Download 7.2 Right Triangle Trigonometry

SECTION 7.2
Right Triangle Trigonometry
51 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 W I revolutions per minute, whereas the pulley with
radius r2 rotates at W2 revolutions per minute. Show that
rl
r2
Source:
W2
WI
www.littleleague.org
Discussion and Writing
118.
Do you prefer to measure angles using degrees or radians?
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?
Or are they equal?
U1.
Explain the difference between linear speed and angular
speed.
U2.
For a circle of radius r, a central angle of (J degrees subtends
an arc whose length s is s
;
123.
124.
125.
r(J. Discuss whether this is a
1 0
true or false statement. Give reasons to defend your position.
=
Discuss why ships and airplanes use nautical miles to measure
distance. Explain the difference between a nautical mile and
a statute mile.
Investigate the way that speed bicycles work. In particular, ex­
plain the differences and similarities between 5-speed and
9-speed derailleurs. Be sure to include a discussion of linear
speed and angular speed.
In Example 6, we found that the distance between Albu­
querque, New Mexico, and Glasgow, Montana, is approxi­
mately 903 miles. According to mapquest.com, the distance
is approximately 1 300 miles. What might account for the
difference?
'Are You Prepared?' Answers
1. C
=
2 7T r
7.2 Right Triangle Trigonometry
PREPARI NG FOR THIS SECTION
•
Before getting started, review the following:
Geometry Essentials (Chapter R, Review, Section R.3,
pp. 30-35)
'\.Now Work
•
Functions (Section 3.1, pp. 208-218)
the 'Are You Prepared?' problems on page 525.
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 i n i n g Trigonometric Functions, Given the
Val u e of One of Them (p. 5 2 1 )
4 Use the Complementary A n g l e Theorem (p. 5 2 3)
1
Figure 1 8
b
a
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, .a nd, 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
518
C H A PTER 7
Trigonometric Functions
Now, suppose that e is an acute angle; that is, 0°
degrees) and 0
< e <
; (if
e
< e <
90° (if e is measured in
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
Figure 1 9
a
I nitial side
b
a
(c) Similar triangles
(b) Right triangle
(a) Acute angle
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.
DEFINITION
The six ratios of a right triangle are called trigonometric functions of acute
angles and are defined as follows:
Function Name
Figure 20
Abbreviation
sine of e
sin e
cosine of e
cos e
tangent of e
tan e
cosecant of e
csc e
secant of e
sec e
cotangent of e
(ot e
Value
b
(
a
(
b
a
(
b
(
a
a
b
Opposite e
a
-.J
b
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:
Adjacent to e
sm e
.
csc e
=
opposite
hypotenuse
b
c
=
hypotenuse
opposite
c
b
cos e
sec e
=
adjacent
hypotenuse
a
c
=
hypotenuse
adjacent
c
a
tan e
cot e
=
opposite
adjacent
b
a
=
adjacent
opposite
a
b
(1 )
Since a, b, and c are positive, each of the trigonometric functions of an acute
angle e is positive.
SECTION 7.2
EXA M P LE 1
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
,n OP "'I
c = hypotenuse = 5
( adjacent)2
32
WA RNING When writing the values of
the trigonometric functions, do not for­
get the argument of the function
sin
=
=
4
correct
5
4
- incorrect
5
a = adjacent = 3
To find the length of the opposite side, we use the Pythagorean Theorem .
3
sin f)
519
Finding the Val ue of Trigonometric F u n ctions
Sol ution
Figure 2 1
Right Triangle Trigonometry
+
+
( oppositef
(opposite)2
(opposite)2
opposite
=
=
=
=
(hypotenuse)2
52
25 - 9 = 1 6
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
opposite 4
adjacent
3
4
. e =
=
tan e =
cos e = ----S1l1
adjacent 3
hypotenuse 5
hypotenuse
5
-
-
•
csc e =
hypotenuse
OpposIte
Ci'1
�-
2
.
hypotenuse
sec e = --=-"'-----adjacent
5
4
5
3
cot e =
adjacent
OpposIte
.
3
4
•
Now Work P R O B L E M 1 1
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
sec e = -­
cos e
1
csc e = -­
sin e
1
cot e = -­
tan e
(2)
Two other fundamental identities that are easy to see are the quotient identities.
Quotient Identities
tan e =
sin 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 val­
ues of the remaining trigonometric functions.
EXA M P LE 2
Find i ng the Val ues of the Remaining Trigonometric
F u n ctions, Given sin 8 and cos 8
Vs
5
trigonometric functions of e.
2Vs
5
G"IV en S1l1
. e = -- and cos e = -- '
Sol ution
B ased on formula (3), we have
f
111
.
d t h e value o f each o f the four rema1l11l1g
.
Vs
5
sin e
1
tan e = -- =
=
2
cos e
2Vs
5
--
-
520
CHA PTER 7
Trigonometric Functions
Then we use the reciprocal identities from formula (2) to get
1
1
csc e = __ = __
sin ()
Vs
5
=
un=;>-
Figure 22
b
Vs
1
5
1
sec e = -- =
= -- = -cos e 2 Vs 2 Vs
2
5
5
__ = Vs
Vs
--
cot e =
1
tan e
--
=
1
1
2
-
=
2
•
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
a
In terms of trigonometric functions of the angle e, this equation states that
(sin e ?
+
(cos e) 2
=
(4)
1
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
=
(5)
1
Another identity can be obtained from equation (5) by dividing each side by
cos2 e.
sin2 e
1
+ 1 = __
2
cos e
cos2 e
Now use formulas (2) and (3) to get
tan2 e
+
Similarly, by dividing each side of equation (5) by sin2 e, we get 1
2
csc e, which we write as
cot2 e
+
(6)
1 = sec2 e
1 = csc2 e
+
cot2 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
csc e
sin2 e
+
=
1
.­
S1l1
cos2 e = 1
e
cot e
=
1
cot e = -­
tan e
cot2 e + 1 = csc2 e
1 = sec2 e
sec e =
tan2 e
+
cos e
sm e
­
.
1
cos e
--
SECTION 7.2
EXA M P LE 3
Right Triangle Trigonometry
52 1
F i n d i n g the Exact Value of a Trigonometric
Expression Using I dentities
Find the exact value of each expression. Do not use a calculator.
( a) tan 200 -
( a) tan 200 -
Solution
--sin 200
cos 200
. 2 7T
(b) sm
12 +
=
sin e
7T
12 +
cos e
1
7T
sec12
?
3
=
tan e
= sin2 � + cos2 � = 1
12
12
r
cas e
�
7T
12
sec-? -
sin 200
tan 200 - tan 200 = a
cos 200 i
--
( b) sin2
1
r
1
= -­
sec e
•
Now Work P R O B L E M 3 9
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.
EXA M P LE 4
F i nding the Values of the Remaining Trigonom etric 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.
Solution
Solution 1
Using the Definition
Figure 23
.d:::1 b = 1
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 + 1 2
a2 + 1
a2
a
a
=
=
=
=
32
9
8
2 V2
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 Ex­
ample 1.) Using a = 2 V2 , b = 1, and c = 3, we have
a
c
cos e = - =
csc e = �
b
=
-2 V2
3
b
a
1
2V2
V2
4
tan e = - = -- = --
3
3 V2
2 V2
l = 3 sec e = � = _- =
= 2 V2
cot e = :: =
1
a 2 V2
b
1
4
•
522
C H A PTER 7
Trigonometric Functions
Solution 2
U sing I dentities
We begin by seeking cos e, which can be found by using the Pythagorean Identity
from equation (5).
sin2 e + cos2 e = 1
Form ula (5)
1
9
- +
cos2 e
=1
cos2 e
=1
sin e
-
1
9
=
8
9
-=-
1
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
.
Now w e know that sm e = - and cos e
3
in Example 2.
1
"3
--
3
1
3
sin e
1
tan e = -- = -- =
cos e
2\12 2\12
sec e
1
=
\12
4
cot e =
3 \12
4
csc e
--
= -- = -- = -- = -2\12
3
cos e
2\12
2 \12
3
= -- , s o we can proceed as we did
1
_
_
tan e
1
=
= -- =
sin e
_1_ = � = 2\12
\12
\12
4
1
-= 3
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
STEP
STEP
1
Using the Definition
1: Draw a right triangle showing the acute angle e.
2: Two of the sides can then be assigned values based on the value of
the given trigonometric function.
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.
STEP 3:
Method 2 Using Identities
Use appropriately selected identities to find the value of each of the remain­
ing trigonometric functions.
E XA M P L E 5
Given One Value of a Trigonometric F u n ction, Find the Val ues
of the Remaining Ones
Given tan e
Solution 1
U sing the D efinition
=
�,
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
2
tan e = - =
opposite
adjacent
b
= -
a
SECTION 7.2
With b = 1 and a
Theorem.
Figure 24
tan
8
=2
1
Right Triangle Trigonometry
2, the hypotenuse c can be found by using the Pythagorean
=
c2
=
a2
+
b2
c = Vs
=
22
+
12 = 5
Now apply the definitions using a = 2, b = 1 , and c
.
SID e
=
b
c
1
Vs
= -5
Vs
= --
csc e = .£ =
b
Sol ution 2
U sing I d entities
Vs
1
523
Vs
=
a
c
cos e
=
=
sec e
= - =
c
a
=
Vs .
2
2 Vs
-- = -5
Vs
Vs
-2
cot e
=
a
b
-
2
= -
1
=2
•
B ecause we know the value of tan e, we use the Pythagorean Identity that involves
tan e:
Form u l a (6)
tan2 e + 1 = sec2 e
tan 8
sec2 e
1
4
-
=
+
5
1 = 4
1
= "2
Proceed to solve for sec 8.
Vs
sec e = 2
Now we know tan e
1
= -
2
and sec e
cos e
=
Vs
.
.
. .
--. U sIDg reClprocaI 1' dentltles, we f'ID d
2
2
1
1
= -- = -sec e
Vs Vs
= --
=
2Vs
-5
2
1
cot e = -tan e
=
1
1
=
2
2
To find sin e, we use the following reasoning:
sin e
.
, so SID e = ( tan e ) ( cos e )
cos e
1
csc e = __ = _1_ = Vs
sin e
Vs
tan e =
--
2
5
5
•
5
,'=z> -
4
Figure 25
A
� Adjacent to A
b opposite B
B
a
Adjacent to B
opposite A
Now Work P R O B L E M 2 5
Use the Complementary Angle Theorem
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
tan B
b
a
a
= - =
cot A
cot B = b = tan A
(8)
524
CHAPTER 7
Trigonometric Functions
Because of these relationships, the functions sine and cosine, tangent and cotan­
gent, and secant and cosecant are called cofunctions of each other. The identities (8)
may be expressed in words as follows:
THEOREM
Complementary Angle Theorem
Cofunctions of complementary angles are equal.
-.J
Here are examples of this theorem.
Complementary angles
Complementary angles
sin 30° = cos 60°
tan 40°
t
J
1-
t
t
Co/unctions
J
1-
= cot
50°
j
Co/uncti ons
Complementary angles
J
sec 80°
!
1-
= csc
10°
t
Co/unctions
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
7T
.
of e, as, for examp Ie, 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.
Table 2
(J ( Degrees)
sin ()
=
(J (Radians)
cos (90° - ())
=
sin e
cos
(f (f (f
(f
(f
(f
()
cos ()
=
s i n ( 90°
e)
cos ()
=
sin
tan e
=
cot(90° - e )
tan ()
=
cot
-
esc e
=
sec(90°
esc e
=
sec
- e
sec e
=
csc( 900 - e )
sec e
=
csc
cot e
=
ta n ( 90° - e )
cot e
=
tan
-
-
8)
()
e
- e
-
)
)
)
)
)
)
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.
EXAM PLE 6
Using the Complementary Angle Theorem
(a) sin 62°
=
cos ( 90° - 62° )
7T
(b) tan 12
=
cot
(
(
(
7T
2
-
7T
12
-- =
)
6) =
7T
. 7T
7T
( c) cos "4 = SlI1 2 - "4
7T
( d) csc 6
=
)
=
7T
7T
sec 2 -
=
cos 28°
57T
cot 12
. 7T
SlI1
"4
7T
sec :3
•
SECTION 7.2
E XA M P L E 7
Right Triangle Trigonometry
52 5
U sing the Complementary Angle Theorem
Find the exact value of each expression. Do not use a calculator.
sin 35°
(b ) cos 55°
(a) sec 28° - csc 62°
(a) sec 28° - csc 62° = csc(90° - 28°) - csc 62°
Solution
= csc 62° - esc 62° = 0
cos( 90° - 3SO )
cos 55°
sin 35°
(b ) cos 550
cm::
=
=
--
cos 55°
= 1
cos 55°
---
•
Now Work P R O B L E M 4 3
l-lisJorical Feature
T
he n a me sine for the s i n e function is due to a medieval confusion.
Figure 26
The n a me comes from the Sanskrit word ji va, (mea n i n g chord ) ,
first used in I ndia by Araybhata the Elder (AD 5 1
0). He really meant
half-chord, but a bbreviated 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, a n d 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 n a me tangent, due to Thomas Finck ( 1 583), can be understood
by looking at Figure
C. If d(O, B)
=
26. The line segment DC is tangent to the circle at
d(O, C)
=
d (0, C)
1 , then the length of the line segment DC is
=
d(D, C)
1
=
d(D, C)
d(O, C)
=
tan
a
The old name for the tangent is u m bra versa (meaning turned
shadow), referring to the use of the t a n gent in solving height problems
The names of the cofunctions came a bout as follows. If A and
are complementary angles, then cos A
=
sin
B
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 n ciation, the co migrated
to the front, and then cosine received a three-letter a b breviation to
match sin, sec, a n d tan.The two other cofu nctions were si milarly treated,
except that the long form s cotan a n d cosec survive to this day in some
cou ntries.
with shadows.
7.2 Assess Your Understanding
'Are You Prepared?' Answers are given at the end of these exercises. If you get a wrong answel; read the pages listed in
1.
In a right triangle with legs a = 6 and b = 10, the Pythagorean Theorem tells us that the hypotenuse c =
.
(pp. 30-35)
___
2.
The value of the function f(x)
(pp. 20S-21S)
= 3x
-
7 at 5 is
red.
___ .
Concepts and Vocabulary
3.
4.
5.
6.
Two acute angles whose sum is a right angle are called
The sine and
tan 2So = cot
functions are cofunctions.
For any angle 8, sin2 8 + cos2 8
7. True or False
tan 8
sin 8
cos 8
= -- .
=
__ ,
__
.
8. True o r False
9. True or False
cos 8
= 3".
1
10. True o r False
1 + tan2 8
=
csc2 8 .
If 8 is an acute angle and sec 8 = 3, then
7T
tan 5
= cot - .
47T
5
526
CHAPTER 7
Trigonometric Functions
Skill Building
1 1-20,
�
In Problems
1 1·
5
13.
12
16.
"
j
�
�
find the value of the six trigonometric functions of the angle () in each figure.
4
3
17.
�
-J2
14 '
3
3
2
18.
tl
15.
3
�
20.
19.
.J3 �
2
2
e
e
2
4
�J"
li
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. Si n (} =
�
,
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 ().
.
25 . Sill ()
v2
v2
=
--
2
"2
1
29. tan () =
v2
33. tan () =
28. Sill
' () =
3
32. csc () = 5
35. csc () = 2
36. cot () = 2
27. cos () =
30. cot ()
31. sec ()
=
34. sec () =
1
"2
2
5
"3
1
"3
26. cos () = --
=
4
V3
In Problems 37-54, use Fundamental Identities andlor the Complementary A ngle Theorem to find the exact value of each expression.
Do not use a calculator.
+
37. sin2 20°
41. tan 50° 45 .
cos2 20°
sin 50°
38. sec2 28° - tan2 28°
42. cot 25° -
­
_
-
cos )0°
COS 10°
-­
46.
. 80°
Sill
49. tan 20° -
COS 70°
cos 20°
53. cos 35° sin 55°
+
Sill 25°
COS 40°
'- 43.
40. tan 1 0° cot 10°
sin 38° - cos 52°
47. 1 - cos2 20° - cos2 70°
.­
Sill 50°
50. cot 40° -
--­
COS 25°
.
­
39. sin 80° csc 80°
sin 50°
-­
s i n 40°
cos 55° sin 35°
51. tan 35° · sec 55° · cos 35°
44. tan 12° - cot 78°
48.
1 +
tan2 5° - csc2 85°
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 60°
(c) csc
(b)
7T
(5
(d) sec
56. G i ven sin 60° =
exact value of
( a ) cos 30°
(c) sec
cos2 30°
V3 , use trigonometric identities to find the
2
(b)
7T
(5
7T
"3
cos2 60°
( d ) csc
7T
"3
57. Given tan () = 4, use trigonometric identities to find the exact
value of
( a ) sec2 ()
(c) cor
(� () )
-
(b)
cot ()
( d ) csc2 ()
(a) cos ()
tan2 ()
(b)
(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 . G i ven t h e approximation sin 38°
�
0.62, use trigonometric
identities to find the approximate value of
tan 38°
(a) cos 38°
(b)
(c) cot 38°
(d) sec 38°
(e) csc 38°
(f) sin 52°
(g) cos 52°
(h) tan 52°
SECTION 7.2
62.
63.
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 2 1 °
( d ) sec 2 1 °
(e) csc 2 1 °
(f) sin 69°
(g) cos 69°
(h) tan 69°
If sin e
= 0.3, find the exact value of sin e + cos
(� - )
e .
64.
65.
66.
If tan e
Right Triangle Trigonometry
= 4, find the exact value of tan e + tan
Find an acute angle e that satisfies the equation
sin e
=
527
(� - )
e .
cos(2e + 30°)
Find an acute angle e that satisfies the equation
tan e
= cot (e + 45° )
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 1 500 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 to­
ward 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 il­
lustration.
(d) Calculate the time T if you walk directly from the park­
ing lot to the house.
500
Hint: tan e =
1 500
[
rt
=
Electrical Engineering A resistor a n d a n inductor con­
nected 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 net­
work 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.
71.
(e) Calculate the time T if you walk 1000 feet along the
paved path and then walk directly to the house.
( f) Graph T = T (e ) . For what angle e is T least? What is
x for this angle? What is the minimum time?
(g) Explain why tan e
possible.
68.
]
69.
Electrical Engineering Refer to Problem 69. A series net­
work 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 imped­
ance Z.
(b) Determine the values of the remaining five trigono­
metric functions of the phase angle cpo
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)
I CDI = sin e and I ODI = cos e
(c) tan
e
2
-
=
sin e
1 + cos e
---
�
1:. gives the smallest angle e that is
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
4 ft
around the corner is
equal to the smallest
value of L.
t
�
72.
73.
A
a
D B
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 an­
gles (see the figure).
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
C
Now, drop the perpendicular from to D
that
tan (ne )
-'---'----,x =
tan e + tan (nB)
= (x, 0) and show
l OBI
cos a
cos 13
(e) sin ( a + 13 ) = sin a cos 13 + cos a sin 13
(d)
-
=
il OAB =
[Hint: Area
y
h
nO
74.
=
+
Area
ilOCB]
..e.=-...l
- :...l...:
-l...I.I ..lA
-- D
x
76.
DB
Geometry Refer to the figure, where a uni t circle is drawn.
The line segment
is tangent to the circle.
y
1
x
-1
Vc;b
--
a+ b
2
-1
ilOBC
C
il 0BD
OBC
(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
ilOAC
o
Geometry Refer to the figure. The smaller circle, whose ra­
dius 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
cas e
Area
OB ]
�e,
(a) Express the area of
in terms of sin e and cos e.
[Hint: Use the altitude from
to the base
= 1.
(b) Express the area of
in terms of sin e and cos e.
= (b - a)j(b + a).]
(c) The area of the sector
of the circle is
where e
is measured in radians. Use the results of parts (a) and
(b) and the fact that
O ���--�-r------�-----+--�
75.
Geometry
(a) Area
(b) Area
(c) Area
Refer to the figure. If
il OAC = �
il O CB = �I OBI2
ilOAB = �I OBI
sin a cos a
1 0AI =
A
Area
ilOBC
< Area of sector
to show that
1 <
1 , show that
77.
If cos a = tan 13 and cos 13
angles, show that
sin 13 cos 13
sin ( a + 13 )
.
< Area
ilOBD
1
e
< -sin e
cos e
--
= tan a , where a and 13 are acute
n
SIl1 a = SIl1 fJ =
.
OBC
)3 - Vs
2
78.
If e is an acute angle and tan e = x, x "* 0, express the re­
maining five trigonometric functions in terms of x.
82.
Look back at Example 5. Which of the two solutions do you
prefer? Explain your reasoning.
Discussion and Writing
79.
80.
81.
If e is an acute angle, explain why sec e
1.
I f e is a n acute angle, explain why 0 < sin e < l .
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