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Unformatted text preview: A22 APPENDIXES D TRIGONbMETRY A23 Notice that in shifting the ellipse, we replaced x by x — h and y by y — k in Equation 4 to 9. 2x2 + 2y2 — x + y = 1 26. 16x2 + 9y2 “ 36y 1 108 obtain Equation 5. We use the same procedure to shift the parabola y : ax2 so that its vertex . . . . 2 2 2 .,':‘ (the origin) becomes the point (h, k) as in Figure 15. Replacing x by x — h and y by y ‘ k, we .L 10 32d:;1::::;c:§$t302nﬁnaihicgjficfgsoa;e%r:::nf foes 27. y z x _ 6x 13 28' x g y _ 4x “ 3 m0 % see that the new e nation is L ' ’ . . _ _ . 29. = 4 ~ 2 30. 2  2 + 6 M 5 =
q ' circle? When that condition 1s satisfied, flnd the center and x y y x y
' " . 31.x2+42~6x+5=0
y _ k 2 “(x ’ hf or y 2 “(X _ hf + k radius of the circle y 2 2 __ 2
11—32 3 Identify the type of curve and sketch the graph. Do 32' 4x + 9y 16x + 54y + 61 O y not plot points. Just use the standard graphs given in Figures 5, ' . b
1 — d d b th .
. 6’ 8, 10’ and 11 and Shift if necessary. 33 34 I Sketch the region oun e y e curves
7 a . .= :2 34.:=4~2, «2:2
17,].2: y=a(x/1)‘+k _ 11_ y = __x 12. yg __ x2 =1 33 y 3x,y x } x x y
j 2 3 _ _ 2 35. Find an equation of the parabola with vertex (1, ~1) that 13' x + 4y _ 16 14' x " 2y passes through the points (—1.3) and (3,3).
\ = ,2 ' 2 _ 2 = .2 2 =
‘l y a“ mic) * 15' 16x 25y 400 15' 25" + 4y 100 36. Find an equation of the ellipse with center at the origin
5 0 x 7‘ 17. 4x2 + y1 = 1 18. y = x2 + 2 that passes through the points (1, —10\/§/3) and
._ FIGURE 15 : 7 a (*2 Sﬁm
g j' 19. x z y — 1 20. 9x _ 25y2 = 225 ’ ' V 7 7
EXAMPLE 7 Sketch the graph ofthe equation y = 2x2 __ 4x + 1’ g, 21. 9y“ ~ x" = 9 22. 212 + 5y2 = 10 37—40 I Sketch the graph of the set.
' . = 24. =2+2 . 2+2<1 38. , 2+2>4
SOLUTION First we complete the square: 23 xy 4 y x x 37 {(x’y) Ix y } {(1 wlx y } 25. 9(x — 1)2 + 4(y ~ 2V : 36 39. {(x,y)y 2 x2 ~ 1} 4o. {(x,y)x2 + 4y2 s 4}
y=2(x2—2x)+1=2(x—1)2—1 In this form we see that the equation represents the parabola obtained by shifting y = 2):2 so   I E that its vertex is at the point (1,—1). The graph is sketched in Figure 16. B TRIGONOMETRY
1 2 3
EXAMPLE 8 Sketch the curve x = '1 — yz. f; ANGLES Angles can be measured in degrees or in radians (abbreviated as rad). The angle given by a
(1"1) ___ complete revolution contains 360°, which is the same as 277 rad. Therefore
SOLUTION This time we start with the parabola): = —y2 (as in Figure 6 with a = ~1) L,
HGURE 16 and shift one unit to the right to get the graph ofx = l — y2 (see Figure 17). ;i
yzzxzw4x+1 (1) wrad= 180°
and
180 ° 0 7r
(2) 1 rad = —— z 573° 1 = ——' rad x 0.017 rad
7T 180
EXAMPLE 1
(a) Find the radian measure of 60°. (b) Express 577/4 rad in degrees.
5 O L U T I O N
FIGURE 17 (a) xz—y2 (mx: 1 “3,2 a (a) From Equation 1 or 2 we see that to convert from degrees to radians we multiply by
77/ 180. Therefore
60° — 60 77 ~ 3: d
EXERCISESC _ 180 “ 3 m
1—4 I Find an equation of a circle that satisfies the given 5—9 I Show that the equation represents a circle and find the (b) To convert from radians to degrees we multiply by 180/77. Thus
conditions. center and radius. 9
1. Center’(3, ~1), radius 5 5. x2 + y2 — 4x + 10y + 13 = 0 5—” rad = z 2250 4 4 7r
2. Center (—2, —8), radius 10 6. x2 + y2 + 6y + 2 = 0
3. Center at the origin, passes through (4,7) 7. x2 + y2 + x = O . ~ . . . .
In calculus we use rad1ans to measure angles except when otherwrse indlcated. The followmg
4. Center (~1, 5), passes through (4, —6) 8. 16):2 + 16y2 + 8x + 32y + 1 = 0 table gives the correspondence between degree and radian measures of some common angles. A24 FIGURE 1 FIGURE 2 Side 6 initial side / FIGURE 3
020 initial side terminal side FIGURE 4
6<0 APPENDIXES
Degrees 0° 30° 45° 60° 90° 120D 135° 150° 180° 270° 360°
I 77 7T 17 7T 27T 317 ST 377
Radlans O —~ —~  — — w — —
6 4 3 2 3 4 6 7T 2 2” Figure 1 shows a sector of a circle with central angle 6 and radius r subtending an arc with
length a. Since the length of the arc is proportional to the size of the angle, and since the entire
circle has circumference 27rr and central angle 277, we have 0 a 277' 2’n'r Solving this equation for 6 and for a, we obtain (3) 6=— a=r6 Remember that Equations 3 are valid only when 9 is measured in radians. In particular, putting a = r in Equation 3, we see that an angle of 1 rad is the angle sub—
tended at the center of a circle by an are equal in length to the radius of the circle (see
Figure 2). E X A M P L E 2
(a) If the radius of a circle is 5 cm, what angle is subtended by an arc of 6 cm? (b) If a circle has radius 3 cm, what is the length of an arc subtended by a central angle
of 37r/8 rad? SO LUTION
(a) Using Equation 3 with a = 6 and r = 5, we see that the angle is 9 = g = 1.2 rad
(b) With r = 3 cm and 0 = 37r/8 rad, the arc length is 3
a=w=3(1):&;m
8 8 The standard position of an angle occurs when we place its vertex at the origin of a coor—
dinate system and its initial side on the positive xaxis as in Figure 3. A positive angle is ob
tained by rotating the initial side counterclockwise until it coincides with the terminal side.
Likewise, negative angles are obtained by clockwise rotation as in Figure 4. Figure 5 shows
several examples of angles in standard position. Notice that different angles can have the same terminal side. For instance, the angles 377/4, 57r/4, and 117r/4 have the same initial and termi
nal sides because 2: 2” 577 377 + 2 _ 117T
4 4 4 7T and 271' rad represents a complete revolution. y y
377
a: —
A, 4
> \‘ >
0 f/ x 0 t
a=—g FIG U R E 5
Angles in standard position THE TRIGONOMETRIC naunuun hypotenuse adjacent FlGURE 6 opposite A: FIGURE 7 FIGURE 8 D TRIGONOlVlETRY A25 For an acute angle 6 the six trigonometric functions are defined as ratios of lengths of sides of a right triangle as follows (see Figure 6): (4) This definition does not apply to obtuse or negative angles, so for a general angle 0 in stan—
dard position we let P(x, y) be any point on the terminal side of 0 and we let r be the distance
IOPI as in Figure 7. Then we define (5) Since division by O is not defined, tan6 and seed are undefined when x = 0 and csc0 and
cote are undefined when y S 0. Notice that the definitions in (4) and (5) are consistent when
6 is an acute angle. If 6 is a number, the convention is that sin 0 means the sine of the angle whose radian mea—
sure is 6. For example, the expression sin 3 implies that we are dealing with an angle of 3 rad.
When finding a calculator approximation to this number we must remember to set our calcula
tor in radian mode, and then we obtain sin 3 2 0.14112 If we want to know the sine of the angle 3D we would write sin 3° and, with our calculator in
degree mode, we find that sin 3° % 0.05234 The exact trigonometric ratios for certain angles can be read from the triangles in Figure 8.
For instance, WWML :_i Smazii
Sll'l — 2 8111 6 2 3 2
W2;_ .:_ii mzzi
COS — 2 C05 6 2 2
1T 7T u
a
H
m
D
l
n tan —"
4 A26 sin I9 > 0 all ratios > 0 tan0>0 cos0>0 FIGURE 9 [)("lti FIGURE 10 FIGURE 12 APPENDIXES The signs of the trigonometric functions for angles in each of the four quadrants can be re
membered by means of the rule “All Students Take Calculus” shown in Figure 9. EXAMPLE 3 Find the exact trigonometric ratios for 6 = 277/3. SOLUTION From Figure 10 we see that a point on the terminal line for 0 = 277/3 is
P(—1, Therefore, taking x=—1 y=ﬁ r=2 in the definitions of the trigonometric ratios, we have Sin 271' = v3 211' __ 1 211' * \/—
""3 “2 COS “*3  2 tan —  3
271' 2 211' 277 1 csc—=—— sec—=—2 cot=—m J3— 3 3 J? a a 0 1 I. .2: 11 37: 371 5W 3W 2
6 4 3 2 3 4 6 7T 2 7T . 1 1 3 3 1 1 stn6 O —— ~— —— 1 —— m— _ _
2 J? 2 2 2 2 0 1 O
3 cost) 1 — —1 i 0 ——1— ——~1 —~——ﬁ —1 0 1
2 2 2 2 J5 2 EXAMPLE 4 If 0050 a and 0 < 6 < 77/2, find the other five trigonometric functions of 0. SOLUTION Since cos6 = 2, we can label the hypotenuse as having length 5 and the adjacent
side as having length 2 in Figure 1]. If the opposite side has length x, then the Pythagorean Theorem gives x2 + 4 = 25 and so x2 = 21, x = V21. We can now use the diagram to write
the other five trigonometric functions: JET J27 sin0 = —~—5 tan0 = ~——2
5 5 2
csc0=~— secB=— cot0=—
v21 2 J21 a EXAMPLE 5 Use a calculator to approximate the value of x in Figure 12. SOLUTION From the diagram we see that tan 40° = 16—
x
Therefore x = —————16 z 19.07
tan 40° TRIGONOMETRIC IDENTITIES D TRIGONOMETRY A27 A trigonometric identity is a relationship among the trigonometric functions. The most elemen—
tary are the following, which are immediate consequences of the definitions of the triggjnomet—
ric functions. (6) For the next identity we refer to Figure 7. The distance formula (or, equivalently, the
Pythagorean Theorem) tells us that x2 + y2 = r2. Therefore 2 2 2 2
. 2 2 y x _ x + y _ r _
sm0+cosO=——+——— _ ..1
r2 r2 r2 1,1
We have therefore proved one of the most useful of all trigonometric identities: (7) sin20 + c0320 = 1 If we now divide both sides of Equation 7 by c0520 and use Equations 6, we get (8) tanZO + 1 = sec26 Similarly, if we divide both sides of Equation 7 by sin20, we get (9) 1 + C0t20 = csc2I9 The identities (10a) sin(—9) = ~sin0
(10b) cos(0) = c050 show that sin is an odd function and cos is an even function. (See Section 1 of Review and Pre—
view.) They are easily proved by drawing a diagram showing 0 and ~19 in standard position (see Exercise 39).
Since the angles 6 and 6 + 277 have the same terminal side, we have (11) sin(9 + 2n) = sin0 cos(9 + 2n) = cosB These identities show that the sine and cosine functions are periodic with period 271‘.
The remaining trigonometric identities are all consequences of two basic identities called the addition formulas: (12a) sin(x + y) = sinxcosy + cosxsiny (12b) cos(x + y) = cosxcosy — sinxsiny A28 APPENDIXES The proofs of these addition formulas are outlined in Exercises 85, 86, and 87. By substituting y for y in Equations 12a and 12b and using Equations 10a and 10b, we
obtain the following subtraction formulas: (13a) sin(x  y) = sinxcosy  cosxsiny (13b) cos(x  y) = cosxcosy + sinxsiny Then, by dividing the formulas in Equations 12 or Equations 13, we obtain the correspond—
ing formulas for tan(x : y): t +
(14a) tan(x + y) = anx tany 1— tanxtany ta —
(14b) tan(x — y) = ﬁx {any 1 + tan x [any If we put y = x in the addition formulas (12), we get the doubleangle formulas: (15a) 2 sin x cos x (15h) coszx — sinzx Then, by using the identity sinzx + coszx = 1, we obtain the following alternate forms of the
double—angle formulas for cos 2x: (16a) cos 2x = 2coszx — 1 (16b) cos 2x = 1 ~ 25inzx If we now solve these equations for coszx and sinzx, we get the following halfangle formulas,
which are useful in integral calculus: (17a) (17b) Finally, we state the product formulas, which can be deduced from Equations 12 and 13. (18a) sinxcosy = %[sin(x + y) + sin(x  y)]
(18b) cosxcosy = %[cos(x + y) + cos(x  y)] (18c) sinxsiny = %[cos(x — y) H cos(x + y)] GRAPHS OF THE
TRIGONOMETRIC FUNCTIONS FIGURE 13 D TRIGONOMETRY A29 There are many other trigonometric identities, but those we have stated are the ones used
most often in calculus. If you forget any of them, remember that they can all be deducedﬁfrom Equations 12a and 12b. :8
EXAMPLE 6 Find all values of x in the interval [0, 277] such that sinx = sin 2x.
SOLUTION Using the double—angle formula (15a), we rewrite the given equation as
sinx = 23inxcosx or sin x(1 ~e Zeosx) = 0
Therefore, there are two possibilities for x:
sinx=0 or l—ZCosx=0
x=0,7r,27r cosx=%
77 571'
X = ?’ “3’
The given equation has five solutions: 0, 77/3, 77, 577/3, and 271. E The graph of the functionﬂx) = sin x, shown in Figure 13(a), is obtained by plotting points for
0 $ x < 277 and then using the periodic nature of the function (from Equation 11) to complete
the graph. Notice that the zeros of the sine function occur at the integer multiples of 77, that is, sinx = 0 whenever x = n77, n an integer 371' Because of the identity _ 77
cosx=s1n x+—2 (which can be verified using Equation 12a), the graph of cosine is obtained by shifting the
graph of sine by an amount 71/2 to the left [see Figure 13(b)]. Note that for both the sine and
cosine functions the domain is (—00, 00) and the range is the closed interval [#1, 1]. Thus, for all
values of x, we have //\
J.
//\
IA —1 £sinx cosx A30 APPENDIXES “
D TRlGONOMETRY A31
The graphs of the remaining four trigonometric functions are shown in Figure 14 and their 23—28 I Find the exact trigonometric ratios for the angle 48. tanza — sinza = tarﬁa Sirﬁa
domains are indicated there. Notice that tangent and cotangent have range (~00,00), whereas whose radian measure is given. 2 2 2 2 0036mm and “can? have range (“00, “1] U [1,00). All four functions are periodic: tangent and 49' wt 6 + sec 9 : tan 0 + C80 6 w}
cotangent have period 7r, whereas cosecant and secant have period 27r. 23_ i7: 24_ i7: 25. 27: 50. 2 csc 22‘ = sec reset
4 3 2
5 ll 51 t 29 2mg
7T . an = ———
W 26. —57r 27. m 28. «3: i  tanzﬁ
6 4
I ' __.__1_.__ + __.._.1__..__ — 2 29
 29—34 I Find the remaining trigonometric ratios. 52' 1 _ Sing 1 + sing _ sec
1 29 “116223. 0<9<1 53. sinxsin2x+cosxcos2x=cosx
 ' ’
54. Sing): — sinzy = sin(x + y) sin(x ~ y)
I
Z 77 33' _7T 7, 0 7 77: 37k: 30. tana=2, 0<oz< singb
2 2 I "f 5 3— 55.m—CSC¢+COtd}
7T
31. sec¢=~15, ——<¢<ar 3111064.”
2 56. tanx + tany =
_ 1 3w cos x cos y
32' COSX_§’ W<x<7 57. sin36+sin6=23in2000s6
(a)y:t3nx (b)y=cotx 33. 0mg = 3, 77 < 3 < 277 58. c0830 = 4c0530  3C0$0
yT M 34 cscez __4_ i71<6<27r 59—64 I Ifsinx=§and secy=%,wherexandylie between
’ 0 and 77/2, evaluate the expression.
l .
I 35—38 I Find, correct to five decimal places, the length of the 59' S‘n(x + y) 60‘ 0030‘ + y)
: y=sinx I Fem. side labeled x. 51. cos(x — y) 62. sin(x ~ y)
l ,
3 l" / \/ 3.1 35 35 x 63. sin 2y 64. c052y
t 04/ . _. 2 —w a / 40° L
\ l . . . . .
/ I \ l ' 'F l‘ l 1" ? 10 65—72 I Find all values ofx in the interval [0, 277] that satisfy
\ x E 7’ / 7; \ 7T / I ,1 cm '
—— —1 2 \ ’ \ x _1 w 5 \ / I x 25 cm the equation.
: I 65. 2cosx—l=0 66. 3cot2x=l
67. 2sin2x =1 68. Itanxl =1
37. 38.
22cm x 69. sin 2x = cosx 70. 2cosx + sin 2x = 0
FIGURE 14 (C)y=CSCA’ (d)y=secx 37, 71. sinx= tanx 72. 2 + c032x=3cosx
x ?
27, 73—76 I Find all values of x in the interval [0, 277] that satisfy
EXERCISES D 5 the inequality.
8cm 73. sinxSé 74. 2cosx+l>0
1—6 E C ' . . _
onvert from degrees to radians. 14. It a Circle has radius 10 cm, what is the length of the arc 39_41 I Prove each equation 75. —l < tanx < l 76. sinx > cosx
1. 210° 2. 300° 3 90 subtended by a central angle of 72°? .
' 15 A Circlc has radius 1 5 Wh t 1 . b d d h 39. (21) Equation 10a (b) Equation 10b 77—82 I Graph the function by starting with the graphs in
o D . .rn. aangeissutene atte . I I
4. —315 5. 900 6 36° . .  Figures 13 and 14 and applying the transformations of
'  t 40. a E uation 14a b E uation 14b . ,
CCU er of the Circle by an are 1 m long? ( ) q . ( ) q ‘ Section 2 of Review and PreView where appropriate.
7.12 I Convert from radians [0 degrees. 16. Find the radius of a circular sector with angle 37T/4 and 41' (3) Equal?“ 183 (b) Equanon 18b 77.
arc length 6 cm. (C) Equatlon 13C 77. y = cos x —~ ~— 78. y = tan 2x
7 47r 3 — 171 57’   3
'  2 9 “I; 17—22 I Draw, in standard position, the angle whose measure 42.58 I Prove web “16an 1 77—
is given. 7T 7T 79. y = w tan x a we 80. y =1 + secx
87]» 3 42. cos ~— ~ x = sinx 43. sin ~—~ + x = cosx 3 2
10 —~ 11 ~ l 12 5 377 2 2
I 3 ' 8 ' 17. 315° 13. 1509 19. —‘—rad _ . _ . 7T
4 44. sin(7T—x)=sinx 45. sin9cot0=cos€ 81Y“5mx 82Y*2+Sm 45+:
13. Find the length of a circular arc subtended by an an 1 f 77 ' 2 — '
£30 46. +CO‘ —l+srn2 . . . .
7T/12 rad if the radius of the Circle is 36 cm. 20. ~3— rad 21. 2 rad 22. —3 rad (smx bx) x 83. Prove the Law of Cosmes: If a triangle has Sides With 47. secy — cosy = tanysiny lengths a, b, and c, and 0 is the angle between the sides A32 APPENDIXES 84. with lengths a and b, then CE = a2 + b2 — Zabcosd P(x,y) [Hint Introduce a coordinate system so that 6 is in stan—
dard position as in the figure. Express x and y in terms of 6
and then use the distance formula to compute c.] In order to find the distance [ABI across a small inlet, a
point C is located as in the figure and the following
measurements were recorded: [.C = 103°, ACI = 820 m,
IBCI = 910 m. Use the Law of Cosines from Exercise 83
to find the required distance. A 85. Use the figure to prove the subtraction formula 86. 87. 88. 89. cos(oz — B) = cosacosB + sinasinB y ‘ A (cos a, sin a) a B(cos ,8, sin ,8) [Him‘: Compute c2 in two ways (using the Law of Cosines
from Exercise 83 and also using the distance formula) and
compare the two expressions] Use the formula in Exercise 85 to prove the addition
formula for cosine (12b). Use the addition formula for cosine and the identities 7T 77
cos(— — 0) = sine sin<— — 6) = c056
2 2 to prove the subtraction formula for the sine function. Show that the area of a triangle with sides of lengths a and
b and with included angle (9 is A=%absin6 Find the area of triangle ABC, correct to five decimal
places, ifIABl = 10 cm, BCI : 3 cm, and
LABC = 107°. MATHEMATICAL INDUCTION The principle of mathematical induction is useful when we need to prove a statement Sn about
the positive integer n. For instance, if S” is the statement then and so on. 1. S1 is true. PRINCIPLE OF MATHEMATICAL INDUCTION Let S” be a statement about the
positive integer n. Suppose that 2. Skid is true whenever 8;, is true. Then 3,, is true for all positive integers n. (abyz = nbn
S. says that ab = ab S; says that (ab)2 = 512172 E MATHEMATICAL INDUCTION A33 This is reasonable because, since SI is true, it follows from condition 2 (with k = 1) that S;
is true. Then, using condition 2 with k = 2, we see that 53 is true. Again using condition}; this
time with k = 3, we have that 54 is true. This procedure can be followed indefinitely. In using the principle of mathematical induction, we follow three steps. Step I: Prove that S" is true when n = l. .
Step 2: Assume that S" is true when n = k and deduce that S,, is true when n = k + 1. Step 3: Conclude that S,, is true for all n by the principle of mathematical induction. EXAMPLE 1 If a and b are real numbers, prove that (ab)" = 51%" for every positive
integer n. SOLUTION Let S, be the given statement. 1. S. is true because (ab)l = ab = a‘b‘. ‘
2. Assume that 5,, is true, that is, (4117)" == akb‘. Then (ab)"+' = (ab)k(ab) = akbkab
= (aka) :2 ak+1bk+l This says that Skﬂ is true. I ~
3. Therefore, by the principle of mathematical induction, S,l 1s true for all n; that is, (611))" = a"b” for every positive integer n. E EXAMPLE 2 Prove that, for every positive integer n, n(n+ l).
1+2+3+~+n=—i—~
SOLUTION Let S,, be the given statement.
1. S, is true because
1_ 10 +1)
_ 2
2. Assume that Sk is true, that is
k(k +1)
1+ 2 + + k = —————
2
Then 1+2+~+(k+1)=(1+2+~+k)+(k+l)
=———~—~—k(k+1) +k+l W
2
(k + 1)(k + 2)
2 II W
2 Thus l+2+~+(k+1) which shows that S1,“ is true. ~ _
3. Therefore, S,l is true for all n by mathematical 1nduct10n,that ls, n(n + l) 1+2++(n+1)= 2 for every positive integer n. ...
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