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Elementary Functions

by: August Feeney

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Elementary Functions MTH 112

Marketplace > Portland Community College > Math > MTH 112 > Elementary Functions
August Feeney
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This 60 page Class Notes was uploaded by August Feeney on Monday October 19, 2015. The Class Notes belongs to MTH 112 at Portland Community College taught by Staff in Fall. Since its upload, it has received 21 views. For similar materials see /class/224645/mth-112-portland-community-college in Math at Portland Community College.

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Date Created: 10/19/15
Haberman MTH112 Section 1 Periodic Functions and Trigonometry u IrI39d m wltro4 nawrgt 0 39vl1P gtlt a Module 4 Part 1 Finding Sine and Cosine Values As we studied in the previous module the equal angles in circles of different radii induce similartriangles see Figure 1 fL T my lt12 2 vquotf Figure 1 The angle 6 in both a unit circle and in a circle of radius r inducing similar right triangles We can use these similartriangles to obtain the following ratios 0056 1 and 51n16 1 r 1 r Solving these ratios for 0056 and 5in6 respectively gives us the following 0056 i and 5in6 l r r A A x ADJ Figure 2 We use the terms opposite er OPP adjacent er ADJ and hypotenuse er HYP m refer m the srues er a ght mang e ngrmmou w a s the ang e gwen m the thtt ang es m Frgure 2 abuve then x AD y OFF d 5 cos 39 an sm 39 H HYF e n s m se rams a ung vvuh the Pythagurean Theurem see be uvv m Earn a great dea abuut nghtmang es THE PYTHAGDREAN THEOREM The swdEJEngths er the rrgm mang e gwen m Frgure 3 sausfy the Equatmn a2 12 2 a Figure EXAMPLE 1 Find the sine and cosine of the angle a given in the right triangle in Fig 4 The triangle might not be drawn to scale 12 Figure4 SOLUTION First we need to use the Pythagorean Theorem to find the length of the hypotenuse c 122 92 c2 2 144 81 c2 We can use this value to label our triangle 615 12 Figures Thus ADJ12i cosa HYP B 5 EXAMPLE 2 Find the sine and cosine of the angle given in the right triangle in Fig 6 The triangle might not be drawn to scale 4 13 5 I a Figure6 SOLUTION First we need to use the Pythagorean Theorem to find the length of the side labeled a a2 52 132 2 a2 25 169 2 a2 144 3 a12 We can use this value to label our triangle 2 12 Figure7 To determine the sine and cosine values of angle imagine standing at angle and looking into the triangle Then AD 5 cos HYP 13 and OPP 12 SW Z EXAMPLE 3 Find the sine and cosine of the angles a and given in the right triangle in Figure 8 The triangle might not be drawn to scale A I 4 Figure8 SOLUTION First we need to use the Pythagorean Theorem to find the length of the side labeled c 42 82 c2 2 16 64 c2 3 c2 80 3 c4 Now we can use the triangle in Figure 8 and the fact that c 43 to find the values of sine and cosine at a and in the triangle in Figure 8 comz zizL comz zizi HYP 4J3 J3 HYP 4J3 J3 ii iL sma H 4 5 J sm H 4 5 Jg Now let s use right triangle trigonometry to determine the sine and cosine of some important angles 30 45 and 60 ie and WE H start Wth 30 and 617quot tu mm the SWE ahu EDSWE at these ang ES 1et s euhstuer ah euut1atera1 thahge t e a thahg1e Wth three stues er euua1 1ehgth As yuu rhay haye stuuteu preytuus1y h a geumetry uursey euut1atera1 thahg1es are a1su Eqmangu ar t e they E ntam three euua1 ahg1es shee the sum er the ahg1erheasures W ahy thahg1e ts 1802 ah euut1atera1 thahg1e has three ahg1es mEaSang 60 1h thure 9 be uvv We ve urayyh ah euut1atera1thahg1e whuse stues are eaeh1uhtt1uhg 1 60 60quot 1 1 60 Figure 9 Ah euut1atera1thahg1e shee thts euut1atera1thahg1e tsh t a hght thahge We eah t use the dE annS tur SWE and Men abuve But We can euhstmm a rtght trtahg1e by WSEmHg a hhe segment that htseets uhe ut the ahg1es t e sphts mm Wu euua1ahg1esahu euhheets thts ahg1e thh the uppuste stue After btseetthg the 60 we u haye twu 30 ahg1es and men spht the 1 WM SwdEJEngth mm twu 1 umt Segments see gure 10 Figure 10 Nuttee that We ve ereateu Wu tuehttea1 hghtthahg1es that E ntam ahg1es 30 and 60 We eah use ett r at these thahg1es tu e ErmmE the SWE ahu EDSWE ya1ues at 30 and 60 Let sfucus uh the upper thahg1e 1h gure 10 We ve uraWh thts thahg1e W gure 11 1 60 L 2 30 b Figure A 30 760 790 thahg1e the eah use the Pythagurean Theurem m nd the ength uf the segment 11 Hip WE can use abe ed b h ngure NuWet s use MS vame m eump ete the abehng uf uur 30 7 60 790 mange see Hg 12 1 so i 2 30 2 Figure 12 A 30260290 mang e We can use the ange m F gure 12 In determme the sme and eusme Elf 30 and 60 J 2i mm A1212 2 HYF 1 2 52 39 1 2 eos30 1 a OFF 60 Sm HYF a OFF 2 1 sm30 1T7 ReeaH the de hmuh unhe sme and eusme fuhetmhs The pmnt P spem ed bythe ang e a unthe umtmm e has eunrdmates x y ease my see gure 13 Figure 13 Thus we now know the coordinates of the points on the circumference of the unit circle specified by the angles 30 and 60 see Figure 14 y if Figure 14 Now let s find the sine and cosine values of a 45 angle Since we ll need a right triangle we need to construct a triangle with a 45 angle and a 90 angle Since the sum of the measures of the angles in a triangle is always 180 the third angle must also be 45 Thus we need to construct a 45 45 90 Since two of the angles are equal this will be an isosceles triangle ie the two sides opposite the two 45 angles have equal length Let s choose this length to be 1 unit Figure 15 We can use Pythagoras to find the length of the segment labeled c in Figure 15 12 12 c2 2 1 1 c2 3 c22 2 cJ Now we can use the triangle in Figure 15 and the fact that c J5 to find the values of sine and cosine at 45 cos45 i and sin45 i HYP J5 HYP J5 Although there is nothing wrong with these expressions in the old daysquot ie the pre calculator era people didn t like to have radicals in the denominator of fractions since the tables they used to help them approximate values didn t contain approximations for expressions with radicals in the denominator As a result there is a procedure known as rationalizing the denominatorquot which allows us to get rid of the radicals in the denominator You should have studied this in an Intermediate Algebra course When you study trigonometry you often see the rationalized form of the number so it s worth taking note of the ratIonalIzed form of J5 Since we know the values of cos45 and sin45 we know the coordinates of the points on the circumference of the unit circle specified by the angle 45 see Figure 16 below gag X Figure 16 Let s put these values for sine and cosine in a table It is important that you learn these values Below we39ll discuss a strategy for remembering them 9 degrees 30 45 60 9 radians 0056 g g sint9 g g EXAMPLE 10 Find the values of A b and c on the right triangle in Figure 17 The triangle might not be drawn to scale Figure 17 SOLUTION First notice that angle A must measure 30 since the sum of the angles in a triangle is 180 and our triangle already has angles measuring 60 and 90 The only thing we know about the sides of the triangle is that the side adjacent to the 60 angle is 7 units long Notice that the cosine of the 60 angle is and we can use this fact to find c cos60 C hlq q1 3 c cos60 L 39 L 3 c 2 7 Since cos60 2 7 3 0 7 2 3 c14 Now that we know the length of two sides of the triangle we could use the Pythagorean Theorem to find the length of the third side b lnstead we ll use the fact that on the triangle the sine of 60 is o L sm60 C 3 sin60 since 6 14 3 14 1 since sin60 3 b 14 3 b7J Let s summarize ourfindings A 30 b 7J5 and c 14 MTH 112 Supplemental Problem Sets SUPPLEMENT TO 63 Degrees Minutes and Seconds When measuring angles in degrees fractions of a degree are often represented in minutes and seconds DEFINITION One minute is h of a degree so 60 minutes equals one degree written 6039 1 L L 60 3600 equals one minute written 60quot139 and 3600 seconds equals one degree written 3600quot 1 One second is h of a minute ie h of a degree so 60 seconds When we write an angle39s measure in the form D M39Squot where D M and S are real numbers then the angle39s measure is D degrees plus M minutes plus S secondsquot EXAMPLE 1 Convert 34 153927quot into decimal form 0 1 u o 1 L N 1 3415 27 34 15 60 27 3600 34 025 00075 342575 EXAMPLE 2 Convert 6172 into D M39Squot form 6172 61 072 61 072 61 43239 61 4339 0239 61 4339 0239 61 4312quot Portland Community College page 1 Revised 012709 MTH 112 Supplemental Problem Sets EXERCISES 1 Convert the angle measure to decimal form round your answers to the nearest thousandth a 243 1039 b 3 2539 c 23 339 d 75 323917quot 2 Convertthe angle measure to D M39Squot form a 124 b 153 c 1449 d 0416 SUPPLEMENT TO 64 EXERCISES 1 Find the value of 6 in degrees 0 lt 6 lt 90 and radians 0 lt 6 lt without a calculator a sin6 l b 0056 E 0 0056 Q d sin6 2 2 2 2 2 Based on the given information determine the quadrant in which 6 lies a sin6 lt 0 and 0056 lt 0 b sin6 gt 0 and 0056 lt 0 c sin6 gt 0 and 0056 gt 0 d sin6 lt 0 and 0056 gt 0 Portland Community College page 2 Revised 012709 MTH 112 Supplemental Problem Sets SUPPLEMENT TO 66 EXERCISES 1 Based on the given information determine the quadrant in which 6 lies a se06 gt 0 and 00t6 lt 0 b se06 lt 0 and tan6 lt 0 c 0506 lt 0 and tan6 gt 0 d se06 gt 0 and 0506 gt 0 2 Find the value of 6 in degrees 0 lt 6 lt 90 and radians 0 lt 6 lt without a calculator a tan61 b tan 9J c se062 d 0506 3 Find the sine cosine and tangent of the given angles without using a calculator a 225 b 225 0 750 d 510 SUPPLEMENT TO 67 EXERCISES 1 Sketch 6 in standard position and find the reference angle 61 a 92o3 b 9127quot c 9 245 d 9 72 e 1927 f 197 9 635 h 658 Portland Community College page 3 Revised 012709 MTH 112 Supplemental Problem Sets 2 Use a calculator to approximate two values of 6 0 S 6 lt 360 that satisfy the given equation Round your answers to the nearest hundredth a sin608191 b sin6 02589 3 Use a calculator to approximate two values of 6 0 S 6 lt 2n that satisfy the given equation Round your answers to the nearest thousandth a cos6 09848 b cos6 05890 c tan6ll92 d tan6 8144 4 Find all of the solutions of the following equations on the interval zr Leave your solutions in exact form a 2cosx1 0 b Jicscoc 2 0 c ZSin2x 1 d 3seczx 4 0 5 Find all of the solutions of the following equations on the interval 0 2n Leave your solutions in exact form a tanxtanx 1 0 b sinxsinx 1 0 c secxcscx 2 cscx 0 d 251nx cscx 0 6 Find all of the solutions of the following polynomial and trigonometric equations Use your solution of the polynomial equation to help you solve the trigonometric equation a ZyZ y 10 b 2y23y10 ZSin2x sinx 1 0 Zeoszx 3cosx 1 0 c y2 1 0 tan2x 1 0 Portland Community College page 4 Revised 012709 MTH ll2 Supplemental Problem Sets 7 Find all of the solutions of the following equations on the interval 0 2n Leave your solutions in exact form See the Example below I a meek g b cos EXAMPLE Solve cos46 on the interval 0 2n i cos46 2 2 49ZT 2m or 494T 2kn keZ M M 47r 216 26 124 or 6124 1 k7r 1 16 2 6 6 2 or 6 3 2 For k 01 2 3 both expressions for 6 result in values in the interval 0 2n but all other values for k result in numbers outside the desired interval Thus the solution set is ZT 5 7 4 57 M 8 The horizontal distance d traveled by a projectile fired with an initial velocity v0 at an 2 elevation angle 6 is given by d V sin26 where g d represents the horizontal distance feet v0 represents the initial velocity feet per second g represents the acceleration due to gravity 32 feet per second per second 6 represents the elevation angle a A fly ball leaves a baseball bat at a velocity of 96 miles per hour and is caught by an outfielder in center field 285 feet away What was the angle of elevation when the ball left the bat Your solution should be accurate to the nearest second b Janeen s rifle has a muzzle velocity of 315 ftsec What angle of elevation should she use in order to hit a small target that is onehalf mile away Your solution should be accurate to nearest tenth of a degree c At what angle will a projectile travel the farthest horizontal distance Explain how you can determine this simply by inspecting the equation and using your knowledge of the range of the sine function Portland Community College page 5 Revised 012709 MTH 112 Supplemental Problem Sets SUPPLEMENT TO 72 Proving Trigonometric Identities There is an important distinction between an equation and an identity 0 An equation is something that we may wish to try and solve for an unknown variable for example 2x 1 3 is an equation An equation will only be true for certain values of the independent variable or perhaps it will not have any solutions at all An identity is an equation that is true for all values of the independent variable for example coszx sin2x 1 is an identity Sometimes we want to prove that a given equation is an identity To do this we start with one side of the equation and must manipulate it until we obtain the other side of the equation When proving an identity we do not solve the equation for the independent variable We demonstrate this in the following example tanx secx EXAMPLE 1 Prove the following identity sinx Depending on the identity you must decide on which side of the equation you wish to start your proof It39s usually best to start with the more complicated side in this case the lefthand side and manipulate it until it equals the more simple side in this case the righthand side So to prove the given identity we will begin with the lefthand side and perform a series of algebraic manipulations and substitutions until we obtain the righthand side sinx tanx cosx f secx m sinx cosx cosx 1 sinx tanx secx side of the identity is equivalent to the right side sinx since we ve shown that the left This is a proof of the identity Portland Community College page 5 Revised 012709 MTH 112 Supplemental Problem Sets EXAMPLE 2 Prove the following identity 5e090509 tan9 00t9 Sometimes it39s not obvious which side of the equation will be better to begin your proof with If you start with one side and hit a deadend don39t hesitate to start over with the other side It39s often a good idea to work on your proof on scratch paper before composing your final draft Sometimes working both sides of the identity will help you figure out how to prove the identity Let39s try to start our proof with the lefthand side of the given identity 5e09 0509 00819 1 00595in9 We seem to be stuck since there aren39t any obvious manipulations we can do to this expression but we still haven39t reached our goal so let39s start over by manipulating the righthand side of the equation and see if we have better luck sin 9 0059 0059 sin 9 sin9 5i119 00590059 0059 sin9 sin9 0059 tan9 00t9 5in29 00529 00595in9 1 now we can use what we learned above 00595in9 when we started with the lefthand side 4 0059 sin9 5e09 0509 This is a proof of the identity 5e090509 tan9 00t9 since we ve shown that the left side of the identity is equivalent to the right side EXERCISES 1 Prove the following identities a tanx050x 5inx 005x b 5e0x tanx 00tx EC 005 x tanw 00mg 5in29 00529 1 25e02t tan9 00t9 1 sint 1 sint Portland Community College page 7 Revised 012709 MTH 112 Supplemental Problem Sets 2 Prove the following identities a cosmcos71 cosm 71 cosm 71 b sinmsin71 cosm 71 cosm 71 c sinm cos71 sinm 71 sinm 71 SUPPLEMENT TO 75 EXERCISES 1 Converteach ofthefollowing points given in polar quot to g39 39 l l a 2 6 b 3 3 Convert each of the following points given in rectangular coordinates to polar coordinates Give two different polar representations of each point a J5 3 b 2 2 N c 2 4 approximatetothe nearesttenth Portland Community College Page 8 Revised 012709 MTH 112 Supplemental Problem Sets SUPPLEMENT TO 76 Roots of Complex Numbers 1 EXAMPLE 1 Find 1 145 using the polarform of 1 1J5 SOLUTION We can associate the number 1 N with the point 1 5 on the coordinate plane and then translate this point into polar coordinates r 9 6 t 7105 we add 7239 since we know that the point is in the 2nd quadrant Therefore we can represent the point 1 5 in polar coordinates as 2 27 2 139 Thus in polarform 1i 2e 3 Therefore 5 mg mug 3 Portland Community College Page 9 Revised 012709 MTH 112 Supplemental Problem Sets 1 EXAMPLE 2 Find 165 162iA using the polarform of 16J 16J i SOLUTION First let39s find the polar form of 16J 16J i ie let39s write 16J 16J2 139 in the form rem Recall that rem rcost9 rsint9i By comparing real and imaginary parts we see that rcosi9 16J2 and rsint9 16J2 which implies that 0056 16 5 r and sint9 Now we can use Pythagoras to find r 1 sin2t9 00526 2 2 3 iM M r r 512 512 3 1 r2 r2 3 121034 r 2 r21024 3 r 32 we reject the negative value since r gt 0 We can substitute this for r in our expressions for sini9 and 0056 0059 M Smog M 32 and 32 2 2 By consulting the unit circle we can see that 9 Thus the polar form of 5r39 1 16J 16J i is 32a 4 We can nowfind 16J2 165139A Portland Community College Page 10 Revised 012709 MTH 112 Supplemental Problem Sets 16J5 16 i 326 g e T i zz i l l 2cos4z s1n4 Q Q 2 2 1 2 J5iJ EXERCISE 1 Find the following by first converting to the polar form of complex number a 18 18J3i b 4J5 51 C J cl 64J 64i SUPPLEMENT TO 102 EXERCISES 1 A cable exerts a force of 350 N at an angle of 382 with the horizontal Resolve this force into vertical and horizontal components 350 N 382 2 A certain escalator travels at a rate of 106 meters per minute and its angle of inclination is 325 What is the vertical component of the velocity How long will it take a passenger to travel 10 meters vertically Portland Community College Page 11 Revised 012709 MTH 112 Supplemental Problem Sets 3 A projectile is launched at an angle of 5560 to the horizontal with a speed of 7550 meters per minute Find the vertical and horizontal components of the projectiles velocity SUPPLEMENT TO 103 EXERCISES 1 Two ropes are holding a crate One is pulling with a force of 994 N in a direction of 1550 with the vertical and the other with a force on 624 N in a direction of 242 with the vertical How much does the crate weigh 624 N 994 N 2 A plane is headed due west with an air speed of 212 kmhr It is driven from its course by a wind from due north blowing at 236 kmhr Find the ground speed of the plane and the actual direction of travel V nd 236 kmhr l l l 212 kmhr North I I M r Actual direction of travel 3 A pilot heading her plane due north finds the actual direction of travel to be N5 1239E The plane39s air speed is 315 mihr and the wind is from due west Find the plane39s ground speed and the speed of the wind Portland Community College Page 12 Revised 012709 MTH 112 Supplemental Problem Sets 4 At what speed with respect to the water should a ship head due north in order to follow a course N 5 1539 E if a current is flowing due east at the rate of 106 mph 5 As an airplane heads west with an air speed of 325 mph a wind with a speed of 35 mph causes the plane to travel slightly south of west with a ground speed of 305 mph In what direction is the wind blowing In what direction does the plane travel 6 A boat heads 815 E on a river that flows due west The boat travels 811 W with speed of 25 km per hour Find the speed of the current and the speed of the boat in still water 7 An airplane is flying N30 W with an airspeed of 500 mph and a 40 mph wind blowing due east What is the groundspeed and direction of the plane 8 An airplane is flying 820 Wwith an airspeed of 750 kmhr is blown off course so that it is actually heading 828 W with a groundspeed of 770 kmhr Find the speed and bearing of the wind 9 A 35 mph wind bearing N 54 E blows a plane so that it has a groundspeed of 475 mph and a heading of 840 E Find the airspeed and original direction of the plane 10 Brandy is on the west bank of a 2 mile wide river with a current of 5 mph She wants to get to a dock that is 12 miles downstream on the east bank In what direction should she row so that she arrives on the other side exactly at the dock 11 An airplane is flying in the direction 832 E with airspeed 875 kmhr Because of the wind its groundspeed and direction are 800 kmhr and 840 E respectively Find the direction and speed of the wind 12 An airplane is flying in the direction N 48 W with airspeed 600 mph A 40 mph wind blows the plane so that it has a groundspeed of 580 mph Find the direction of the wind and the actual bearing of the plane There are 2 possibilities Portland Community College Page 13 Revised 012709 MTH 112 Supplemental Problem Sets SUPPLEMENT TO 104 EXERCISES 1 A cable running from the top of a telephone pole creates a horizontal pull of 850 N A support cable running to the ground is inclined 7150 from the horizontal Find the tension in the support cable 850 N 2 A MTH 112 instructor curiously represented by an donkey weighing 175 pounds is parked on a 50 foot ramp with an angle of inclination of 15 Assuming the only force to overcome is that due to gravity find the magnitude of force required to keep the Math instructor from sliding down the ramp and the work required to push the Math instructor up the ramp 15 3 Find the work done by a force of magnitude of 3 pounds applied in the direction of 7 2F2f to move an object 5 feet from 0 0 to 3 4 Portland Community College Page 14 Revised 012709 MTH 112 Supplemental Problem Sets SUPPLEMENT TO 121 EXERCISES 1 The position of a projectile fired with an initial velocity v0 feet per second at an elevation angle of 9 after t seconds is given by the parametric equations xt v0 cost9t yt 16t2 v0 sint9t a Obtain the rectangular equation of the trajectory and identify the curve b Show that the projectile hits the ground when I ve sint9 c How far has the projectile traveled horizontally when it hits the ground N Killer Kidoguchi shoots an arrow which leaves a bow 4 feet above the ground with an initial velocity of 88 feet per second at an elevation angle of 48 a Graph the arrow s path b V the arrow go over a 40foot school wall that is 200 feet from The Killer Portland Community College Page 15 Revised 012709 Haberman MTH 112 Section II Trigonometric Identities Module 2 The Laws of Sines and Cosines In Section 1 Module 4 we studied right triangle trigonometry and learned how we can use the sine and cosine functions to obtain information about right triangles In this section we ll study how we can use sine and cosine to obtain information about nonright triangles The triangle in Figure 1 below is a nonright triangle since none of it s angles measure 90 c Figure 1 Let s derive the Laws of Sines and Cosines so that we can study nonright triangles The laws are identities since they are true for alltriangles The Law of Sines To derive the Law of Sines let s construct a segment h in the triangle given in Figure 1 that connects the vertex of angle C to the side c this segment should be perpendicular to side c and is called a height of the triangle See Figure 2 Figure 2 The segment h splits the triangle into two right triangles on which we can apply what we know about right triangle trigonometry See Figure 3 below Figure We ean use the Me ngm mang es m ngure 3 m ubtam Expressmns fur bum smA and n 51 3 smA and sm8 We ean nuvv sewe bum ufthese Equatmns fur h smA and sm8 2 h AsmA and h asm8 New smce bum ef the h s represent the wean ef the same segment they are Equa By semng the h s Equa m Each uthervve ubtam the ququg AsmA A sm8 Tm Equatmn pmwues usthh Mat 5 knuvvn asthe Law elf Smes TypmaHy the avv s vv tten m terms elf rams WWE de2 mm swdes by A A we ubtam the ququg AsmA asm8 AsmA asm8 A A A A smA sm8 A A c and we a su mm M we fucus an m ang e and swde THE LAW OF SINES If a triangle s sides and angles are labeled like the triangle in Figure 4 then 5 a sinA sinB a b A 3 Figure4 This is an identity since it is true for alltriangles EXAMPLE 1 Find all of the missing angles and sidelengths of the triangle below The triangle is not necessarily drawn to scale 8 Figure 5 SOLUTION We can easily find 9 since we know that the sum of the anglemeasures in a triangle is always 180 So 9 55 75 180 3 t9 50 Now we can use the Law of Sines to find m and n Be sure your calculator is in degree mode to approximate these values sint9sin55 3 m 8 m 8 sint9 sin55 m 8 3 sin50 sin55 8sin50 sin55 m 748 5m 5 7 sm55 n 8 n 8 mosquot sm55 n 78 75 s 9 43 sm55 n Frgure a we ve drawn the gwen ange and nduded 3 er us ang e anu sruexengtn rneasures 8 Figures Nunee thatthe unges t srue rs uppusne the argesi ang e and the snenes1 srue rs uppusne the smaHEs L ang e ans rs awaystrue fur a ange and a very he pfu fact m keep m mmd su thatyuu ean make sure that yuur answer rs Even pessrmy eunem n urder m use the Law er Smes m nnu a rnrssrng pan er rnrssrng pan er a Hammer we need m nuvvthree er these furthmgs 5 the Law er Smes un y rs he pfu r we knuvvthe Ength Drown er he sees and the rneasure er the ang e uppesue ene er these srues ur r we knuvvthe rneasure er Wu makes and the Ength er the see uppesue ene er these ang es n the EXamp E abeve ang es and the Ength enne srue uppesue ene ennese ang es Censruertne mang e m Frgure 7 be uvv 12 Figure7 In this triangle we are not given enough information to use the Law of Sines since we aren t given any angle and side oppositequot combination In other words if we know a side s length we don t know the opposite angle s measure and if we know the angle s measure we don t know the opposite side s length In order to find the missing measurements of this triangle we need another law the Law of Cosines Let s derive the Law of Cosines just as we derived the Law of Sines earlier in this module The Law of Cosines Let s start with a generic triangle and draw the height h just as we did at the previously in this module See Figure 8 below c Figure 8 Again we want to consider the two right triangles induced by constructing the line h on the triangle given in Figure 8 This time we want to use the two pieces that the side c is split into Let s call the segment on the left the one closest to angle A x and then the segments on the right must be c x units long We ve emphasized the two right triangles and labeled the two pieces of side c in Figure 9 below First notice that the following is true We ll use this fact later Now let s apply the Pythagorean Theorem to each of the two triangles in Figure 9 The pink triangle on the left gives us x2 h2 b2 and we can solve this for h2 and obtain The blue triangle on the right gives us c x2hza2 and we can use the fact that h2 b2 x2 to eliminate h fromthis equation c x2hza2 3 c x2 b2 x2 a2 Finally we can simplify the left side of this equation and use the fact that x bcosA to eliminate x c x2bz x2a2 3 cz Zcxx2 bz x2a2 c2 20x b2a2 3 c2 20bcosA b2 a2 U This last equation is known as the Law of Cosines Below we ve rewritten the law in its standard form THE LAW OF COSINES If a triangle s sides and angles are labeled like the triangle in Figure 10then a2 b2 c2 2bccosA This is an identity since it is true for all triangles c Figure 10 Notice that when A 90 the Law of Cosines is equivalent to the Pythagorean theorem Verify this by substituting 90 for A For this reason the Law of Cosines is considered the generalization of the Pythagorean Theorem EXAMPLE 2 Find all of the missing angles and sidelengths of the triangle below The triangle is not necessarily drawn to scale 10 w 12 Figure 11 SOLUTION We can use the Law of Cosines to find w Be sure your calculator is in degree mode to approximate this value wz122 102 21210cos70 2 w2 144 100 240cos70 3 w 144 100 240cos70 3 w m 12725 Now that we know w we now know an angle and the side opposite that angle so we can use the Law of Sines to find either of the other angles Let s find sin6 sin70 10 3 W 3 sin 1sin sin 11039SiTv1 70 3 sin 11039511 70 3 m 476 using the fact that w m 12725 nauu we ean nd a by usmg the famthatthe sum ufthe ang1erneasures1n a mang1e 1s a1ways 180 a 70 180 a476 70 180 2 a624 1n gure 12 We ve drawn the gwen mang1e and me1ugeg aH gr 1 ang1e and s1ge1engm measures 12 Figure 12 Nemee that as thh the mang1e gwen 1n gure 61am s1de1s uppusne me 1unges1 ang1e and the smaHEs L 5MB 1 111m 1 ge a wuh a11r1ang1es the ungest uppusue the smaHEs L ang1e Weren tthe ease we wuu1g knuvvthat uur ansvvervvas ncurrect and knuvvthatvve ve ma rmstake g EXAMPLE 3 11 a mang1e has a sme gr 1engm s umts a sme gr 1engm 8 umts and the ang1e 35 and the rmssmg ang1erneasures SOLUT ON rst we need 1 1rans1a1e the mfurmatmn an a drawn mange see gure 13 bemw e Tr1ang1e nutdravvntu sea1 A 35 b Figure 11 arnmguuus enuugn that there 1 an enure1y mfferent Way we euu1u draw a 1nang1e W1ththe B 51nee tn1s segment 1 snuner than tne 8 mm segment 11 puss1b1e tnat 11 n anmnermeatmn e1userm the 35 ang1e Th1 w111 create a seeunu puss1b1e ang1e A and a currespundmg seeunu pussm E ang1e 3 and seeunu puss1b1e s1ue 1n gure 14 be1uw We ve snuwn the Wu puss1b1e 1nang1es thh the same gwen mfurmatmn 51 Figure 14 WE H 51am vth the 1nang1e cuntammg A 3 and A Part uf 1m 1nang1e 1s punue 1n F1gure 14 F051 1e1 s use the Law uf Smestu nnu A smA 7 smlt35 8 7 6 2 510 2 5mquot smA 5mquot 2 A 5mquot W s 2 A u 49 9 We ean nuvv nnu 3 by usmg the factthat the sum uf the ang1erneasures1n a 1nang1e 1s a1ways 180 A 3 35 180 2 49 9 3 35 8 180 2 3 u 95 1 Nuvv that we knDW the measure uf the ahgte uppeste a r we can use the Law uf shes tn the t j smlt3 5m 35 s 7 sm951 sm35 b 39 b 6 b j L sm951 sm35 N s sm95 1 2 a sm35 042 These vames ter A 3 a u grve us pessrhte thahgte wrth the gwen measurements We ve urawh thrs thahgte hette seate m Frgure 15 be uvv 1U 42 Figure 15 We ve uehe Everymmg currec y ahu teuhu one thahgte usmg aH uf the gwen thfur aUEI 2 2r rm m HHaH r m Frgure 147th urdertu hhu thrs ether pessrhte Hang E WE havetu remember the prehterh A functer has range thus when we use the mvErSE SWE functch we wm only Dbtam aeute ahgtes t e ahgtes that measure has than 90 hr urder tn the the ubtuse te greaterthah 90 pusstbmty we need tn use the tuehttty sm5 sm7r7 a that we rst hutteeu W Muuute 3 s e we re eurrehtty Wurkmg wtth degrees thsteau at ramahs Et s use sm5 smaw 7 a thsteau Abuve we tuuhu that smA e 9 85m35 s The ruehtrtytehs usthat smA sm180 7 A1 Sn we can tet A2 180 e A and we wt have another ang E Azrsueh that smA2 gsmi Thus A2 180 7 A 180 e 49 9 e1301 We ean nuvvfmd 32 by usmg the fact thatthe sum ufthe angemeasures m a ange 15 away 180 A2 32 3 180 21301 3 35 180 32 14 9 FmaHy we ean use the Law 0 Smestu nd A 32 7 me sm14 9 52 5 52 L N sm14 9 smlt35 sz65m149N269 smlt35 These vames fur AT 32 and A gwe me urements us anuther pess1b1e mange vth the gwen We ve drawn th1str1ang e nut 1e sea1e1n gure 1B bemw Figure 16 Smes Haberman MTH 112 Section 1 Periodic Functions and Trigonometry Module 6 Other Trigonometric Functions DEFINITION The tangent function denoted tant is the ratio of the sine and sint cosine functions Thus tant cost Let s graph y tant by plotting some ordered pairs It is easy to calculate values of tangent since it s is the ratio of sine and cosine and we already know the values of sine and cosine t tant t tant l unde no 2 fined point y 1 L 1 L I 4 l I 6 3 6 6 I o 0 0 0 2 1 1 r 6 E 11i 2 mil in 311ix 2 21 e 6 ea 5 I 2 I I l unde n0 4 2 fined point 277 27 Graph of ytant 5 7 L 57r 6 J3 6 J3 7r 0 7r 0 Note that the period of the tangent function is 7239 units unlike the sine and cosine functions whose periods are both 27239 units Also note that tant is undefined for I t t etc since cost equals 0 at these values and division by 0 is undefined As you can see in the graph y tant has a vertical asymptote at t t ie y tant has a vertical asymptote at each t value that makes cost 0 The fact that there are vertical asymptotes tells us that there are no maximum or minimum outputs for the tangent function Since the definitions of midline and amplitude involve maximum and minimum outputs the tangent function has no midline or amplitude Still the t axis ie the line y 0 behaves like a midline for y tant 37r t 2 etc We can define three other trigonometric functions by forming the reciprocal of the three trigonometric functions we ve studied thus far DEFINITIONS The secant function denoted sect is defined by sect 0 The cosecant function denoted csct is defined by csct The cotangent function denoted cott is defined by cott 10 Below we ve graphed these three functions 5 1 NotIce that sect has w x2 21 5 y 0 0050 l l vertical asymptotes at all t values for Which cost 0 The graph of y sect y 0050 is graphed in orange Notice that y csct has 1 sint vertical asymptotes at all t values for which sint 0 The graph of y csct y sinr is graphed in orange K lyk r 39 z I NotIce that y cott tan has 4quot 1 2 4 3 2 51532 vertical asymptotes at all t values for l which tant 0 The graph of y cott NA ReeaH thatvve can de ne the sme and eusme funenuns m terms r uf the swdeJEngths uf ngm mangxes Based an the mang e gwen be uvv an the rwght We knuvvthat cos HYF HY and OPP 5 E HYF A I ADJ we Varman mannerquot Pram m m H m u H 5 OFF m H OFF Land 39 cusS 39 ADJ 39 ADJ em 5355 1 1 HYF cusS ADJ ADJ HYF csc5 1 1 LY smS OFF OFF HYF a 1 Us HYF ADJ 939sm 5 OFF OFF HYF More Trigonometric Identities Recall the Pythagorean Identity that we first studied in Module 4 2 2 sm 9 cos 9 1 We can use this identity to find identities involving the new trigonometric functions we ve studied in this module If we divide both sides of the Pythagorean identity by 00520 we can find another identity sin2t9 00526 1 sin2t9 00526 1 2 00526 00526 00526 3 tan2t9 1 5630209 Alternatively we can divide both sides of the Pythagorean identity by sin20 and find another identity 2 2 sm 9 cos 9 1 sin2t9 00526 1 sin2t9 sin2t9 sin2t9 3 1 0096 05026 We ll This gives us two more identities that are also considered Pythagorean Identitiesquot study more identities in Section II Pythagorean Identities sin2t9 00526 1 tan2t91 56026 1 cot2t9 05026 Haberman MTH 112 Section 1 Periodic Functions and Trigonometry Module 0 Sets and Numbers DEFINITION A set is a collection of objects specified in a manner that enables one to determine if a given object is or is not in the set In other words a set is a welldefined collection of objects EXAMPLE Which of the following represent a set a The students registered for MTH 112 at PCC this quarter b The good students registered for MTH 112 at PCC this quarter SOLUTIONS a This represents a set since it is well definedquot We all know what it means to be registered for a class b This does NOT represent a set since it is not well defined There are many different understandings of what it means to be a good student get an A or pass the class or attend class or avoid falling asleep in class EXAMPLE Which of the following represent a set a All of the really big numbers b All the whole numbers between 3 and 10 SOLUTION a It should be obvious why this does NOT represent a set What does it mean to be a big numberquot b This represents a set We can represent sets like b in roster notation see box at top of next page quotAll the whole numbers between 3 and 10quot 4 5 6 7 8 9 ll Roster Notation involves listing the elements in a set within curly brackets quot DEFINITION An object in a set is called an element of the set symbol quot6quot EXAMPLE 5 is an element of the set 4 5 6 7 8 9 We can express this symbolically Se4 5 6 7 8 9 DEFINITION Two sets are considered equal if they have the same elements We used this definition earlier when we wrote quotAll the whole numbers between 3 and 10quot 4 5 6 7 8 9 DEFINITION A set S is a subset of a set T denoted S g T if all elements of S are also elements of T If S and T are sets and S T then S g T Sometimes it is useful to consider a subset S of a set T that is not equal to T In such a case we write S c T and say that S is a proper subset of T EXAMPLE 4 7 8 is a subset of the set 4 5 6 7 8 9 We can express this fact symbolically by 4 7 8 g 4 5 6 7 8 9 Since these two sets are not equal 47 8 is a proper subset of 4 5 6 7 8 9 so we can write 4 7 8 c4 5 6 7 89 DEFINITION The empty set denoted 0 is the set with no elements 0 There are NO elements in 0 The empty set is a subset of all sets Note that 0 at 0 DEFINITION The union of two sets A and B denoted AUB is the set containing all of the elements in either A or B or both A and B EXAMPLE Considerthe sets 47 8 0 2 4 6 8 and 1 3 5 7 Then aZU lin 14iZQ bZUmZampQQampz4QZQ a mzaagu iinmLzaaiaza DEFINITION The intersection of two sets A and B denoted AnB is the set containing all of the elements in both A and B Q EXAMPLE Considerthe sets 47 8 0 2 4 6 8 and 1 3 5 7 Then aZnmzampQQ bzanaginw c 0 2 4 6 8nl 3 5 7 0 These sets have no elements in common so their intersection is the empty set 4 EXAMPLE All of the whole numbers positive and negative form a set This set is called the integers and is represented by the symbol Z We can express the set of integers in roster notation Z 3 2 10123 Note that Z is used to represent the integers because the German word for quotnumberquot is quotzahlenquot Now that we have the integers we can represent sets like All of the whole numbers between 3 and 10quot using setbuilder notation SETBUILDER NOTATION quotAll the whole numbers between3 and10quot xler and 3 lt x lt10 This vertical line means quotsuch thatquot Armed with setbuilder notation we can define important sets of numbers DEFINITIONS The set of natural numbers N 1 2 3 4 5 The set of integers Z 3 2 1 0 1 2 The set of rational numbers Q f n q e Z and q at 0 This set is sometimes described as the set of fractions The set of real numbers R All the numbers on the number line abeR and 139 The set of complex numbers C a bi Note that NC Z c Q c R c C Le the set of natural numbers is a subset of the set of integers which is a subset of the set of rational numbers which is a subset of the real numbers which is a subset of the set of complex numbers Throughout this course we will assume that the numberset in question is the real numbers R unless we are specifically asked to consider an alternative set Since we use the real numbers so often we have special notation for subsets of the real numbers interval notation Interval notation involves square or round brackets Use the examples below to understand how interval notation works EXAMPLE a x x 6 R and 2 S x S 3 2 3 We use square brackets here since the endpoints are included Set builder Notation Interval Notation b xx e R and 2 lt x lt 3 2 3 We use round brackets here since the endpoints are NOT included Set builder Notation Interval Notation c x x 6 R and 2 lt x S 3 2 3 We use a round bracket on the left since 2 is NOT included Set builder Notation Interval Notation d xlxe R and 2 S x lt 3 2 3 We use a round bracket on the right T since 3 is NOT included Setbuilder Notation Interval Notation EXAMPLE When the interval has no upper or lower bound the symbol 00 or 00 is used a xlxe R and x S 4 00 4 We ALWAYS use a round bracket with 00 T since it is NOT a number in the set Setbuilder Interval Notation Notation b xlxe R and x Z 4 4 00 We ALWAYS use a round bracket with 00 T T since it is NOT a number in the set Setbuilder Interval Notation Notation 4 EXAMPLE Simplify the following expressions a 400U 8 3 b 4 oou oo 2 c 4 oom oo 2 d 4 oon 10 5 SOLUTION a 4 oOU 8 3 8 00 b 4 00U 00 2 00 00 R c 4 oom oo 2 4 2 d 4 oon 10 5 0 Haberman MTH 112 Section II Trigonometric Identities Module 5 Sum and Difference Identities In this module we ll study identities that allow us to change the form of expressions involving sine and cosine As we ll discover by using these identities to change the form of an expression we ll make it possible to calculate values that we wouldn t otherwise be able to calculate and we ll be able to solve realworld problems The identities we ll study in this module will allow us to change expressions of the form a sina i and cosa i b A1 sinBt A2 cosBt and c sint9 i sin and 0056 i cos Although we have the tools necessary to prove these identities we won t bother proving them here You can see proofs or sketches of proofs in Section 73 of the textbook We ll just state the identities and then see some examples of how these identities can be used Let s start with identities involving expressions of the form sinai and cosai These identities allow us to calculate the sine and cosine of the sum and difference of angles if we know the sine and cosine of the angles THE SUM AND DIFFERENCEOFANGLES IDENTITIES sine sina sinacos sin cosa sina sinacos sin cosa cosine cosa cosacos sinasin cosa cosacos sinasin 4 EXAMPLE 1 Use the sumofangles or the differenceofangles identities to calculate the following a cos75 b sin 15 c tanG g SOLUTIONS a Since 75 45 30 we can use the cosine of a sumof angles identity to calculate cos75 cos75 cos45 30 cos45 cos30 sin45 sin30 1 2 2 2 2 4 4 JEn5 4 b Since 15 30 45 we can use the sine of a differenceofangles identity to calculate sin 15 sin 15 sin30 45 sin30 cos45 sin45 cos30 1 2 2 2 2 4 4 JimE 77r s1n c To calculate tanG g we need to use the fact that tanG g along with the COS 12 sumof angles identities since 77r 37r 47r 12 12 12 1 4 339 Thus l 2 2 2 2 72l7273 2 2 2 2 J5J3 4 J57J3 4 J5 J3 J5 J3 use the conjugate of the denominator J5 7 J3 J5 J3 to rationalize the denominator 22J 6 276 84J 4 2J Now let s study an identity involving expressions of the form A1 sinBt A2 cosBt Before we state the identity let s investigate Consider the function ft 3sin2t 4cos2t it should be clear that this function has form AlsinBt AzcosBt where B 2 A1 3 and A2 4 We ve graphed ft 3sin2t 4cos2t in Figure 1 Notice that it looks like a single sinusoidal function It turns out that functions of the form A1 sinBt A2 cosBt can be written as single sine functions See the box at the top of the next page lt wwwu Figure 1 The graph of ft 3sin2t 4cos2t THE SUM OF SINE AND COSINE IDENTITY A1 sinBt A2 cosBt can be written as AsinBt where A JA12 A and tan 2 1 and sin we can find the signs of cos and sin and use these signs to determine the quadrant of Using the fact that cos EXAMPLE 2 Find A B and so that we can write ft 3sin2t 4cos2t as ft AsinBt SOLUTION As we noted above ft 3sin2t 4cos2t has form A1 sinBt A2 cosBt where B 2 A1 3 and A2 4 which immediately tells us that B 2 We can find A using the formula A IA A A JAI2 A 3 42 JR 5 Using B 2 and A 5 we know that ft Ssin2t To find let s first determine what quadrant it lies in Since ii i cos A 5 lt0 and sm A 5 gt0 we see that cos is negative and sin is positive so must be in Quadrant II To find the actual value of we can use the fact that tan 1 A2 3 tan i tan A1 3 Ustng tne mVErSE tangent functtun We see tnat apptuxttnatety 70 93 tagtans nas tangent 7 s tn QuadrantIV We need tet ttng an angte tn Quadrant H tnat nas tne satne tangent as 70 93 tagtans Stnee tne penetg etttangent ts 7r tagtanst We ean add 7r 70 93 and ubtatn an angte vntn tangen t ts tn Quadrant 11 Thus 93 s 2 21 and m s 55m21 2 2 Let s gtapn tnts mnettetn and vertiy tnat tt nas tne satne gtapn as m e3sm2z 420521 tapn ett S the t e 35m21 Figure 2 Tne g m s 55m21 2 21 Wnten satne as tne gtapn ett 4eos2z see th t FtnaHy tet s study tgenttttes tnvettvtng Expresstuns ett tne tetnn sm t5m0 and 2055 cos Tnese tgenttttes have a vanety ett rEaLWDrd apptteattetns THE SUM AND DIFFERENCE 0F SINE AND CDSINE IDENTITIES sm sm 25m a a 539 3 52 05 9 a 05 a 47 a z 5 e sm 205 2 1 3 52 J eos5eos 2eos e 5 5m 1 4 2 2055 7 cos 7 25m 4 EXAMPLE 3 Use identities given above to calculate the following a cos75 cos15 b sin75 sin15 SOLUTIONS a cos75 cos15 25in Jsim 25in45 sin30 5 1 2 7 7 2 b sin75 sin15 Zeos sin 2 cos45 sin30 IS m 2 2 2 2 Be sure to study the example on pp 333 334 in our textbook to see how these identities can be used to explain acoustic beats that are used for such things as tuning pianos Haberman MTH 112 Section II Trigonometric Identities Module 4 Proving Trigonometric Identities This quarter we ve studied many important trigonometric identities Because these identities are so useful it is worthwhile to learn or memorize most of them But there are many other identities that aren39t particularly important so they aren t worth memorizing but they exist and they offer us an opportunity to learn another skill proving mathematical statements In this module we will prove that some equations are in fact identities Recall that an identity is an equation that is true for all values in the domains of the involved expressions Thus to prove an identity we need to show that the two sides of the equation are always equal To accomplish this we need to start with the expression on one side of the equation and use the rules of algebra as well as the identities that we ve already studied to manipulate the expression until it s identical to the expression on the other side of the equation Let s look at a few examples to help you make sense of this procedure tanx secx I EXAMPLE 1 Prove the identity sinx As we mentioned above we prove identities by manipulating the expression on one side of the equation using the rules of algebra and the identities we already know until it looks like the expression on the other side of the equation We can start with either side of the equation but it s usually most sensible to start with the more complicatedquot side since it will be easier to manipulate it In this example tanoc secx is more complicatedquot than sinx so let s start with tanoc and try to manipulate it until it looks like sinx secx sinx tanoc 06 Since tanx 811106 and secx secx 1 cosx cosx cosx sinx cosx 00506 1 usmg the rules ofalgebra sinx more algebra tanx secx This is a proof of the identity sinx since we ve shown that the right side is equivalent to the left side EXAMPLE 2 Prove the identity cotx tanx cscx secx Here both sides are equally complicated so it s not obvious which side we should start with In such a case just start with either side and see what happens If you get stuck start over using the other side Let s start with the left side cotx tanx Since cotx 38 and tanx cosxCOSx sinx sinx using the rules ofalgebra to obtain a common denominator sinx cosx cosx sinx cos2 x sin2 x sinx cosx 1 2 2 Sinxcosx Slnce cos x 111 x 1 1 1 Sinx 00806 usmg more algebra cscxsecx Since cscx and secx m This is a proof that cotx tanx cscxsecx since we ve shown that the left side is equivalent to the right side For the next Examples 3 and 4 we ll need to use variations of the Pythagorean ldentity discussed at the end of Section 11 Module l 1 cost1 cost EXAMPLE 3 Prove the identity sint smt The left side is more complicated so we ll start with it 1 cost1 cost 1 0050 cost coszt using algebra sint sint 2 w using algebra smt 2 Since 1 coszt sin2t smt sint using more algebra 1 005t1 005t This is a proof that 51nt 5int since we ve shown that the left side is equivalent to the right side 0056 1 5in6 4 EXAMPLE 4 Prove the identity 1 51n6 0056 The proof of this identity requires a relatively commonly employed trick that is important to become familiar with Sometimes it is useful to use the conjugate of some part often the denominator of the expression The conjugate of the expression a b is the expression a b and vice versa As you can see in the proof below by multiplying the denominator by its conjugate allows us to manipulate the left side so that it is equivalent to the right side 0056 2 0056 1 5in6 1 5in6 1 5in6 1 5in6 00561 5in6 1 5in61 5in6 00561 5in6 1 5in6 5in6 5in26 00561 5in6 1 5in26 00561 5in6 Since 1 5in2 6 0052 6 00826 926950 5in6 00516 1 5in6 0056 using the conjugate of 1 5in6 0056 1 5in6 1 5in6 0056 equivalent to the right side This is a proof that since we ve shown that the left side is For the Example 5 we ll need to use a variation of the identity tan2t9 1 seczt9 discussed at the end of Section 11 Module 1 EXAMPLE 5 Prove the identity secu tanu secu tanu 2tanu To prove this identity we will again use the conjugate of the denominators of the expressions on the left side of the equation 1 1 1 secu tanu 1 secu7 tanu secu tanu secutanu secu tanu secu tanu secutanu secu7tanu secu tanu secu tanu secu tanusecu tanu secu tanu secu tanu seczu secutanu secutanu tanzu 2tanu seczu tan2u 2tanu 1 2tanu Since seczu tan2u 1 18 December 2008 MTH112 Elementary Functions 6 Trigonometric Functions FL 1 FLH39MMJH s txhtmehthhhnmhepmtm The London EyeA1ems Wheel on the Thames The Wudd slzrgest femswheel the Lehaeh Eye has a memetet emu feet andcumpletesunerevuluhunevery39KUmnutes Letwhemeheghtetthe mthutes The passehgehheetas the wheel at gnuha level 5 that u u sketeh a gaph ett Ever the eemplete cycle 12 u 5 I in 56 I Immdvctmnm Penman Fmthhs e Fems Wheel on the Thames The Wed argestfemswheel the Lehaeh Eye has a memetet emu feet and eempletes ehe revuluuun every 3n mthutes Lett he the hetght ufthe mthutes Thepassengerbuards the wheel at guundlevel 5 that u u sketeh a gaph ett Ever the eemplete cycle 12 u 5 I in PD smve Rntanun 1 thoqucm Kenneth 18 December 2008 55 I Immductmnm Pemdm New The Fems Wheel on the Thames The wudd39slzrgest femswheel Ihe Lundun Eye has a mameIer emu fee and cumpletes ehe revuluhun every 3n mmutes LetI he Ihe heIghI quhe mmutes Thepassengerbuardsthewheel atguundlevel suthat u Sketch a gaph efI uver ehe eempleIe cycle I2 I 5 I in 55 I Immehmhh tn Pemdm Functh The Fems wheel an the Thames n mdlme m m 7mpiltudlt l penea I I I I I e I I I I I I I I I 15 6D 55 I Immamhhh Pemdm Functh P r o Ic MoLIon A weIghIIs suspendedfmm a cedmg bye spring Letdrepxesenlthe nancemcennmeues am Ihe calm ta39hewexghl thnthe weIghz Is mauanless I1 25 mhe weIgIIIs mmhee xtbegns ta bub up and den ax osmllwz me Lb gaph deurmmz Ihe Imam heme amphmde and Ihe mxmmum maxxmum values m I as a fuman ufume Imerpxet hese qumuues phymny I a m mese quanuuesta desenhe Ihe mauan af39he weIghz omx l 4 cm 2 KIquuchL Kenneth 611ntxoductlonto Periodic Funcuons Periodic Function De nitions 18 December 2008 A function fis periodic ifits values repeat at regular intervals Hence fis periodic ifthere exists some constant c suc th t c f for all t in the domain off The smallest value ofl c l forwhich this relationship is true is called the poriod of f The midline ofaperiodic function is the horizontal line midvmy between the function s minimum d maximum values The amplimde is the vertical distance between the function s maximum or minimum value and the midline and fmm and ii fnix minimum values of frespectively then the midline equation is yMfmax fmm2 and the amplitude is A f m ifmmlZ 611ntxoductlonto Periodic Funcuons Periodic amp Aperiodic Functions 7 Three Examples 1 Determine ifqr might be periodic 1fthe function is periodic d etennine its peno n n n n 1 Prob Function Pen39odic 1 611ntxoductlonto Periodic Funcuons Periodic amp Aperiodic Functions 7 Three Examples 2 Determine ifvt might be periodic Lfthe function is periodic determine its period Prob Function Pen39odic Aperiodic PeriodEstimate 1 97 X NA 2 vt x 2 Kidoguchi Kenneth 18 December 2008 6 1 Intmdumunto Penudm Funmunx Pen39odic amp ApeIiodic Functions rThree Examples 3 Determine 1f y r mlght be penodle If Lhe funeuon is penodle detenmne 115 penod Prob Functlon Perlodle Apenodw PerlodEsumate 1 NA 2 2 2 t 2 unmmmm m mmmm 4 Kidoguchi Kenneth

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