers. (This is when thesolutions are special angles.)

Π6

Π31

3

2

Π4

Π4

1

1

2

1. 2 sin θ − 1 = 0

2. 2 cos θ +√

3 = 0

3. tan θ − 1 = 0

4. 1.5 sin θ + 2.3 = 0

5. 3 cos θ − 2 = 0

6. 5 tan θ + 4 = 0

7. 4 cos θ − 2 = 0

8.√

2 sin θ − 1 = 0

9. tan θ −√

3 = 0

Final Review

Review1. Convert the following angles to radian

measure. Give exact-value answers.

(a) 15

(b) −225

(c) 330

(d) 270

(e) 60

(f) 120

2. Convert the following angles to degreemeasure.

(a) 3π8

(b) 1.5

(c) 5.25

(d) π

6(e) −2.7

(f) 4π3

3. A 16-foot ladder is resting against a wall.The top of the ladder is 15 feet from theground. What angle does the ladder makewith the wall?

4. From a point 200 feet from the base of aRoman aqueduct in northern Spain the an-gle of elevation to the top of the aqueductis 19. Find the height of the aqueduct.

5. Find the values of the six trigonometricfunctions on an angle whose terminal raypasses throug the point P (7,−4).

6. Find the exact value of sin θ, cos θ, andtan θ for the following angles.

(a) θ = π

(b) θ = 3π4

(c) θ = 5π6

(d) θ = 3π2

(e) θ = 7π4

(f) θ = 4π3

7. Use a calculator to evaluate the followingexpressions. Give your answer accurate to4 decimal places.

(a) sin 127

(b) cos(−116)(c) sec(−4.45)(d) csc 0.34

8. Find two values θ with 0 ≤ θ < 2π thatsatisfy the trigonometric equation. Giveexact-value answers.

(a) sin θ = 12

(b) cos θ = −√

32

(c) tan θ = − 1√3

(d) tan θ = 1

(e) sin θ = −12

(f) cos θ = 12

9. Graph the following functions. State thedomain, range, amplitude and period.

(a) y = 2 sin θ (b) y = 12 cos θ

10. Solve the following equations in the inter-val 0 ≤ θ < 2π.

(a) 2 sin θ−√

2 = 0(b) cos θ = 0.6725

(c) 2 cos θ + 1 = 0(d) 2 tan θ + 5 = 0

11. Use the definitions of sin θ = y

r, cos θ =

x

r, and tan θ = y

xto prove the following

trigonometric identities.

(a) sin2 θ + cos2 θ = 1

(b) tan θ = sin θcos θ

(c) 1 + tan2 θ = sec2 θ

(d) 1 + cot2 θ = sec2 θ

12. Multiply and simplify:

(a) (sin θ − cos θ)(sin θ + cos θ)(b) (sin θ − cos θ)2

(c) (1 + tan θ)2

(d) tan θ(cos θ − csc θ)

47


Practice Test1. Convert 36 to radian measure.

Give an exact-value answer.

2. Convert 5π8 to degree measure.

Give your answer accurate to twodecimal places.

3. At a point 180 feet away from thebase of a building, the angle ofelevation to the top of the build-ing is 38. What is the height ofthe building? Give your answercorrect to two decimal places.

4. Use a calculator to evaluate thefollowing. Give your answers cor-rect to four decimal places. (Besure to check your mode.)

(a) sin 72

(b) cos(−3.4)

5. Evaluate the following. Giveexact-value answers.

(a) sin π3(b) cos 2π

3(c) tan 5π

4

(d) sin π2(e) tan π

(f) cos 11π6

6. Find the exact value in radi-ans of

cos−1(−1

2

)7. Find the exact value in radi-

ans oftan−1 (1)

8. Sketch the graph of

y = 3 sin θ

for 0 ≤ θ ≤ 2π. Label the θ-axiswith multiples of π2 and state theamplitude of the function.

9. Multiply and simplify fully:

sin θ(cot θ + tan θ)

10. Find the value of the six trigono-metric functions on the angleshown in the diagram below

Θ

x

y

-4 -3 -2 -1 1 2 3 4

-6

-5

-4

-3

-2

-1

1

2

3

4

5

6

Problem 10

Appendix A

The Greek Alphabet

alpha α A nu ν N

beta β B xi ξ Ξ

gamma γ Γ omicron o O

delta δ ∆ pi π Π

epsilon ε E rho ρ P

zeta ζ Z sigma σ Σ

eta η E tau τ T

theta θ Θ upsilon υ Υ

iota ι I phi ϕ Φ

kappa κ K chi χ X

lambda λ Λ psi ψ Ψ

mu µ M omega ω Ω

49

Appendix B

Midcourse Review Answers

1. Find the values of all six trigonometric functionson angle A in the triangle below.

11

b

6

A

B

C

Using Pythagoras, b = 5.

sinA =√

116 cscA = 6√

11

cosA = 56 secA = 6

5

tanA =√

115 cotA = 5√

11

2. Find the reference angles for each of the followingangles.

(a) 60

(b) 32(c) 10

(d) 83(e) 30

(f) 45

3. Given that α is an acute angle and sinα = 2/3,find the values of the remaining 5 trigonometricfunctions on θ.

2

5

3

Θ

Using Pythagoras, theremaining side is

√5.


2

cos θ =√

53 sec θ = 3√

5

tan θ = 2√5 cot θ =

√5

2

4. Given that 90 < θ < 180 and sin θ = 2/3, findthe values of the remaining 5 trigonometric func-tions on θ.

0°

360°

90°

180°

270°

Θ

Θ' = Α

sin θ = 2/3 ⇒ θ′ = αfrom problem 3. Since θlies in Quad IV,


2

cos θ = −√

53 sec θ = − 3√

5

tan θ = − 2√5 cot θ = −

√5

2


c

3.5

a

33°A

B

C

B = 90− 33 = 57

tan 33 = a

3.5a = 3.5 tan 33 = 2.3

Using Pythagoras,c =√

3.52 + 2.32 = 4.2

6. Evaluate sin2 23 + cos2 23.sin2 23 + cos2 23 = 1

7. Given that sin 23 = 0.3907, find:Note that 90 − 23 = 67, so cos 67 = sin 23.After finding the reference angle for each anglebelow, and considering the quadrant in which theangle lies, we find:

(a) sin 157 = .3907 (d) cos 67 = .3907(b) sin 203 = −.3907 (e) cos 113 = −.3907(c) sin 337 = −.3907 (f) cos(−113) = −.3907

8. Use the special triangles to evaluate:

(a) cos 120 = − 12

(b) tan 210 = 1√3

(c) sin 330 = − 12

(d) cos 135 = − 1√2

(e) tan(−60) = −√

3(f) sin 150 = 1

2

9. Evaluate:

(a) cos 90 = 0(b) tan 180 = 0(c) sin 0 = 0

(d) sin 270 = −1(e) tan 270: undefined(f) cos 0 = 1

(See Figure B.0.1)

50

APPENDIX B. MIDCOURSE REVIEW ANSWERS 51

H1,0LH-1,0L

H0,1L

H0,-1L

0°

360°

90°

180°

270°

Figure B.0.1

10. Given sin θ = −.225.

(a) Find θ′, the referance angle for θ.sin−1 .225 = 13

(b) Given that θ lies in Quad. IV, evaluate:i. cos θ = cos 13 = 0.974ii. tan θ = tan 13 = −0.231


c

4.7

3.1

A

B

C

Using Pythagoras,c =√

3.12 + 4.72 = 5.6A = tan−1 3.1

4.7 = 33.4

B = 90 = 33.4

= 56.6

12. Prove the following trigonometric identities:

(a) cot θcsc θ = cos θ

cot θcsc θ = cos θ/ sin θ

1/ sin θ

= cos θsin θ ·

sin θ1

= cos θ

(b) cotα+ tanα = secα cscα

cotα+ tanα = cosαsinα + sinα

cosα

= cos2 α+ sin2 α

cosα sinα= 1

cosα sinα= 1

cosα ·1

sinα= secα cscα

(c) tan β sin β + cosβ = secβ

tan β sin β + cosβ = sin βcosβ sin β + cosβ

= sin2 β

cosβ + cosβ1

= sin2 β

cosβ + cos2 β

cosβ

= sin2 β + cos2 β

cosβ

= 1cosβ = secβ

(d) 11− sin θ −

11 + sin θ = 2 tan θ sec θ

11− sin θ −

11 + sin θ = 1

1− sin θ ·1 + sin θ1 + sin θ −

11 + sin θ ·

1− sin θ1− sin θ

= 1 + sin θ − (1− sin θ)1− sin2 θ

= 2 sin θcos2 θ

= 2 sin θcos θ ·

1cos θ = 2 tan θ sec θ

Appendix C

Final Review Answers

1. Convert the following angles to radian measure.Give exact-value answers.

(a) π

12

(b) −5π4

(c) 11π6

(d) 3π2

(e) π

3

(f) 2π3

2. Convert the following angles to degree measure.

(a) 67.5

(b) 85.9

(c) 300.8

(d) 30

(e) −154.7

(f) 240

3. cos θ = 1516 ⇒ θ = cos−1 ( 15

16)

= 20.4

4. tan 19 = h200 ⇒ h = 200 tan 19 = 68.9ft

5. x = 7, y = −4. r2 = 72 + (−4)2 =√

65

(a) sin θ = − 4√65

(b) cos θ = 7√65

(c) tan θ = −47

(d) csc θ = −√

654

(e) sec θ =√

657

(f) cos θ = −74

6. Find the exact value of sin θ, cos θ, and tan θ forthe following angles.

(a) sin π = 0cosπ = −1tan π = 0

(b) sin 3π4 = 1√

2cos 3π

4 = − 1√2

tan 3π4 = −1

(c) sin 5π6 = 1

2

cos 5π6 = −

√3

2tan 5π

6 = 1√3

(d) sin 3π2 = −1

cos 3π2 = 0

tan 3π2 : undef.

(e) sin 7π4 = − 1√

2

cos 7π4 = 1√

2

tan 7π4 = −1

(f) sin 4π3 = −

√3

2cos 4π

3 = −12

tan 4π3 = −

√3

7. Use a calculator to evaluate the following expres-sions. Give your answer accurate to 4 decimalplaces.

(a) 0.7986(b) −0.4384

(c) −3.8552(d) 2.9986

8. Find two values θ with 0 ≤ θ < 2π that satisfythe trigonometric equation. Give exact-valueanswers.

(a) θ = π

6 ,5π6

(b) θ = 5π6 ,

7π6

(c) θ = 5π6 ,

11π6

(d) θ = π

4 ,5π4

(e) θ = 7π6 ,

11π6

(f) θ = π

3 ,5π3

9. Graph the following functions. State the domain,range, amplitude and period.

52

APPENDIX C. FINAL REVIEW ANSWERS 53

Π

2Π

3 Π

2

Θ

-2

2

y = 2sin Θ

amplitude: 2; period: 2π.

Π

2Π

3 Π

2

Θ

-1

2

1

2

y =1

2cos Θ

amplitude: 12 ; period: 2π.

10. Solve the following equations in the interval 0 ≤θ < 2π.

(a) θ = π

4 ,3π4

(b) θ = 0.8332, 5.4500

(c) θ = 2π3 ,

4π3

(d) θ = 1.9513, 5.0929

11. Use the definitions of sin θ = y

r, cos θ = x

r, and

tan θ = y

xto prove the following trigonometric

identities.

(a)

sin2 θ + cos2 θ =(yr

)2+(xr

)2

= y2 + x2

r2

= r2

r2 = 1

(b)

sin θcos θ = y/r

x/r

= y

r· rx

= y

x= tan θ

(c)

1 + tan2 θ = 1 + y2

x2

= x2 + y2

x2

= r2

x2

= sec2 θ

(d)

1 + cot2 θ = 1 + x2

y2

= y2 + x2

y2

= r2

y2

= sec2 θ

12. Multiply and simplify:

(a) (sin θ − cos θ)(sin θ + cos θ) = sin2 θ − cos2 θ

(b)

(sin θ − cos θ)2 = (sin θ − cos θ) (sin θ − cos θ)= sin2 θ − 2 sin θ cos θ + cos2 θ

= 1− 2 sin θ cos θ

(c)

(1 + tan θ)2 = (1 + tan θ) (1 + tan θ)= 1 + 2 tan θ + tan2 θ

= sec2 θ + 2 tan θ

(d)

tan θ(cos θ − csc θ) = sin θcos θ

(cos θ − 1

sin θ

)= sin θ − 1

cos θ= sin θ − sec θ

Appendix D

Answers to Odd-NumberedProblems

Section 1.1

1. 23

3. 52

5.√

13

7.√

5

9. 102

11. 40

13. 63

Section 1.2

1. B = 68b = 21.5c = 23.2

3. A = 38a = 5.0b = 6.4

5. A = 34.5B = 55.5c = 7.6

7. A = 43a = 7.4c = 10.8

9. A = 37a = 3.2b = 4.5

11.sin θ = 12

13 csc θ = 1312

cos θ = 513 sec θ = 13

5

tan θ = 125 cot θ = 5

12

13.sin θ = 1

2 csc θ = 2cos θ =

√3

2 sec θ = 2√3

tan θ = 1√3 cot θ =

√3

15.sin θ =

√3

2 csc θ = 2√3

cos θ = 12 sec θ = 2

tan θ =√

3 cot θ = 1√3

17. 24.1 ft.

19. 33.0

21. 77.9 ft

23. 13.5 ft

Section 1.3

1. 60

3. 65

54

APPENDIX D. ANSWERS TO ODD-NUMBERED PROBLEMS 55

5. 27

7. 45

9. 50

11. 560, 920, −160.

13. 140, 500, −580.

15. 670, 1030, −50.

17.sin θ = − 4

5 csc θ = − 54

cos θ = − 35 sec θ = − 5

3

tan θ = 43 cot θ = 3

4

19.sin θ = 0 csc θ : undef.cos θ = −1 sec θ = −1tan θ = 0 cot θ : undef.

21.

sin θ = −1 csc θ = −1cos θ = 0 sec θ : undef.tan θ : undefined cot θ = 0

23.sin 120 =

√3

2 csc 120 = 2√3

cos 120 = − 12 sec 120 = −2

tan 120 = −√

3 cot 120 = − 1√3

25.sin 180 = 0 csc 180 : undef.cos 180 = −1 180θ = −1tan 180 = 0 cot : undefined

27.sin 270 = −1 csc 270 = −1cos 270 = 0 sec 270 : undef.tan 270 : undefined cot 270 = 0

Section 2.1

1.

tan θsin θ = sin θ/ cos θ

sin θ=

sin θcos θ ·

1sin θ

= 1cos θ

= sec θ

3.

sin θ + sin θ cot2 θ = sin θ(1 + cot2 θ)= sin θ csc2 θ

= sin θ 1sin2 θ

= 1sin θ

= csc θ

5.

tan θ cos θ = sin θ/cos θcos θ

= sin θ

7.

1 + tan2 θ

1− tan2 θ=

1 + sin2 θ

cos2 θ

1− sin2 θ

cos2 θ

=

cos2 θ + sin2 θ

cos2 θcos2 θ − sin2 θ

cos2 θ

= cos2 θ + sin2 θ

cos2 θ·

cos2 θ

cos2 θ − sin2 θ

= cos2 θ + sin2 θ

cos2 θ − sin2 θ

= 1cos2 θ − sin2 θ


9.

sec θ − 1sec2 θ + 1 =

1cos θ − 1

1cos θ + 1

=

1− cos θcos θ

1 + cos θcos θ

= 1− cos θcos θ ·

cos θ1 + cos θ

= 1− cos θ1 + cos θ

Section 3.1

1. 2π3

3. π4

5. 5π3

7. 3π4

9. 7π4

11. 135

13. 225

15. 300

17. θ′ = π

4

19. θ′ = π

6

21. θ′ = π

323. −1

25. 12

27. 0

29. 0

Section 4.1

1.(

1√2 ,

1√2

)3.(− 1

2 ,√

32

)5. (−1, 0)

Section 4.2

Π

2Π

3 Π

2

Θ

-3

y = 3sin Θ

Problem 1

Π

2Π

3 Π

2

Θ

-4

4

y = 4cos Θ

Problem 3

Π

2Π

3 Π

2

Θ

-1

2

1

2

y =1

2cos Θ

Problem 5


Section 5.1

1. π4

3. π4

5. 0

7. −π39. π

3

Section 5.2

1. θ = π

6

θ = 5π6

3. θ = π

4

θ = 5π4

5. θ = 0.8411θ = 5.4421

7. θ = π

3

θ = 5π3

9. θ = π

3

θ = 4π3

Index

amplitude, 37angle, 12

boundary angles, 13

coterminal, 14

initial ray, 12

period, 35periodic, 35Pythagoras, 3

radian measure, 27reference angle, 13reference triangle, 15

similar, 2solve a triangle, 6standard position, 12

terminal ray, 12

58

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