SECTION 7.2 Right Triangle Trigonometry 5 1 7

[Note: There is a 900 angle between the two foul lines. Then there are two 3°angles between the foul lines and the dotted lines shown. The angle between the two dotted lines outside the 200 foot foul lines is 96°.]

117. Pulleys Two pulleys, one with radius rl and the other with radius r2 , are connected by a belt. The pulley with radius r 1 rotates at WI revolutions per minute, whereas the pulley with radius r2 rotates at W2 revolutions per minute. Show that r l W2

r2 W I

Source: www. littleleague.org

Discussion and Writing

118. Do you prefer to measure angles using degrees or radians? 123. Discuss why ships and airplanes use nautical miles to measure distance. Explain the difference between a nautical mile and a statute mile.

Provide justification and a rationale for your choice.

119. What is 1 radian?

UO. Which angle has the larger measure: 1 degree or 1 radian? 124. Investigate the way that speed bicycles work. In particular, ex­plain the differences and similarities between 5-speed and 9-speed derailleurs. Be sure to include a discussion of linear speed and angular speed.

Or are they equal?

U1. Explain the difference between linear speed and angular speed. 125. In Example 6, we found that the distance between Albu­

querque, New Mexico, and Glasgow, Montana, is approxi­mately 903 miles. According to mapquest. com, the distance is approximately 1 300 miles. What might account for the difference?

U2. For a circle of radius r, a central angle of (J degrees subtends

an arc whose length s is s = 1;

0 r(J. Discuss whether this is a

true or false statement. Give reasons to defend your position.

'Are You Prepared?' Answers

1. C = 27Tr

7.2 Right Triangle Trigonometry

Figure 1 8

PREPARI NG FOR THIS SECTION Before getting started, review the following: • Geometry Essentials (Chapter R, Review, Section R.3, • Functions (Section 3 .1 , pp. 208-218)

pp. 30-35)

'\. Now Work the 'Are You Prepared?' problems on page 525.

b

a

OBJECTIVES 1 Find the Va l ues of Trigonometric Fu nctions of Acute Ang les (p. 5 1 7)

2 Use the Fundamenta l Identities (p. 5 1 9)

3 Find the Va lues of the Rema in ing Trigonometric Functions, Given the

Value of One of Them (p. 52 1 )

4 Use the Complementary Angle Theorem (p. 523)

1 Find the Values of Trigonometric Functions of Acute Angles

A triangle in which one angle is a right angle (90°) is called a right triangle. Recall that the side opposite the right angle is called the hypotenuse, and the remaining two sides are called the legs of the triangle. In Figure 18 we have labeled the hypotenuse as c to indicate that its length is c units, .and, in a like manner, we have labeled the legs as a and b . Because the triangle is a right triangle, the Pythagorean Theorem tells us that

5 1 8 CHAPTER 7 Trigonometric Functions

Figure 20

a Adjacent to e

Figure 19

DEFINITION

Opposite e b

Now, suppose that e is an acute angle; that is, 0° < e < 90° (if e is measured in

degrees) and 0 < e < ; (if e is measured in radians). See Figure 19(a). Using this

acute angle e, we can form a right triangle, like the one illustrated in Figure 19(b), with hypotenuse of length c and legs of lengths a and b . Using the three sides of this triangle, we can form exactly six ratios:

b a b c c a c c a b ' a b

b

I nitial side a a (a) Acute angle (b) Right triangle (c) Simi lar triangles

In fact, these ratios depend only on the size of the angle e and not on the triangle formed. To see why, look at Figure 19( c). Any two right triangles formed using the angle e will be similar and, hence, corresponding ratios will be equal. As a result,

b b' a a' b b' c c' c c' a a ' c c' c c' a a' b b' a a' b b '

Because the ratios depend only on the angle e and not on the triangle itself, we give each ratio a name that involves e: sine of e, cosine of e, tangent of e, cosecant of e , secant of e, and cotangent of e.

The six ratios of a right triangle are called trigonometric functions of acute angles and are defined as follows:

Function Name Abbreviation Value

sine of e s in e b

( cosine of e cos e

a ( tangent of e tan e

b a

cosecant of e csc e ( b

secant of e sec e ( a

cotangent of e (ot e a b

-.J As an aid to remembering these definitions, it may be helpful to refer to the

lengths of the sides of the triangle by the names hypotenuse (c), opposite (b), and adjacent (a) . See Figure 20. In terms of these names, we have the following ratios:

. opposite sm e = b

cos e = adjacent a tan e = opposite b

csc e = adjacent hypotenuse c hypotenuse c a

(1) hypotenuse c hypotenuse c adjacent a

sec e = adjacent

cot e = opposite b opposite b a

Since a, b, and c are positive, each of the trigonometric functions of an acute angle e is positive.

EXA M P LE 1

Solution

Figure 2 1 ,nOPP"'I' 3

WARNING When writing the values of the trigonometric functions, do not for­get the argument of the function

4 sin f) = - correct 5

4 sin = - incorrect 5

EXA M P LE 2

•

Solution

SECTION 7.2 Right Triangle Trigonometry 5 1 9

Finding the Value of Trigonometric Functions

Find the value of each of the six trigonometric functions of the angle e in Figure 21 .

We see in Figure 21 that the two given sides of the triangle are c = hypotenuse = 5 a = adjacent = 3

To find the length of the opposite side, we use the Pythagorean Theorem. ( adjacent ) 2 + ( oppositef = (hypotenuse)2

32 + (opposite)2 = 52 (opposite) 2 = 25 - 9 = 1 6

opposite = 4

Now that we know the lengths of the three sides, we use the ratios in (1) to find the value of each of the six trigonometric functions:

. opposite 4 adjacent

S1l1 e = cos e = ----- 3 opposite 4 tan e = = -

hypotenuse 5 hypotenuse 5 adjacent 3 hypotenuse 5

csc e = . hypotenuse 5 sec e = --=-"'------

adjacent 3 cot e = . OpposIte 4 adjacent 3 OpposIte 4

Ci'1� - Now Work P R O B L E M 1 1

2 Use the Fundamental Identities

•

You may have observed some relationships that exist among the six trigonometric functions of acute angles. For example, the reciprocal identities are

Reciprocal Identities

1 csc e = -­

sin e 1

sec e = -­cos e 1

cot e = -­tan e (2)

Two other fundamental identities that are easy to see are the quotient identities.

Quotient Identities

sin e tan e = -­

cos e cos e

cot e = -­sin e (3)

If sin e and cos e are known, formulas (2) and (3) make it easy to find the val­ues of the remaining trigonometric functions.

Findi ng the Values of the Remaining Trigonometric Functions, Given sin 8 and cos 8

G" . e Vs d 2Vs f d h . .

IV en S1l1 = -5- an cos e = -

5-' 111 t e value o f each of the four rema1l11l1g

trigonometric functions of e.

Based on formula (3), we have

Vs

sin e 5 1 tan e = -- = -- = -

cos e 2Vs 2

5

520 CHAPTER 7 Trigonometric Functions

Figure 22

Then we use the reciprocal identities from formula (2) to get

csc e = _1_ = _1_ = _

5_ = Vs

sin () Vs Vs 1 1 5 Vs

sec e = -- = -- = -- = --cos e 2Vs 2Vs 2

1 1 cot e = -- = - = 2

tan e 1

5

b

a

5 2 •

un=;>- Now Work P R O B L E M 2 1

Refer now to the right triangle in Figure 22. The Pythagorean Theorem states that a2 + b2 = c2, which we can write as

Dividing each side by c2, we get

In terms of trigonometric functions of the angle e, this equation states that

(sin e? + (cos e )2 = 1 (4)

Equation (4) is, in fact, an identity, since the equation is true for any acute angle e. It is customary to write sin2 e instead of (sin e )2, cos2 e instead of (cos e?, and

so on. With this notation, we can rewrite equation (4) as

sin2 e + cos2 e = 1 (5)

Another identity can be obtained from equation (5) by dividing each side by cos2 e.

sin2 e + 1 = _1_

cos2 e cos2 e

Now use formulas (2) and (3) to get

tan2 e + 1 = sec2 e (6)

Similarly, by dividing each side of equation (5) by sin2 e, we get 1 + cot2 e = csc2 e, which we write as

cot2 e + 1 = csc2 e (7)

Collectively, the identities in equations (5), (6), and (7) are referred to as the Pythagorean Identities.

Let's pause here to summarize the fundamental identities.

Fundamental Identities

sin e tan e = -­

cos e

1 csc e = -.­

S1l1 e

sin2 e + cos2 e = 1

cos e cot e = -.­

sm e

1 sec e = --

e cos

tan2 e + 1 = sec2 e

1 cot e = -­

tan e

cot2 e + 1 = csc2 e

EXA M P LE 3

Solution

EXA M P LE 4

Solution

Solution 1 Using the Definition

Figure 23 .d:::1b= 1 a

SECTION 7.2 Right Triangle Trigonometry 52 1

Finding the Exact Value of a Trigonometric Expression Using Identities

Find the exact value of each expression. Do not use a calculator.

(a) 200 sin 200

tan - --- . 2

7T 1 (b) sm

12 +

? 7T cos 200

sin 200 (a) tan 200 - = tan 200 - tan 200 = a

cos 200 i sin e -- = tan e cos e

(b) sin2 1

7T

2 +

1 = sin2 � + cos2 � = 1

? 7T 12 12 sec-

12 r r 1

cas e = -­sec e

� Now Work P R O B L E M 3 9

sec- -12

3 Find the Values of the Remaining Trigonometric

Functions, Given the Value of One of Them

•

Once the value of one trigonometric function is known, it is possible to find the value of each of the remaining five trigonometric functions.

F inding the Values of the Remaining Trigonometric Functions, G iven sin 8, 8 Acute

1 Given that sin e = '3 and e is an acute angle, find the exact value of each of the re-

maining five trigonometric functions of e.

We solve this problem in two ways: The first way uses the definition of the trigonometric functions; the second method uses the fundamental identities.

We draw a right triangle with acute angle e, opposite side of length b = 1, and hy­

potenuse of length c = 3 ( because sin e = � = �) . See Figure 23. The adjacent

side a can be found by using the Pythagorean Theorem .

a2 + 12 = 32

a2 + 1 = 9

a2 = 8 a = 2V2

Now the definitions given in equation (1 ) can be used to find the value of each of the remaining five trigonometric functions. (Refer back to the method used in Ex­ample 1 . ) Using a = 2V2, b = 1 , and c = 3, we have

a 2 V2 cos e = - = --

c 3

b 1 V2 tan e = - = -- = --

a 2V2 4

csc e = � = l = 3 sec e = � = _3 - =

3 V2 cot e = :: =

2 V2 = 2 V2 b 1 a 2V2 4 b 1

•

522 CHAPTER 7 Trigonometric Functions

Solution 2 Using I dentities

EXA M P LE 5

Solution 1 Using the Definition

We begin by seeking cos e, which can be found by using the Pythagorean Identity from equation (5).

sin2 e + cos2 e = 1

1 - + cos2 e = 1 9

1 8 cos2 e = 1 - - = -

9 9

Formula (5)

1 sin e = -3

Recall that the trigonometric functions of an acute angle are positive. In particular, cos e > 0 for an acute angle e, so we have

cos e = � = 2\12 \/ "9 3

. 1 2\12 Now w e know that sm e = - and cos e = -- , s o we can proceed as we did

in Example 2. 3 3

1 sin e "3 1 \12

tan e = -- = -- = -- = --cos e 2\12 2\12 4 cot e = _

1_ = _1_ = � = 2\12

tan e \12 \12 3

1 1 3 3\12 sec e = -- = -- = -- = --

cos e 2\12 2\12 4 3

1 1

4

csc e = -- = - = 3 sin e 1

3

Finding the Values of the Trigonometric Functions When One Is Known

Given the value of one trigonometric function of an acute angle e, the exact value of each of the remaining five trigonometric functions of e can be found in either of two ways.

Method 1 Using the Definition

STEP 1: Draw a right triangle showing the acute angle e. STEP 2: Two of the sides can then be assigned values based on the value of

the given trigonometric function. STEP 3: Find the length of the third side by using the Pythagorean Theorem. STEP 4: Use the definitions in equation (1) to find the value of each of the

remaining trigonometric functions.

Method 2 Using Identities

Use appropriately selected identities to find the value of each of the remain­ing trigonometric functions.

Given One Value of a Trigonometric Function, Find the Values of the Remaining Ones

•

Given tan e = � , e an acute angle, find the exact value of each of the remaining

five trigonometric functions of e.

Figure 24 shows a right triangle with acute angle e, where

1 opposite b tan e = - = = -2 adjacent a

Figure 24

1 tan 8 = 2

Figure 25

Solution 2 U sing Identities

�A

Adjacent to A b opposite B

B a Adjacent to B opposite A

4

SECTION 7.2 Right Triangle Trigonometry 523

With b = 1 and a = 2, the hypotenuse c can be found by using the Pythagorean Theorem.

c2 = a2 + b2 = 22 + 12 = 5 c = Vs

Now apply the definitions using a = 2, b = 1 , and c = Vs .

. b 1 Vs SID e = - = -- = --c Vs 5

csc e = .£ = Vs = Vs

b 1

a 2 2Vs cos e = - = -- = --c Vs 5

c Vs sec e = - = --

a 2 a 2

cot e = - = - = 2 b 1

•

Because we know the value of tan e, we use the Pythagorean Identity that involves tan e:

tan2 e + 1 = sec2 e

1 5 sec2 e = - + 1 = -

4 4

Vs sec e = 2

Formu la (6) 1

tan 8 = "2

Proceed to solve for sec 8.

1 Vs U

. . I ' d . . f' d Now we know tan e = - and sec e = --. sIDg reClproca 1 entltles, we ID 2 2

1 1 2 2Vs cos e = - - = -- = -- = --

sec e Vs Vs 5 2

1 1 cot e = -- = - = 2

tan e 1

2 To find sin e, we use the following reasoning:

sin e . tan e = -- , so SID e = ( tan e ) ( cos e )

cos e

csc e = _1_ = _1_ = Vs

sin e Vs 5

,'=z> - Now Work P R O B L E M 2 5

Use the Complementary Angle Theorem

2 5 5

•

Two acute angles are called complementary if their sum is a right angle. Because the sum of the angles of any triangle is 1800, it follows that, for a right triangle, the two acute angles are complementary.

Refer now to Figure 25; we have labeled the angle opposite side b as B and the angle opposite side a as A. Notice that side a is adjacent to angle B and is opposite angle A. Similarly, side b is opposite angle B and is adjacent to angle A. As a result,

. b SID B = - = cos A c

c csc B = b = sec A

a . cos B = - = SID A c

c sec B = - = csc A

a

b tan B = - = cot A

a a

cot B = b = tan A (8)

524 CHAPTER 7 Trigonometric Functions

THEOREM

Table 2

EXAM PLE 6

Because of these relationships, the functions sine and cosine, tangent and cotan­gent, and secant and cosecant are called cofunctions of each other. The identities (8) may be expressed in words as follows:

Complementary Angle Theorem

Cofunctions of complementary angles are equal.

Here are examples of this theorem.

Complementary angles J 1-

sin 30° = cos 60° t t

Co/unctions

Complementary ang les J 1-

tan 40° = cot 50° t j

Co/uncti ons

Complementary ang les J 1-

sec 80° = csc 10° ! t

Co/unctions

-.J

If an angle e is measured in degrees, we will use the degree symbol when writing a trigonometric function of e, as, for example, in sin 30° and tan 45°. If an angle e is measured in radians, then no symbol is used when writing a trigonometric function

I . 7T of e, as, for examp e, 1I1 cos 7T and sec :3 '

7T If e is an acute angle measured in degrees, the angle 90° - e (or 2 - e, if e is

in radians) is the angle complementary to e. Table 2 restates the preceding theorem on cofunctions.

(J ( Degrees) (J (Radians)

sin () = cos (90° - ()) sin e = cos( f - () ) cos () = s i n (90° - e) cos () = s in( f - () ) tan e = cot(90° - e ) tan () = cot( f - e) esc e = sec(90° - 8) esc e = sec( f - e) sec e = csc(900 - e ) sec e = csc( f - e) cot e = tan (90° - e ) cot e = tan( f - e)

The angle e in Table 2 is acute. We will see later (Section 8.4) that these results are valid for any angle e.

Using the Complementary Angle Theorem

(a) sin 62° = cos(90° - 62° ) = cos 28°

7T (7T 7T ) 57T (b) tan - = cot - - - = cot -

12 2 12 12

7T . (7T 7T) . 7T ( c) cos "4 = SlI1 2 - "4 = SlI1 "4

7T (7T 7T) 7T ( d) csc 6 = sec 2 - 6 = sec :3

•

SECTION 7.2 Right Triangle Trigonometry 525

E XA M P L E 7 Using the Complementary Angle Theorem

Find the exact value of each expression. Do not use a calculator.

sin 35°

cos 55° (a) sec 28° - csc 62° (b)

Solution (a) sec 28° - csc 62° = csc(90° - 28°) - csc 62°

= csc 62° - esc 62° = 0

sin 35° (b) cos 550

cos(90° - 3SO )

cos 55°

cos 55° --- = 1 cos 55°

•

cm::==-- Now Work P R O B L E M 4 3

l-lisJorical Feature

The name sine for the s ine function is due to a medieval confusion.

The name comes from the Sanskrit word jiva, (meaning chord), first used in I ndia by Araybhata the Elder (AD 5 1 0). He really meant

half-chord, but abbreviated it. This was brought into Arabic as ji ba,

which was meaningless. Because the proper Arabic word jaib would be

written the same way (short vowels are not written out in Ara bic),jiba,

was pronounced as jaib, which meant bosom or hollow, and jaib remains

as the Arabic word for sine to this day. Scholars translating the Arabic

works into Latin found that the word sinus also meant bosom or hollow,

and from sinus we get the word sine. The name tangent, due to Thomas Finck ( 1 583), can be understood

by looking at Figure 26. The line segment DC is tangent to the circle at

C. If d(O, B) = d(O, C) = 1 , then the length of the line segment DC is

d d(D, C) d(D, C) (0, C) = -1- = d(O, C) = tan a

The old name for the tangent is umbra versa (meaning turned

shadow), referring to the use of the tangent in solving height problems

with shadows.

7.2 Assess Your Understanding

Figure 26

The names of the cofunctions came a bout as follows. If A and B

are complementary angles, then cos A = sin B. Because B is the

complement of A. it was natural to write the cosine of A as sin co A. Probably for reasons involving ease of pronu nciation, the co migrated

to the front, and then cosine received a three-letter abbreviation to

match sin, sec, and tan.The two other cofu nctions were similarly treated,

except that the long forms cotan and cosec survive to this day in some

cou ntries.

'Are You Prepared?' Answers are given at the end of these exercises. If you get a wrong answel; read the pages listed in red.

1. In a right triangle with legs a = 6 and b = 10, the Pythago-rean Theorem tells us that the hypotenuse c = ___ . (pp. 30-35)

Concepts and Vocabulary

3. Two acute angles whose sum is a right angle are called __ . 4. The sine and functions are cofunctions.

5. tan 2So = cot

6. For any angle 8, sin2 8 + cos2 8 = __ , sin 8

7. True or False tan 8 = -- . cos 8

2. The value of the function f(x) = 3x - 7 at 5 is ___ . (pp. 20S-21S)

8. True or False 1 + tan2 8 = csc2 8.

9. True or False If 8 is an acute angle and sec 8 = 3, then 1

cos 8 = 3".

10. True o r False 7T 47T tan - = cot -.

5 5

526 CHAPTER 7 Trigonometric Functions

Skill Building

In Problems 1 1-20, find the value of the six trigonometric functions of the angle () in each figure. 1 1 · � 5 13. j

3 14'

3� 15. �

e 4 2

1 2 4 2 3

16. 3

� 17.

tl 19. 20. �J" 18. �

.J3� 2 li -J2 e 2

In Problems 21-24, use identities to find the exact value of each of the four remaining trigonometric functions of the acute angle (). "

. 21 . Sin (} = � , COS (} = � 22. sin (} = �, COS (} = � 23. Sin (} = � , COS (} = v: 24. Sin (} = � , COS (} =2�

In Problems 25-36, use the definition or identities to find the exact value of each of the remaining five trigonometric functions of the acute angle ().

. v2 25 Sill () = --. 2

1 29. tan () = "2 33. tan () = v2

v2 26. cos () = -2-

1 30. cot () = "2

5 34. sec () = "3

1 27. cos () = "3

31. sec () = 3

35. csc () = 2

8 ' V3 2 . Sill () = 4

32. csc () = 5

36. cot () = 2

In Problems 37-54, use Fundamental Identities andlor the Complementary Angle Theorem to find the exact value of each expression. Do not use a calculator.

37. sin2 20° + cos2 20°

sin 50° 41. tan 50° - --_­cos )0°

COS 10° 45 -­.

. 80° Sill

COS 70° 49. tan 20° - --­cos 20°

38. sec2 28° - tan2 28°

COS 25° 42. cot 25° - -.-­Sill 25°

COS 40° 46. -.-­Sill 50°

sin 50° 50. cot 40° - -­sin 40°

53. cos 35° sin 55° + cos 55° sin 35°

(a) cos 60° (b) cos2 30° 7T

(c) csc (5 7T (d) sec "3

56. Given sin 60° = V3 , use trigonometric identities to find the exact value of 2

(a) cos 30° 7T

(c) sec (5 (b) cos2 60°

(d) 7T

csc "3

57. Given tan () = 4, use trigonometric identities to find the exact value of

(a) sec2 ()

(c) cor(� - ()) (b) cot ()

(d) csc2 ()

39. sin 80° csc 80° 40. tan 10° cot 10°

'- 43. sin 38° - cos 52° 44. tan 12° - cot 78°

47. 1 - cos2 20° - cos2 70° 48. 1 + tan2 5° - csc2 85°

51. tan 35° · sec 55° · cos 35° 52. cot 25° · csc 65° · sin 25°

54. sec 35° csc 55° - tan 35° cot 55°

58. Given sec () = 3, use trigonometric identities to find the exact value of (a) cos () (b) tan2 () (c) csc(90° - (}) (d) sin2 ()

59. Given csc () = 4, use trigonometric identities to find the exact value of (a) sin () (b) cot2 () (c) sec(90° - (}) (d) sec2 ()

60. Given cot () = 2, use trigonometric identities to find the exact value of (a) tan ()

(c) tan(� - (}) (b) csc2 ()

(d) sec2 ()

61. Given the approximation sin 38° � 0.62, use trigonometric identities to find the approximate value of (a) cos 38° (b) tan 38° (c) cot 38° (d) sec 38° (e) csc 38° (f) sin 52° (g) cos 52° (h) tan 52°

62. Given the approximation cos 2 1 ° � 0.93, use trigonometric identities to find the approximate value of

(a) sin 21° (b) tan 21°

(c) cot 21° (d) sec 21 °

(e) csc 2 1 ° (f) sin 69°

(g) cos 69° (h) tan 69°

63. If sin e = 0.3, find the exact value of sin e + cos( � - e ) .

Applications and Extensions

67. Calculating the Time of a Trip From a parking lot you want to walk to a house on the ocean. The house is located 1500 feet down a paved path that parallels the beach, which is 500 feet wide. Along the path you can walk 300 feet per minute, but in the sand on the beach you can only walk 100 feet per minute. See the illustration.

(a) Calculate the time T if you walk 1500 feet along the paved path and then 500 feet in the sand to the house.

(b) Calculate the time T if you walk in the sand directly to­ward the ocean for 500 feet and then turn left and walk along the beach for 1500 feet to the house.

(c) Express the time T to get from the parking lot to the beach house as a function of the angle e shown in the il­lustration.

(d) Calculate the time T if you walk directly from the park­ing lot to the house.

[ 500 ] Hint: tan e =

1500

(e) Calculate the time T if you walk 1000 feet along the paved path and then walk directly to the house.

rt ( f) Graph T = T(e) . For what angle e is T least? What is x for this angle? What is the minimum time?

68.

(g) Explain why tan e = 1:. gives the smallest angle e that is possible. 3

Carrying a Ladder around a Corner Two hallways, one of width 3 feet, the other of width 4 feet, meet at a right angle. See the illustration.

(a) Express the length L of the line segment shown as a function of the angle e.

(b) Discuss why the length of the longest ladder that can be carried around the corner is equal to the smallest value of L.

t 4 ft

�

SECTION 7.2 Right Triangle Trigonometry 527

64. If tan e = 4, find the exact value of tan e + tan( � - e ) .

65. Find an acute angle e that satisfies the equation

sin e = cos(2e + 30°)

66. Find an acute angle e that satisfies the equation

tan e = cot ( e + 45° )

69. Electrical Engineering A resistor and an inductor con­nected in a series network impede the flow of an alternating current. This impedance Z is determined by the reactance X of the inductor and the resistance R of the resistor. The three quantities, all measured in ohms, can be represented by the sides of a right triangle as illustrated, so Z2 = X2 + R2 The angle cp is called the phase angle. Suppose a series net­work has an inductive reactance of X = 400 ohms and a resistance of R = 600 ohms.

(a) Find the impedance Z. (b) Find the values of the six trigonometric functions of the

phase angle cpo

�, R

70. Electrical Engineering Refer to Problem 69. A series net­work has a resistance of R = 588 ohms. The phase angle cp is

5 such that tan cp =

12 '

( a ) Determine the inductive reactance X and the imped­ance Z.

(b) Determine the values of the remaining five trigono­metric functions of the phase angle cpo

71. Geometry Suppose that the angle e is a central angle of a circle of radius 1 (see the figure). Show that

e (a) Angle OAC = 2 (b) ICD I = sin e and IODI = cos e

e sin e (c) tan - = ---

2 1 + cos e

� A a D B

72. Geomeh,)' Show that the area A of an isosceles triangle is A = a2 sin e cos e, where a is the length of one of the two equal sides and e is the measure of one of the two equal an­gles (see the figure).

73. Geometry Let n � 1 be any real number and let e be any

angle for which 0 < ne < �. Then we can draw a triangle

with the angles e and ne and included side of length 1 (do you see why?) and place it on the unit circle as illustrated.

528 CHAPTER 7 Trigonometric Functions

Now, drop the perpendicular from C to D = (x, 0) and show that

tan(ne ) x = ---'---'------, tan e + tan(nB)

y

h nO

x

74. Geometry Refer to the figure. The smaller circle, whose ra­dius is a, is tangent to the larger circle, whose radius is b. The ray OA contains a diameter of each circle, and the ray OB is tangent to each circle. Show that

Vc;b cas e = --

b a + 2

(This shows that cos e equals the ratio of the geometric mean of a and b to the arithmetic mean of a and b.) [Hint: First show that sin e = (b - a)j(b + a) . ]

O ���--�-r------�-----+--� A

75. Geometry Refer to the figure. If 1 0A I = 1 , show that

(a) Area ilOAC = � sin a cos a

(b) Area ilOCB = � IOBI2 sin 13 cos 13

(c) Area ilOAB = � IOBI sin (a + 13 )

Discussion and Writing

79. If e is an acute angle, explain why sec e > 1 .

80. I f e is a n acute angle, explain why 0 < sin e < l .

81. How would you explain the meaning o f the sine function to a fellow student who has just completed college algebra?

'Are You Prepared?' Answers

1. 2V34 2 . 1(5) = 8

(d) l OB I = cos a

cos 13

(e) sin (a + 13 ) = sin a cos 13 + cos a sin 13

[Hint: Area ilOAB = Area ilOAC + Area ilOCB]

..e.=--...ll...::...---

-l....II-..lA o D

76. Geometry Refer to the figure, where a uni t circle is drawn. The line segment DB is tangent to the circle.

-1

y 1

-1

x

(a) Express the area of ilOBC in terms of sin e and cos e.

[Hint: Use the altitude from C to the base OB = 1 . ] (b) Express the area of il 0 BD in terms of sin e and cos e.

(c) The area of the sector OBC of the circle is �e, where e

is measured in radians. Use the results of parts (a) and (b) and the fact that

Area ilOBC < Area of sector OBC < Area ilOBD to show that

e 1 1 < -- < --

sin e cos e

77. If cos a = tan 13 and cos 13 = tan a, where a and 13 are acute angles, show that

. . n )3 - Vs SIl1 a = SIl1 fJ = 2

78. If e is an acute angle and tan e = x, x "* 0, express the re­maining five trigonometric functions in terms of x.

82. Look back at Example 5. Which of the two solutions do you prefer? Explain your reasoning.