Section 4.3 Trigonometry Extended: The Circular Functions

Transcription

1 Section. Trigonometr Etended: The Circular Functions 9 7. False. This is onl true if is an acute angle in a right triangle. (Then it is true b definition.) 8. False. The larger the angle of a triangle, the smaller its cosine. 9. sec 90 = is undefined. The answer is E. cos 90 = 0 o 70. sin = The answer is A. h =. 7. If the unknown sloe is m, then m sin =,so m = - = -csc u. The answer is D. sin u 7. For all, cos. The answer is B. 7. For angles in the first quadrant, sine values will be increasing, cosine values will be decreasing and onl tangent values can be greater than. Therefore, the first column is tangent, the second column is sine, and the third column is cosine. 7. For angles in the first quadrant, secant values will be increasing, and cosecant and cotangent values will be decreasing. We recognize that csc (0 )=. Therefore, the first column is secant, the second column is cotangent, and the third column is cosecant. 7. The distance d A from A to the mirror is cos 0 ; the distance from B to the mirror is d B =d A -. Then PB = = - L.9 m. 7. Let P be the oint at which we should aim; let Å and ı be the angles as labeled in #7. Since Å=ı, tan Å=tan ı. P should be inches to the right of C, where is chosen 0 - so that tan Å= tan ı=. Then = 0 0=(0-), so =0, which gives =8. Aim 8 in. to the right of C (or in. to the left of D). 77. One ossible roof: sin + cos = a a c b + a b c b = a c = a + b c = c c d B cos ı = d A - cos 0 = - cos 0 + b c (Pthagorean theorem: a +b =c.) = 78. Let h be the length of the altitude to base b and denote the area of the triangle b A. Then h a = sin h=a sin Since A=, we can substitute h=a sin to get bh A = ab sin. Section. Trigonometr Etended: The Circular Functions Eloration. The side oosite in the triangle has length and the hotenuse has length r. Therefore sin = o. h = r. cos = adj h = r. tan = o = adj r r. cot = ; sec = ; csc = Eloration. The -coordinates on the unit circle lie between and, and cos t is alwas an -coordinate on the unit circle.. The -coordinates on the unit circle lie between and, and sin t is alwas a -coordinate on the unit circle.. The oints corresonding to t and t on the number line are wraed to oints above and below the -ais with the same -coordinates. Therefore cos t and cos ( t) are equal.. The oints corresonding to t and t on the number line are wraed to oints above and below the -ais with eactl oosite -coordinates. Therefore sin t and sin ( t) are oosites.. Since is the distance around the unit circle, both t and t+ get wraed to the same oint.. The oints corresonding to t and t+ get wraed to oints on either end of a diameter on the unit circle. These oints are smmetric with resect to the origin and therefore have coordinates (, ) and (, ). Therefore sin t and sin (t+) are oosites, as are cos t and cos (t+). 7. B the observation in (), tan t and tan(t+) are ratios - of the form and, which are either equal to each - other or both undefined. 8. The sum is alwas of the form + for some (, ) on the unit circle. Since the equation of the unit circle is + =, the sum is alwas. 9. Answers will var. For eamle, there are similar statements that can be made about the functions cot, sec, and csc. Quick Review tan = =. cot =

2 70 Chater Trigonometric Functions 7. csc = 8. sec = 9. Using a right triangle with hotenuse and legs (oosite) and - = (adjacent), we have sin =, cos =, tan = ; csc =, sec =, cot =. 0. Using a right triangle with hotenuse 7 and legs (adjacent) and 7 - = 8 (oosite), we have sin =, cos =, tan = ; csc =, sec =, cot =. 8 Section. Eercises. The 0 angle lies on the ositive ais (0-0 =90 ), while the others are all coterminal in Quadrant II.. The - angle lies in Quadrant I -, + = while the others are all coterminal in Quadrant IV. In #, recall that the distance from the origin is r= +.. sin =, cos =, tan = ; csc =, sec =, cot =.. sin =, cos =, tan = ; csc =, sec =, cot =.. sin =, cos =, tan =; csc =, sec =, cot =.. sin =, cos =, tan = ; csc =, sec =, cot =. 7. sin =, cos =, tan = ; csc =, sec =, cot =. 8. sin =, cos =, tan = ; csc =, sec =, cot =. 9. sin =, cos =0, tan undefined; csc =, sec undefined, cot =0.. sin =, cos =, tan = ; csc =, sec =, cot =.. sin =, cos =, tan = ; csc =, sec =, cot =. For #, determine the quadrant(s) of angles with the given measures, and then use the fact that sin t is ositive when the terminal side of the angle is above the -ais (in Quadrants I and II) and cos t is ositive when the terminal side of the angle is to the right of the -ais (in quadrants I and IV). Note that since tan t= sin t>cos t, the sign of tan t can be determined from the signs of sin t and cos t: if sin t and cos t have the same sign, the answer to (c) will be + ; otherwise it will be. Thus tan t is ositive in Quadrants I and III.. These angles are in Quadrant I. (a)+(i.e., sin t 7 0). (b)+(i.e., cos t 7 0). (c)+(i.e., tan t 7 0).. These angles are in Quadrant II. (a) +. (b). (c).. These angles are in Quadrant III. (a). (b). (c) +.. These angles are in Quadrant IV. (a). (b) +. (c). For #7 0, use strategies similar to those for the revious roblem set. 7. is in Quadrant II, so cos is negative is in Quadrant III, so tan 9 is ositive rad is in Quadrant II, so cos is negative rad is in Quadrant II, so tan is negative.. A (, ); tan = = =.. B (, ); tan is in Quadrant II, = = -. so is negative. 7. C (, ); is in Quadrant III, so and are both negative. tan 7 =.. D (, ); 0º is in Quadrant IV, so is ositive while is negative. tan -0 = -. For #, recall that the reference angle is the acute angle formed b the terminal side of the angle in standard osition and the -ais.. The reference angle is 0. A right triangle with a 0 angle at the origin has the oint P(, ) as one verte, with hotenuse length r=, so cos 0 = =. r. The reference angle is 0. A right triangle with a 0 angle at the origin has the oint P(, ) as one verte, so tan 00 = =. 0. sin =0, cos =, tan =0; csc undefined, sec =, cot undefined.

3 Section. Trigonometr Etended: The Circular Functions 7 7. The reference angle is the given angle,. A right triangle with a radian angle at the origin has the oint P(, ) as one verte, with hotenuse length r=, so r sec = =. 8. The reference angle is. A right triangle with a radian angle at the origin has the oint P(, ) as one verte, r with hotenuse length r=, so csc = =. 9. The reference angle is (in fact, the given angle is coterminal with ). A right triangle with a radian angle at the origin has the oint P(,) as one verte, with hotenuse length r=, so sin = =. r 0. The reference angle is (in fact, the given angle is coterminal with ). A right triangle with a radian angle at the origin has the oint P(, ) as one verte, 7 with hotenuse length r=, so cos = =. r. The reference angle is (in fact, the given angle is coterminal with ). A right triangle with a radian angle at the origin has the oint P(, ) as one verte, - so tan = =.. The reference angle is. A right triangle with a radian angle at the origin has the oint P(, ) as one verte, so cot = =.. cos =cos = 7. cos =cos =. sin =sin = 9 7. cot =cot = 7. 0 is coterminal with 70, on the negative -ais. (a) (b) 0 (c) Undefined is coterminal with 90, on the ositive -ais. (a) (b) 0 (c) Undefined 9. 7 radians is coterminal with radians, on the negative -ais. (a) 0 (b) (c) 0 0. radians is coterminal with radians, on the negative -ais. (a) (b) 0 (c) Undefined -7. radians is coterminal with radians, on the ositive -ais. (a) (b) 0 (c) Undefined. radians is coterminal with 0 radians, on the ositive -ais. (a) 0 (b) (c) 0. Since cot 7 0, sin and cos have the same sign, so sin sin = + - cos =, and tan =. cos =. Since tan 0, sin and cos have oosite signs, so cos = - sin =, and cos u cot = =. sin u sin. cos = + - sin =, so tan = cos = and sec = =. cos. sec has the same sign as cos, and since cot 7 0, sin must also be negative. With =, = 7, and 7 r= + 7 = 8, we have sin = and 8 cos = Since cos 0 and cot 0, sin must be ositive. With =, =, and r= + =, we have sec = and csc =. 8. Since sin 7 0 and tan 0, cos must be negative. With =, =, and r= + =, we have csc = and cot =. 9. sin a =sin a b = + 9,000 b 0. tan (,,7)-tan (7,,) =tan ()-tan ()=0. cos a,, b = cos a b = 0-70,000. tan a b =tan a =undefined. b. The calculator s value of the irrational number is necessaril an aroimation. When multilied b a ver large number, the slight error of the original aroimation is magnified sufficientl to throw the trigonometric functions off.. sin t is the -coordinate of the oint on the unit circle after measuring counterclockwise t units from (, 0). This will reeat ever units (and not before), since the distance around the circle is. sin 8. Â=.9 sin sin. sin = (a) When t=0, d=0. in. (b) When t=, d=0.e 0. cos 0.8 in.

4 7 Chater Trigonometric Functions 8. When t=0, =0. (rad). When t=., =0. cos rad. 9. The difference in the elevations is 00 ft, so d=00/sin. Then: (a) d= ft. (b) d=00 ft. (c) d 9. ft. 0. Januar (t=): sin =0.. Aril (t=): sin.8. June (t=): sin =7.. October (t=0): sin December (t=): sin =7.. June and December are the same; erhas b June most eole have suits for the summer, and b December the are beginning to urchase them for net summer (or as Christmas resents, or for mid-winter vacations).. True. An angle in a triangle measures between 0 and 80. Acute angles (<90 ) determine reference triangles in Quadrant I, where the cosine is ositive, while obtuse angles (>90 ) determine reference triangles in Quadrant II, where the cosine is negative.. True. The oint determines a reference triangle in Quadrant IV, with r = = 0. Thus sin =/= /0 = 0.. If sin =0., then sin ( )+csc = sin + sin u = 0.+ =.. The answer is E. 0.. If cos =0., then cos ( +)= cos = 0.. The answer is B.. (sin t) +(cos t) = for all t. The answer is A.. sin = - - cos, because tan =(sin )/(cos )>0. So sin = - - = -. B 9 The answer is A. 7. Since sin 7 0 and tan 0, the terminal side must be in Quadrant II, so =. 8. Since cos 7 0 and sin 0, the terminal side must be in Quadrant IV, so =. 9. Since tan 0 and sin 0, the terminal side must be 7 in Quadrant IV, so =. 70. Since sin 0 and tan 7 0, the terminal side must be in Quadrant III, so =. 7. The two triangles are congruent: both have hotenuse, and the corresonding angles are congruent the smaller acute angle has measure t in both triangles, and the two acute angles in a right triangle add u to >. 7. These coordinates give the lengths of the legs of the triangles from #7, and these triangles are congruent. For eamle, the length of the horizontal leg of the triangle with verte P is given b the (absolute value of the) - coordinate of P; this must be the same as the (absolute value of the) -coordinate of Q. Q( b, a) 7. One ossible answer: Starting from the oint (a, b) on the unit circle at an angle of t, so that cos t=a then measuring a quarter of the wa around the circle (which corresonds to adding > to the angle), we end at ( b, a), so that sin t + > = a. For (a, b) in Quadrant I, this is shown in the figure above; similar illustrations can be drawn for the other quadrants. 7. One ossible answer: Starting from the oint (a, b) on the unit circle at an angle of t, so that sin t=b then measuring a quarter of the wa around the circle (which corresonds to adding > to the angle), we end at ( b, a),so that cos t + >= b= sin t. For (a, b) in Quadrant I, this is shown in the figure above; similar illustrations can be drawn for the other quadrants. 7. Starting from the oint (a, b) on the unit circle at an angle of t, so that cos t=a then measuring a quarter of the wa around the circle (which corresonds to adding > to the angle), we end at ( b, a), so that sin t + >= a. This holds true when (a, b) is in Quadrant II, just as it did for Quadrant I. P(a, b) Q( b, a) t + π t + π P(a, b) t t t (, 0) (, 0) 7. (a) Both triangles are right triangles with hotenuse, and the angles at the origin are both t (for the triangle on the left, the angle is the sulement of -t). Therefore the vertical legs are also congruent; their lengths corresond to the sines of t and -t. (b) The oints P and Q are reflections of each other across the -ais, so the are the same distance (but oosite directions) from the -ais. Alternativel, use the congruent triangles argument from art (a).

5 ` ` Section. Grahs of Sine and Cosine: Sinusoids Seven decimal laces are shown so that the slight differences can be seen. The magnitude of the relative error is less than % when œ œ 0. (aroimatel). This can be seen b etending the table to larger values sin - of, or b grahing sin 78. Let (, ) be the coordinates of the oint that corresonds to t under the wraing. Then + +(tan t) =+ a = = =(sec t). b (Note that + = because (, ) is on the unit circle.) 79. This Talor olnomial is generall a ver good aroimation for sin in fact, the relative error (see #77) is less than % for 79. œ œ (aro.). It is better for close to 0; it is slightl larger than sin when 0 and slightl smaller when sin sin sin - sin sin - sin This Talor olnomial is generall a ver good aroimation for cos in fact, the relative error (see #77) is less than % for œ œ. (aro.). It is better for close to 0; it is slightl larger than cos when Z 0. cos - + cos -(- + ) Section. Grahs of Sine and Cosine: Sinusoids Eloration. / (at the oint (0, )). / (at the oint (0, )). Both grahs cross the -ais when the -coordinate on the unit circle is 0.. (Calculator eloration). The sine function tracks the -coordinate of the oint as it moves around the unit circle. After the oint has gone comletel around the unit circle (a distance of ), the same attern of -coordinates starts over again.. Leave all the settings as the are shown at the start of the Eloration, ecet change Y T to cos(t). Quick Review.. In order: +,+,-,-. In order: +,-,-,+. In order: +,-,+,-. # 80 =. -0 # 80 = -. 0 # 80 = 7. Starting with the grah of, verticall stretch b to obtain the grah of. 8. Starting with the grah of, reflect across -ais to obtain the grah of. 9. Starting with the grah of, verticall shrink b 0. to obtain the grah of.

6 7 Chater Trigonometric Functions 0. Starting with the grah of, translate down units to obtain the grah of. Section. Eercises In #, for =a sin, the amlitude is a. If a >, there is a vertical stretch b a factor of a, and if a <, there is a vertical shrink b a factor of a. When a<0, there is also a reflection across the -ais.. Amlitude ; vertical stretch b a factor of.. Amlitude /; vertical shrink b a factor of /.. Amlitude ; vertical stretch b a factor of, reflection across the -ais.. Amlitude 7/; vertical stretch b a factor of 7/, reflection across the -ais.. Amlitude 0.7; vertical shrink b a factor of Amlitude.; vertical stretch b a factor of., reflection across the -ais. In #7, for =cos b, the eriod is / b. If b >, there is a horizontal shrink b a factor of / b, and if b <, there is a horizontal stretch b a factor of / b. When b<0, there is also a reflection across the -ais. For =a cos b, ahas the same effects as in #. 7. Period /; horizontal shrink b a factor of /. 8. Period /(/)=0; horizontal stretch b a factor of /(/)=. 9. Period /7; horizontal shrink b a factor of /7, reflection across the -ais. 0. Period /0.=; horizontal stretch b a factor of /0.=., reflection across the -ais.. Period /=; horizontal shrink b a factor of /. Also a vertical stretch b a factor of.. Period /(/)=; horizontal stretch b a factor of /(/)=/. Also a vertical shrink b a factor of /. In #, the amlitudes of the grahs for =a sin b and =a cos b are governed b a, while the eriod is governed b b, just as in #. The frequenc is /eriod.. For = sin (/), the amlitude is, the eriod is /(/)=, and the frequenc is /().. For = (/) sin, the amlitude is /, the eriod is /=, and the frequenc is /.. For = sin (/), the amlitude is, the eriod is /(/)=, and the frequenc is /(). Note: the frequenc for each grah in #7 is /(). 7. Period, amlitude= 8. Period, amlitude=. 9. Period, amlitude= 0. Period, amlitude= [, ] b [, ] [, ] b [, ]... Period, amlitude=0.. Period, amlitude= 0. [, ] b [, ]. For = cos (/), the amlitude is, the eriod is /(/)=, and the frequenc is /(). 0. [, ] b [, ]

7 Section. Grahs of Sine and Cosine: Sinusoids 7. Period, amlitude=,. Period, frequenc=/ amlitude=, frequenc = /().. Period /,. Period /, amlitude=0., amlitude=0, frequenc=/ frequenc = / Period 8, 8. Period /, amlitude=, amlitude=8, frequenc=/(8) frequenc = /() Maimum: at, ; minimum: (at 0,, ). 7 Zeros:,,,. 0. Maimum: at -, ; minimum: at,. Zeros: 0,,.. =sin has to be translated left or right b an odd multile of. One ossibilit is =sin ( + ).. =sin has to be translated right b lus an even multile of. One ossibilit is =sin (-/).. Starting from =sin, horizontall shrink b and verticall shrink b 0.. The eriod is /. Possible window: c - b c -.,, d d. Starting from =cos, horizontall shrink b and verticall stretch b.. The eriod is /. Possible window: c - b [, ]., d Period ; amlitude.; [, ] b [, ] 0. Period /; amlitude ; c - b [, ], d. Period ; amlitude ; [, ] b [, ]. Period ; amlitude ; [, ] b [ 0, 0]. Period ; amlitude ; [, ] b [, ]. Period ; amlitude ; [, ] b [, ]. Maimum: at and ; minimum: at and. Zeros: 0,,.. Maimum: (at 0); minimum: (at ). Zeros:. 7. Maimum: (at 0,, ); minimum: 7 at and. Zeros:,,,. 8. Maimum: at and ; minimum: - at - and. Zeros: 0,,. [, ] b [ 0.7, 0.7] [, ] b [ 0.7, 0.7] For # For #. Starting from =cos, horizontall stretch b, verticall shrink b, reflect across -ais. The eriod is. Possible window: [, ] b [, ].. Starting from =sin, horizontall stretch b and verticall shrink b. The eriod is 0. Possible window: [ 0, 0] b [, ]. [, ] b [, ] [ 0, 0] b [, ] For # For # 7. Starting from =cos, horizontall shrink b and verticall stretch b. The eriod is. Possible window: [, ] b [.,.].

8 7 Chater Trigonometric Functions 8. Starting from =sin, horizontall stretch b, verticall stretch b, and reflect across -ais. The eriod is 8. Possible window: [ 8, 8] b [, ]. [, ] b [.,.] [ 8, 8] b [, ] For #7 For #8 9. Starting with, verticall stretch b. 0. Starting with, translate right units and verticall shrink b.. Starting with, horizontall shrink b.. Starting with, horizontall stretch b and verticall shrink b. For #, grah the functions or use facts about sine and cosine learned to this oint.. (a) and (b). (a) and (b). (a) and (b) both functions equal cos. (a) and (c) sin a + b =sin ca - =cos a - b + d b In #7 0, for =a sin (b(-h)), the amlitude is a, the eriod is / b, and the hase shift is h. 7. One ossibilit is = sin. 8. One ossibilit is = sin (/). 9. One ossibilit is =. sin (-). 0. One ossibilit is =. sin (-). Amlitude, eriod, hase shift, vertical translation unit u.. Rewrite as =. sin c a -. bd - Amlitude., eriod, hase shift, vertical translation unit down.. Rewrite as = cos c a - bd Amlitude, eriod, hase shift, vertical 8 translation units u.. Amlitude, eriod, hase shift, vertical translation units down.. Amlitude, eriod, hase shift 0, vertical translation unit u.. Amlitude, eriod, hase shift 0, vertical translation units down Amlitude, eriod, hase shift -, vertical translation unit down. 8. Amlitude, eriod 8, hase shift, vertical translation unit u. 9. = sin (a=, b=, h=0, k=0). 70. = sin[(+0.)] (a=, b=, h=0., k=0). 7. (a) There are two oints of intersection in that interval. (b) The coordinates are (0, ) and (,. ) (.8, 0.9). In general, two functions intersect where cos =, i.e., =n, n an integer. 7. a= and b =.. = 7 7. The height of the rider is modeled b h=0- cos a, where t=0 corresonds 0 t b to the time when the rider is at the low oint. h=0 - when =cos a. Then, so t 0 t b 0 t L sec. 7. The length L must be the distance traveled in 0 min b an object traveling at 0 ft/sec: L=800 sec # ft 0 97,000 ft, or about 8 miles. sec = 7. (a) A model of the deth of the tide is d= cos c t - 7. d + 9, where t is hours since. midnight. The first low tide is at :00 A.M. (t=). (b) At :00 A.M. (t=): about 8.90 ft. At 9:00 P.M. (t=): about 0. ft. (c) :0 A.M. (t=. halfwa between :00 A.M. and 7: A.M.). 7. (a) second. (b) Each eak corresonds to a heartbeat there are 0 er minute. (c) [0, 0] b [80, 0]

9 Section. Grahs of Sine and Cosine: Sinusoids (a) The maimum d is aroimatel.. The amlitude is (.-7.)/=7.. Scatterlot: [0, ] b [0, 80] [0,.] b [7, ] (b) The eriod aears to be slightl greater than 0.8, sa 0.8. (c) Since the function has a minimum at t=0, we use an inverted cosine model: d(t)= 7. cos (t/0.8)+.. (d) [0,.] b [7, ] 78. (a) The amlitude is.7, half the diameter of the turntable. (b) The eriod is.8, as can be seen b measuring from minimum to minimum. (c) Since the function has a minimum at t=0, we use an inverted cosine model: d(t)=.7 cos (t/.8)+7.7. (d) [0,.] b [9, 8] 79. One ossible answer is T =. cos a - 7 b Start with the general form sinusoidal function = a cosb - h + k, and find the variables a, b, h, and k as follows: 79 - The amlitude is 0 a 0 = =.. We can arbitraril choose to use the ositive value, so a=.. The eriod is months. = 0 b 0 =. 0 b 0 = Again, we can arbitraril choose to use the ositive value, so b =. The maimum is at month 7, so the hase shift h= The vertical shift k = = One ossible answer is = cos a - 7 b +. Start with the general form sinusoidal function = a cosb - h + k, and find the variables a, b, h, and k as follows: 8-0 The amlitude is 0 a 0 = =. We can arbitraril choose to use the ositive value, so a=. The eriod is months. = 0 b 0 =. 0 b 0 = Again, we can arbitraril choose to use the ositive value, so b =. The maimum is at month 7, so the hase shift h= The vertical shift k = =. [0, ] b [0, 80] 8. False. Since =sin is a horizontal stretch of =sin b a factor of, =sin has twice the eriod, not half. Remember, the eriod of =sin b is / b. 8. True. An cosine curve can be converted to a sine curve of the same amlitude and frequenc b a hase shift, which can be accomlished b an aroriate choice of C (a multile of /). 8. The minimum and maimum values differ b twice the amlitude. The answer is D. 8. Because the grah asses through (, 0), f()=0. And lus eactl two eriods equals 9, so f(9)=0 also. But f(0) deends on hase and amlitude, which are unknown. The answer is D. 8. For f()=a sin (b+c), the eriod is / b, which here equals /0=/0. The answer is C. 8. There are solutions er ccle, and 000 ccles in the interval. The answer is C. 87. (a) [, ] b [.,.]

10 78 Chater Trigonometric Functions (b) cos The coefficients given as 0 here ma show u as ver small numbers (e.g.,.*0 ) on some calculators. Note that cos is an even function, and onl the even owers of have nonzero (or a least non-small ) coefficients. (c) The Talor olnomial is - ; the + = coefficients are fairl similar. 88. (a) For #9 9, note that A and C are one eriod aart. Meanwhile, B is located one-fourth of a eriod to the right of A, and the -coordinate of B is the amlitude of the sinusoid. 9. The eriod of this function is and the amlitude is. B and C are located (resectivel) units and units to the right of A. Therefore, B=(0, ) and C= a., 0 b 9. The eriod of this function is and the amlitude is.. B and C are located (resectivel) units and units to the right of A. Therefore, B= a and,. b C = a 9, 0 b [, ] b [.,.] (b) sin The coefficients given as 0 here ma show u as ver small numbers (e.g.,.*0 ) on some calculators. Note that sin is an odd function, and onl the odd owers of have nonzero (or a least non-small ) coefficients. (c) The Talor olnomial is - ; = the coefficients are somewhat similar. 89. (a) = sec = (b) f = ( ccles er sec ), or Hertz (Hz). sec (c) [0, 0.0] b [, ] 90. Since the cursor moves at a constant rate, its distance from the center must be made u of linear ieces as shown (the sloe of the line is the rate of motion). A grah of a sinusoid is included for comarison. [0, ] b [.,.] 9. (a) a-b must equal. (b) a-b must equal. (c) a-b must equal k. 9. (a) a-b must equal. (b) a-b must equal. (c) a-b must equal k. 9. The eriod of this function is and the amlitude is. B and C are located resectivel units and units to the right of A. Therefore, B= a and C= a, b, 0 b 9. The first coordinate of A is the smallest ositive such n + that -=n, n and integer, so = must equal. The eriod of this function is and the amlitude is. B and C are located (resectivel) units and units to the right of A. Therefore, A= a,, 0 b B = a., b and C = a, 0 b 97. (a) Since sin ( )= sin (because sine is an odd function) a sin [ B(-h)]+k= a sin[b(-h)]+k. Then an eression with a negative value of b can be rewritten as an eression of the same general form but with a ositive coefficient in lace of b. (b) A sine grah can be translated a quarter of a eriod to the left to become a cosine grah of the same sinusoid. Thus = a sin c b a - h + # b bd + k = a sin c b a - a h - has the same b bbd + k grah as = a cosb - h + k. We therefore choose H = h -. b (c) The angles + and determine diametricall oosite oints on the unit circle, so the have oint smmetr with resect to the origin. The -coordinates are therefore oosites, so sin( +)= sin. (d) B the identit in (c). = a sinb - h + + k = -a sinb - h + k. We therefore choose H = h -. b (e) Part (b) shows how to convert = a cosb - h + k to = a sinb - H + k, and arts (a) and (d) show how to ensure that a and b are ositive.

11 Section. Grahs of Tangent, Cotangent, Secant, and Cosecant Section. Grahs of Tangent, Cotangent, Secant, and Cosecant 79 Eloration. The grahs do not seem to intersect.. Set the eressions equal and solve for : k cos =sec k cos =/cos k(cos ) = (cos ) = /k Since k 7 0, this requires that the square of cos be negative, which is imossible. This roves that there is no value of for which the two functions are equal, so the grahs do not intersect. Quick Review.. Period. Period. Period. Period For # 8, recall that zeros of rational functions are zeros of the numerator, and vertical asmtotes are found at zeros of the denominator (rovided the numerator and denominator have no common zeros).. Zero:. Asmtote: =. Zero:. Asmtote: = 7. Zero:. Asmtotes: = and = 8. Zero:. Asmtotes: =0 and = For #9 0, eamine grahs to suggest the answer. Confirm b checking f( )=f() for even functions and f( )= f() for odd functions. 9. Even: ( ) += + 0. Odd: = Section. Eercises. The grah of = csc must be verticall stretched b comared to =csc, so = csc and =csc.. The grah of = tan must be verticall stretched b 0 comared to =0. tan, so = tan and =0. tan.. The grah of = csc must be verticall stretched b and horizontall shrunk b comared to =csc,so = csc and =csc.. The grah of =cot(-0.)+ must be translated units u and 0. units right comared to =cot,so =cot(-0.)+ and =cot.. The grah of =tan results from shrinking the grah of = tan horizontall b a factor of. There are vertical asmtotes at =..., -....,,, [, ] b [, ]. The grah of = cot results from shrinking the grah of =cot horizontall b a factor of and reflecting it across the -ais. There are vertical asmtotes at =....,-...., -, 0,, b [, ], 7. The grah of =sec results from shrinking the grah of =sec horizontall b a factor of. There are vertical asmtotes at odd multiles of., b [, ] 8. The grah of =csc results from shrinking the grah of =csc horizontall b a factor of. There are vertical asmtotes at =....,-, -...., 0,, [, ] b [, ]

12 80 Chater Trigonometric Functions 9. The grah of = cot results from shrinking the grah of =cot horizontall b a factor of and stretching it verticall b a factor of. There are vertical asmtotes at =....,-, -...., 0,, [, ] b [, ] 0. The grah of = tan results from stretching the grah of =tan horizontall b a factor of and stretching it verticall b a factor of. There are vertical asmtotes at =....,-,,,.... [, ] b [, ]. The grah of =csc results from horizontall stretching the grah of =csc b a factor of. There are vertical asmtotes at =....,-, -, 0,,.... [, ] b [, ]. The grah of = sec results from horizontall shrinking the grah of =sec b a factor of and stretching it verticall b a factor of. There are vertical asmtotes at odd multiles of. 8 [, ] b [, ]. Grah (a); Xmin= and Xma=. Grah (d); Xmin= and Xma=. Grah (c); Xmin= and Xma=. Grah (b); Xmin= and Xma= 7. Domain: All reals ecet integer multiles of Range: ( q, q) Continuous on its domain Decreasing on each interval in its domain Smmetric with resect to the origin (odd) Not bounded above or below No local etrema No horizontal asmtotes Vertical asmtotes = k for all integers k End behavior: lim cot and lim cot do not eist. Sq S-q 8. Domain: All reals ecet odd multiles of Range: ( q, ] [, q) Continuous on its domain On each interval centered at an even multile of : decreasing on the left half of the interval and increasing on the right half On each interval centered at an odd multile of : increasing on the left half of the interval and decreasing on the right half Smmetric with resect to the -ais (even) Not bounded above or below Local minimum at each even multile of, local maimum at each odd multile of No horizontal asmtotes Vertical asmtotes = k/ for all odd integers k End behavior: lim sec and lim sec do not eist. Sq S-q 9. Domain: All reals ecet integer multiles of Range: ( q, ] [, q) Continuous on its domain On each interval centered at = (k an integer): + k decreasing on the left half of the interval and increasing on the right half On each interval centered at : increasing on the + k left half of the interval and decreasing on the right half Smmetric with resect to the origin (odd) Not bounded above or below Local minimum at each =, local maimum + k at each =, where k is an even integer in + k both cases No horizontal asmtotes Vertical asmtotes: = k for all integers k End behavior: lim csc and lim csc do not eist. Sq S-q 0. Domain: All reals ecet odd multiles of Range: ( q, q) Continuous on its domain Increasing on each interval in its domain Smmetric with resect to the origin (odd) Not bounded above or below No local etrema No horizontal asmtotes Vertical asmtotes =k for all odd integers k End behavior: lim tan> and lim tan> Sq S-q do not eist.

13 Section. Grahs of Tangent, Cotangent, Secant, and Cosecant 8. Starting with =tan, verticall stretch b.. Starting with =tan, reflect across -ais.. Starting with =csc, verticall stretch b.. Starting with =tan, verticall stretch b.. Starting with =cot, horizontall stretch b, verticall stretch b, and reflect across -ais.. Starting with =sec, horizontall stretch b, verticall stretch b, and reflect across -ais. 7. Starting with =tan, horizontall shrink b and reflect across -ais and shift u b units. 8. Starting with =tan, horizontall shrink b and verticall stretch b and shift down b units. 9. sec = 0. csc =.. cos = = sin = = cot = - tan = - = sec = - cos = - =. csc = sin = =. cot = tan = = -. tan =. L 0.9. sec =. cos =. L. 7. cot = 0. tan = csc =. sin = -. -( 0.7) csc = sin = 0. or tan = or (a) One elanation: If O is the origin, the right triangles with hotenuses OP and OP, and one leg (each) on the -ais, are congruent, so the legs have the same lengths. These lengths give the magnitudes of the coordinates of P and P ; therefore, these coordinates differ onl in sign. Another elanation: The reflection of oint (a, b) across the origin is ( a, b). sin t (b) tan t=. cos t = b a sint - -b (c) tan(t-)= =. -a = b cost - a = tan t (d) Since oints on oosite sides of the unit circle determine the same tangent ratio, tan(t_)=tan t for all numbers t in the domain. Other oints on the unit circle ield triangles with different tangent ratios, so no smaller eriod is ossible. (e) The tangent function reeats ever units; therefore, so does its recirocal, the cotangent (see also #).. The terminal side asses through (0, 0) and (cos, sin ); sin - 0 sin the sloe is therefore m= = =tan. cos - 0 cos. For an, a (+)= = = a (). f b f + f f b This is not true for an smaller value of, since this is the smallest value that works for f.. (a), (b) The angles t and t+ determine oints (cos t, sin t) and (cos(t+), sin(t+)), resectivel. These oints are on oosite sides of the unit circle, so the are reflections of each other about the origin. The reflection of an oint (a, b) about the origin is ( a, b), so cos(t+)= cos t and sin(t+)= sin t. sint + -sin t sin t (c) tan(t+)= = = =tan t. cost + -cos t cos t In order to determine that the eriod of tan t is,we would need to show that no satisfies tan(t+)= tan t for all t. 0. (a) d=0 sec = ft cos (b) d,8 ft

14 8 Chater Trigonometric Functions 800. (a) =800 cot = ft tan (b),0 ft (c) # 80 0 = 9 For #7 0, the equations can be rewritten (as shown), but generall are easiest to solve grahicall. 7. sin =cos ; ; cos =sin ; 0. or.7 9. cos = ; ;.07 or ;.0 0. cos =sin ;.08 or.00. False. f()=tan is increasing onl over intervals on which it is defined, that is, intervals bounded b consecutive asmtotes.. True. Asmtotes of the secant function, sec =/cos, occur at all odd multiles of > (where cos =0), and these are eactl the zeros of the cotangent function, cot =cos /sin.. The cotangent curves are shaed like the tangent curves, but the are mirror images. The reflection of tan in the -ais is tan. The answer is A.. sec just barel intersects its inverse, cos, and when cos is shifted to roduce sin, that curve and the curve of sec do not intersect at all. The answer is E.. =k/sin and the range of sin is [, ]. The answer is D.. =csc =/sin has the same asmtotes as =cos /sin =cot. The answer is C. 7. On the interval [, ], f 7 g on about ( 0., 0) (0., ). 0. The look similar on this window, but the are noticeabl different at the edges (near 0 and ). Also, if f were equal to g, then it would follow that = cos = =on this interval, which we know to be f g false.. csc =sec a - (or csc =sec a - a + n bb b for an integer n) This is a translation to the right of [, ] b [ 0, 0] [0, ] b [ 0, 0] a or units. + n b. cot = tan a - (or cot = b tan - a for an integer n). + n b This is a translation to the right of a or units, + n b and a reflection in the -ais, in either order. 0. d=0 sec = cos [, ] b [ 0, 0] 8. On the interval [, ], f 7 g on about (,.) a - a., 0 b,. b [, ] b [ 0, 0] 9. cot is not defined at 0; the definition of increasing on (a, b) requires that the function be defined everwhere in (a, b). Also, choosing a= > and b=>, we have a b but f(a)= 7 f(b)=. [ 0., 0.] b [0, 00]. (a) For an acute angle, cos =sin a the sine - b of the comlement of. This can be seen from the right-triangle definition of sine and cosine: if one of the acute angles is, then the other acute angle is, since all three angles in a triangle must add - to. The side oosite the angle is the side adjacent to the other acute angle. (b) (cos t, sin t)