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# Lecture 16, 17, 18, and 19 Math 240

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This 31 page Class Notes was uploaded by AnnMarie on Friday October 23, 2015. The Class Notes belongs to Math 240 at Louisiana Tech University taught by Jonathan B Walters in Fall 2015. Since its upload, it has received 56 views. For similar materials see Precalculus in Mathematics (M) at Louisiana Tech University.

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Date Created: 10/23/15

52 Trigonometric Functions of Real Numbers The circumference of a circle is C 2Hr If r1 the distance around the circle is 2H units If you went half way it would be H units If you went 34 of the way it would be 3H4 units Now Every point in the xy plane is some distance away from the origin V x The pythagorean theorem r xix2y2 Where r gt O The line from origin to xy creates an angle 9 from positive x axis Definition The six trigonometric function are as follows 0056 3 Sin 2r tan sec 3 csc g cot i Each Quadrant has its own negative andor positive xy points which determine if the Trig Function is either positive or negative I 11 in I all x and y values are positive in II x is negative and y is positive in 111 all x and y values are negative 111 IV in IV x is positive and y is negative Example If cos and 9 is in quadrant IV Find the other trigonometric values Solution First we will want to find y by using the pythagorean theorem z 52 Z 32 y2 25 9 y2 y j y j 4 Since we know that 9 is in QIV we will know that y will be negative Sin tan sec csc cot Notice that sec csc and cot look like cosh Sink and tan lag which is true because if you perform L4 you get a 3 You will want to remember this later on when having to solve trigonometric functions and identities Trigonometric Function Values of Some Special Angles 0 a a a Vquot 51716 V70 0 O 21 01 if if V74 0 l 0056 if 07 1 if l ZZ 71 01 l ZO or O You will want to remember the table above for solving questions such as the next example However we will need to go over angle references 97 first In General If 9 is known 0 If 9 is in Q1 then calculate the trig function values directly from the chart above other methods will discussed later 0 If 9 is in QII we will use or appropriately and take trig function values of 9 H where 9 is the reference angle 0 If 9 is in QIII we will use or appropriately and take trig function values of 9 9 H where 9 is the reference angle 0 If 9 is in QIV we will use or appropriately and take trig function values of 9 2H where 9 is the reference angle Example Find a COS b Six4 c tam 41 Solution a COS is in QII because it39s not at H but past Hmn mmn 4 4 W 4 4 4 004 72 Since we are in QII cos 6 is negative b Sin1 is in QIII because it39s not at I but past H 7H E 6 6 6 SIM 71 Since we are in QIII sin is negative c tam 41 is in QIV because negative angles go clockwise 1 I which means you go around once and then stop at tam 41 1 Since we are in QIV tan6 is negative due to y Example Find the value of 564 654 and cot 4I Solution 564 125 since sec 6 m then you can take l g and ip it 654 2 since sec 6 sin 1amp9 then you can take l g and ip it cot 1 since sec 6 m then you can take 1 and ip it EvenOdd Function cos 9 0066 3 even Sin 9 Sin 3 odd tan 9 tan 3 odd Pythagorean Identities Recall x2 y2 r2 If you divide by r2 ff 92 1 00526 Sin2 1 If you divide by COS2 00529 Sin2 1 00529 00529 00529 1 tan2 S602 If you divide by Sin2 00529 Sin2 1 Sin2 Sin2 Sin2 cot2 1 cs02 If you can remember just the first identity you can easily proof the other two identities Please accept my apologies if you have some issue with reading these notes because my backpack which contained my MATH 240 notebook got soaked Thus after drying the notebook in the oven at 150 degrees for half a day I noticed that some of the ink bleed through the page If you have issues reading them you are more than welcome to send me a message via mth039latechedu and I will be more than happy to type my notes up to resubmit I will also advise that Lecture 17 should have contained 54 but that will be covered in Lecture 18 I have also included the Suggested Assignments to assist with understanding how to complete the homework Thank you for your time and happy studying Mariah Thomas 1M 39 ril 53 5 Z Tri n ME39jr L Cmmpka r e 36 E Pager 1 W n um 15 L135 139 tPf imr Win32 1 NEW3i Ma k93 HMSE CUM Him b pa o Ma 3 Lve a Baha ier 55quot a1v1 Casi 5 39E Lrgbr 39E39i39 N mi aE F art quotW quotfi jaairfih r Cris I ErHr 111 34 will aregear 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wquot 3631M an 7 I r A D Erial Wt 1iquot 1 A MEMS 95 Lt 39 ng tutY 39 i EJEGLMEPJ E tlis39rampkh 395 quot E 03 39 Iquot 5A r 33quot 4L V W V ll A a 9 EWT H 1 r1 hasa hv f Z A in e Sjr39a h39 aima nmh Ch g3 56134119111ch T Ep nwh Emma1 Ewan 3 191 r a m C imgh f T IM1 FL Graph 1135quot quot39 1 Ein M121 7117 rn fhm 35 3 L 3 417 1M 393 7 an ET San 1m 1777 l r 7 w 7 7 7 z 1 i I 15 my mill1 ch 3 quotE 3 L03 L 14115 3 39 finamp 4kg CaMPI i lfm t Mm quot39rigj 417 WEE quotgiantHaw Smith 73139s r Phdv 3 mati A1 211mm 2L Fi RMWiMQEQML P ie i TE EHMJEWJ 3 339C39E4h51quot 3 Yang1 393 Fl Fin lard CLMPIHHEE nail parasl EVE ELE i hwhmj RAIL SI39WJ L i413 LannyL Y5 iig if ee39l og P red2 p gnaw E g tnmi 1qu aw iMh 21 sz Yarigl c i E 7 M 3le g Earnpin 1 53h ah I E Aquot p at 1n mi away 1 W man 47 Kim 5 H Mquot 1 323 1 l l u 93 II 331 5 Fi PM mM P H u trire m h r39Hznge g hik39l a mnr ti m 011 mi ash mm c mpira a IEr r 55 L X 1 13311 har MMX hw H 25 a H swim4r M H 3n392 A h rrigarHal 3M4 T I a 44 4 r44 4 4 A4 wm i ugg Hm ms L M 21 F 3 f39 5 Hm tnhli thi Hi i E 9 9m Etit gl l SHE quot39 13 wrw39 ri wmaSHE 7 1 3 J L 5 39 Sm in 41 73x Jami V mammal 5M 1 Li39s r eh all HEM CQMP ltM wrm a 1 Sim if mam4L mr c RE ai m r Ein s 4L2 ung th ti anLhatiw SH w an haH npmsm 4L1 Lur d i 1H Carla xi 4 g 3 Mat 15 er 3 a ma quot1 9 A h Tquot Qn kiss in 5quot m1 9 qrHamm an 91 7quot 11 E1 11 k 1 11quot 25w bu 393 L222 33 a z 9 1 We 1U 11 W 13 5 I M 3 37a Wei m3 31w mquot 3 5 39We hrm gunr h w an a tammm Enff fh Haw are ying Stwufs h f Eiahg i at am a 35 a sax g awe 11 511 and WEEKE g3 uni f Va 1 m 1 may October 28 2015 MATH 240 Lecture 18 Chapter 54 More Trigonometric Functions and Graphs Note that tan H tan The period of tan is H w You will notice in the graph bellow as X approaches zero and 9 approaches L tan approaches 00 and as X approaches zero and 9 approaches 13 tan approaches oo fx tankx and the period is You can shift the graph of tangent horizontally vertically and change its period Graoh of cot Recall that cot MT and that tan F1 Anywhere tan is zero cot is undefined These undefined parts are the asym ptotes of cot 42 d l E This similar for sec and csc because sec F1 and csc aim 19 To graph sec and csc you will want to graph the cosine or sine function first and anywhere that the cosine or sine function equal zero there will be an asymptote 4 Er CSC X SEC X Chapter 55 Inverse Trigonometric Functions and Their Graphs sin X cos X and tan X are not onetoone but if the domain where to be restricted to get a piece that is onetoone then the function would be onetoone On l21 1 sinX looks like the following On 011 cosX looks like the following On l21 13 tanX looks like the following Defin The inverse trigonometric functions are defined by 1 For all y in 3921 we define y arcsinX sin391X lt3 X siny 2 For all y in 011 we define y arccosX cos391X Q X cosy 3 for all y in 43 13 we define y arctanX tan391X lt9 X tany Note Inverse Trigonometric Functions output angles Graphs of sin391X cos 1X and tan391X F rill ii is 3 j i sin391 X cos39l X tan391 X Example Evaluate a sin391 bsin391 ccos 1 d cosh E e tan3911 f tan391 13 Solution a Atwhatangle m 4313 do we use for sine to get 2 39 l 1 H snn 2 6 b Atwhatangle m 4313 do we use forsine to get 39 l 1 H snn 2 6 c At what ange m 011 do we use for cosine to get 1 cos g E3 d At what ange m 011 do we use for cosine to get 1 cos 1 e At what angle m 4313 do we use for tangent to get 1 tan3911 34 33 391 1 H an E 6 f At what angle m H H do we use for tangent to get Example Find sincos 1 L Solution Let 9 cos K lt9 cos which gives us X 1 and r 4 and tells us that 9 is in QI since a gt O X2y2r2 1y216 y215 y i 15 we use since we are in QI sincos 1 g Example Find tancos 1 Solution Let 9 cos 1 Igt cos which gives us x and r 4and 9 is in Qllsince VT4 lt O X2y2r2 3y216 y213 y i 13 we use since we are in QII tancos 1i 7V1 Example Find arctantan1 Solution Since tan 2 1 and the restricted domain has to be between we find the angle reference 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