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# Elementary Functions MTH 112

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This 23 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 28 views. For similar materials see /class/224645/mth-112-portland-community-college in Math at Portland Community College.

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Haberman MTH 112 Section II Trigonometric Identities Module 1 Review of Identities Recall the following definition that we first studied in Section 1 Module 3 DEFINITION An identity is an equation that is true for all values in the domains of the involved expressions In this module we ll list the identities that we studied in Section I If you haven t yet learned these identities you should learn by understanding and memorizing all of them In Section I Module 3 we noticed the following identities Some Identities sint9 sint9 27239 0056 cos6 27239 0056 sin6 sint9 0056 cos t9 0056 sin t9 sint9 sint9 sin7r t9 In Section 1 Module 6 we defined the other trigonometric functions and obtained identities in the process Other Trig Functions 39 6 6 tanltegti9 tlt9gtifi ei 5006 m 0506 Also in Section 1 Module 6 we summarized the Pythagorean identities Pythagorean Identities sin26 00526 1 tan26 1 50026 1 00t26 05026 Sometimes we need to use different forms of the Pythagorean Identities By subtracting expressions from both sides of the different Pythagorean Identities we can obtain the following identities Be sure to convince yourself that these are all valid by adjusting the appropriate Pythagorean Identities More Pythagorean Identities sin26 1 00526 00526 1 sin26 tan26 50026 1 50026 tan26 1 00t26 05026 1 05026 00t26 1 Haberman MTH 112 Section IV Parametric and Implicit Equations Module 2 Introduction to Implicit Equations When we describe curves on the coordinate plane with algebraic equations we can define the relationship between the x and ycoordinates of the curve either explicitly or implicitly To explain the difference let s consider the equation y 3x 7 o The equation y 3x 7 establishes a relationship between x and y We say that y is defined explicitly in terms of x since the equation gives us a very clear description of the relationship between x and y no matter what the xvalue is the corresponding y value is always 3x 7 If we subtract 3x from each side of this equation we obtain the equation 3x y 7 Clearly this equation describes the same relationship between x and y as the equation y 3x 7 but when y isn t isolated the relationship between the variables isn t explicit Instead this relationship is implied by the equation so we say that it is an implicit equation An equation involving two variables is considered explicit if one of the variables is isolated on one side of the equation while and equation is considered implicit if neither of the variables is isolated on one side of the equation As we ve seen equations like y 3x 7 can be written either explicitly or implicitly but many equations can only be written implicitly For example it s not possible to solve the implicit equation x2 y2 1 for either variable If we try to solve the equation for y we need to use two explicit equations to communicate the information in the implicit equation 4 EXAMPLE 1 Recall from Example 4 in the previous module that the system of parametric equations below describe a unit circle centered at the point 0 0 x cost y sint As we learned in the previous module we use the Pythagorean Theorem to eliminate the parameter t and transform this system into a single implicit equation coszt sin2 I 1 3x2 y2 1 Thus the implicit equation x2 y2 1 describes a unit circle centered at the point 0 0 Since the parametric system x cost y sint describes a circle of radius 1 unit we should expect that if we multiply both the x and y coordinate by the factor r we will stretch or compress the circle so that its radius will change from 1 unit to r units and if we add constants h and k to the x and y coordinates respectively we will shift the center of the circle from the point 0 0 to the point h k A generalized parameterization of a circle is given in the box below If h k reR and rgt0 then the system of parametric equations below defines a circle of radius r centered at the point h k x h rcost y k rsint EXAMPLE 2 Eliminate the parameter r from the system of parametric equations given in the box above to obtain an explicit equation that defines a circle of radius r centered at the point h k SOLUTION We can use the Pythagorean Theorem to eliminate the parameter just as we did in Example 1 The Pythagorean Theorem involves sint and cost so we need to first we need to isolate sint and cost in the equations in our system x h rcost y k rsint 3 rcost x h and 3 rsint y k 3 cost xih 3 Sing yik r r x 7 h y 7 k Now we can substitute the expressIons and for cost and smt In the Pythagorean Theorem and obtain an implicit equation for the circle coszt sin2t 1 2 WWW quoti1 MY 1 r r2 If h k r e R and r gt 0 then the implicit equation x h2 y k2 r2 defines a circle of radius r centered at the point h k 4 EXAMPLE 3 Determine the center and radius of the circle defined by the implicit equation x72 y 62 9 SOLUTION Notice that the equation x 72 y 62 9 has the form x h2 y k2 r2 where h 7 k 6 and r 3 Thus the center of the circle is at the point 7 6 and the radius is 3 units We could check if we are correct by graphing the circle on our graphing calculator but our calculators aren t able to graph implicit equations If we transform our implicit equation into a system of parametric equations we could use our graphing calculator to graph the circle To transform the implicit equation into a system of parametric equations we mimic what we did in Example 2 but do everything in the opposite order First let s get 1 on the right side of the equation by dividing both sides by 32 Le 9 x72 y 62 9 x72y 6232 x72 ye62 3 23 2 1 7i 6i1 ltagt We now have an equation in which the sum of two squares is equal to 1 Since U U Pythagorean Theorem has this same form ie since 0052t sin2t1 we can let x g 7 and sint Notice that if we substitute these values for cost and sint in the Pythagorean Theorem we obtain the equation labeled a above We cost can solve these equations for x and y to obtain a parameterization of the given implicit equation x 7 y 7 6 t t cos 3 and sm 3 3 x73cost 3 y 63sint 3 x 7 3cost 3 y 6 3sint Thus the system of parametric equations below describes the same circle as the implicit equation x 72 y 62 9 x 7 3cost y 6 3sint Notice that from the generalization given at the beginning of Example 2 we can see that this parameterization describes a circle of radius 3 units with center 7 6 which is exactly what we were looking for 4 EXAMPLE 4 The graph of the equation x2 y2 6x 4y 3 is a circle Identify the center and radius by using appropriate algebra SOLUTION In order to determine the center and radius of the circle we need to get our implicit equation into the form x h2 y kzrz To accomplish this we can complete the square twice If you need to review the procedure for completing the square see 55 in our text x2y2 6x4y3 x2 6xy24y3 x2 6x9 9y24y4 43 x2 6x9y24y4 9 43 x 32x22 133 x 32x2216 x 32x2242 UUUUUU Thus the center of the circle is at the point 3 2 and the radius of the circle is 4 units Based on what we observed in Examples 3 and 4 we can conclude that the system of parametric equations below describes this same circle x 3 4cost y 2 4sint Graph the system on your graphing calculator to check Haberman MTH 112 Section 1 Periodic Functions and Trigonometry Module 5 Sinusoidal Functions DEFINITION A sinusoidal function is function f of the form y AsinBt h k A B h k 6 1R NOTE 1 Recall from Module 3 that cost sint ie since the cosine function is a transformation of the sine function Therefore we can write any function of the form y AcosBt h k in the form given in the definition of a sinusoidal function Thus functions of the form y AcosBt h k are also sinusoidal functions NOTE 2 Based what we know about graph transformations which are studied in the previous course we should recognize that sinusoidal functions are transformations of the function y sint Below is a summary of what is studied in the previous course about graph transformations SUMMARY OF GRAPH TRANSFORMATIONS Suppose that f and g are functions such that gtAfBt hk and A B h k e R In order to transform the graph ofthe function f into the graph of g 1 horizontally stretchcompress the graph of f by a factor of liq and if B lt 0 reflect it about the yaxis 2quot shift the graph horizontally h units shift right if h is positive and left if h is negative 3quot vertically stretchcompress the graph by a factor of IAI and if A lt 0 reflect it about the t axis 4 shift the graph vertically k units shift up if k is positive and down if k is negative The orderin which these transformations are performed matters Examples 1 4 below will provide a review of the graph transformations as well as an investigation of the affect of the constants A B h and k on the period midline amplitude and horizontal shift of a sinusoidal function You may want to follow along by graphing the functions on your graphing calculator Don t forget to change the mode of the calculator to the radian setting under the heading angle EXAMPLE 1 Describe how we can transform the graph of ft sint into the graph of gt 2sint 3 State the period midline and amplitude of y gx SOLUTION Our goal is to use Examples 1 4 to determine how the constants A B h and k affect the period midline amplitude and horizontal shift of a sinusoidal function so let s start by observing what the values of A B h and k are in gt 2sint 3 It should be clear that function g is a sinusoidal function of the form yAsinBt hk where A2 B1 h0and k 3 After inspecting the rules for the functions f and g we should notice that we could construct the function gt 2sint 3 by multiplying the outputs of the function ftsint by 2 and then subtracting 3 from the result We can express this algebraically with the equation below g0 2m 3 Based on what we know about graph transformations we can conclude that we can obtain graph of g by starting with the graph of f and first stretching it vertically by a factor of 2 and then shifting it down 3 units Since ft sint has amplitude 1 unit if we stretch it vertically by a factor of 2 then we ll double the amplitude so we should expect that the amplitude of g to be 2 units Also since ft sint has midline y 0 when we shift it down 3 units to draw the graph of g the resulting midline will be y 3 Note that since graphing g required no horizontal transformations of ft sint the graph of g must have the same period as the graph of ft sint 27239 units Let s summarize what we ve learned about gt 2sint 3 period 27239 units midline y 3 amplitude 2 units horizontal shift 0 units The graphs of ft sint and gt 2sint 3 are given in Figure 1 below 2y Figure 1 The graphs of ft sint and gt 2sint 3 EXAMPLE 2 Describe how we can transform the graph of ft sint into the graph of nt sint state the period midline amplitude and horizontal shift of y nx SOLUTION Notice that the function n is a sinusoidal function of the form y AsinBt h k whereA1B1h andk0 After inspecting the rules for the functions f and n it should be clear that we can write n in terms of f as follows nt ft Based on what we know about graph transformations we can conclude that we can obtain graph of n by starting with the graph of f and shifting it left units Since a horizontal shift won t affect the period midline or amplitude we should expect that the period midline and amplitude of nt sin I are the same as ft sint period 27239 units midline y 0 amplitude 1 unit horizontal shift units The graphs of ft sint and nt sint are given in Figure 2 39 f 1r 31rx 4 39nx392 15M 14 15x 31x 1 513 4 315x 15M 39 1 Figure 2 The graphs of ft sint blue and nt sint purple EXAMPLE 3 Describe how we can transform the graph of pt cost into the graph of qt cos2t and find the period midline amplitude and horizontal shift of y qx SOLUTION Notice that the function q is a sinusoidal function of the form y AcosBt h k where A 1B2 h0and k0 After inspecting the rules for the functions p and q it should be clear that we can write q in terms of p as follows qt p2t Based on what we know about graph transformations we can conclude that we can obtain graph of q by starting with the graph of p and first stretching it horizontally by a factor of ie compressing the graph by a factor of 2 and then reflecting it about the t axis Since pt cost has period 27239 units if we compress the graph by a factor of 2 then the period will be shrunk to 7239 units Since we aren t stretching the graph of p vertically we should expect that the amplitude of q is the same as the amplitude of p 1 unit Also since we aren t shifting the graph of p vertically we should expect that the midline of q is the same as the midline of p y 0 Let s summarize what we ve learned about qt cos2t period 7239 units midline y 0 amplitude 1 unit horizontal shift 0 units The graphs of pt cost and qt cos2t are given in Figure 3 i om 0m WV Figure3 The graphs of ptcost blue and qt cos2t purple Notice that the graph of qr cos2r completes two periods in the interval 0 2n In general the number B in a sinusoidal function of the form y AsinBt h k or y AcosBt h k represents the number of periods or cycles that the function completes on an interval of length 27239 This number B is called the angular frequency of a sinusoidal function When we use sinusoidal functions to represent reallife situations we often take the input variable to be a unit of time Suppose that in the function qt cos2t t represents seconds Since the input of the cosine function must be radians the units of B 2 must be radians per secondquot This way 2 radians t seecrn d 2r radians which has the appropriate units for the input of the cosine function So if t represents seconds the angular frequency of qt cos2t is 2 radians per secondquot Another way to obtain the unit of the angular frequency is to use what we noticed above the number 2 in qt cos2t represents the number of cycles that the function completes on an interval of length 27239 Since a cycle is equivalent to a complete rotation around a circle or 27239 radians two cycles is equivalent to 47239 radians If the input variable t represents seconds then the angularfrequency is 47r radians 2radsec 27r seconds EXAMPLE 4 Describe how we can transform the graph of pt cost into the graph wt 3pt 5 State the period midline amplitude and horizontal shift of y wx SOLUTION Notice that the function w is a sinusoidal function of the form y AcosBt h k whereA3B handk5 L 2 After inspecting the rules for the functions p and w it should be clear that we can write w in terms of p as follows wt 3pt 5 Based on what we know about graph transformations we can conclude that we can obtain graph of q by starting with the graph of p and first stretching it horizontally by a factor of 2 then shifting it right units then stretching it vertically by a factor 3 and finally shifting it up 5 units Since pt cost has period 27239 units if we stretch the graph by a factor of 2 then the period will be stretched to 47239 units Similarly if we stretch the graph of pt cost vertically by a factor of 3 then we ll triple the amplitude so we should expect the amplitude of w to be 3 units Also since pt cost has midline y 0 when we shift it up 5 units to draw the graph of w the resulting midline will be y 5 Since we are shifting the graph hght umts the huhzuhtai shift is El units Let s summarize What we ve iearhed abuut y W0 period An units mid ine y 7 5 amplitude 3 units horizontal shin units 5 are given in Figure 4 The graphs uf pt c050 and Wu 7 3p 7 8 6 777774 2 72nv Vzitvzwv piue and 2 Figured The gr phs uf ptcost z 7 5 purpie wt 305 Nutice that the graph uf wt 3cos 7t 7 31 5 eumpietes uhehait uf a periud Dr it We iet the input vanabie 1 represent SEEDndS theh eyeie h the intervai 0 Zn W 314 7 a 5 eumpietes uhehait uf a eyeie every 27r secunds shee uhe hait uf a eyeie is Equivaient tn hait uf a rutatiun aruund a cimie ur 7r radian then the s angularfrequency utthemhetph w i humans Insecuhus T39adS Based on what we learned in the examples above we can summarize the affect of the constants A B h and k on the period midline amplitude and horizontal shift of functions of the form y AsinBt h k and y AcosBt h k SUMMARY Graphs of Sinusoidal Functions The graphs of the sinusoidal functions y AsinBt h k and y AcosBt h k where A B h k e R have the following properties period f units midline y k amplitude IAI units horizontal shift h units angular frequency B radians per unit of t EXAMPLE 5 Sketch a graph of ft 25in7rt 3 SOLUTION In order to use what we ve just studied about functions of the form y AsinBt h k we need to write the given function in this form ie we need to factor 7239 which is playing the role of quot3quot out of the input expression m Tquot ft 25in7rt 3 25in7rt 3 It should be clear that ft 25in 7rltt 3 is a sinusoidal function of the form yAsinBt hk where A2 B7r h and k 3 Using whatwe L 4 found above we can find the period midline amplitude and horizontal shift of y ft period 1 7 7 2 units midline y amplitude 2 2 unit horizontal shi AL uf a unit We ean use this inturmatiun tn sketch a graph uf m 7 25m 7r 1 7 7 3 see Figure 5 beiqu Nate that the hurizuntai shiftteiis usvvhere tn start uur usuai sine Wave iy uzs usu D75 iuu izs isu 175 mm 225 zsu 275 mu Figure5 The graph uf m 25m7rt 7 7 7 3 The piue puint represents where We start uur sine Wave sinee the hurizuntai shift is uf a unit Nuts that aeeurding In what we diseussed in Exarnpies 3 and 4 it We iet 1 represent i y i e K A 3 is 7r radians per seeund sinee 7r radians represents unehait uf a rutatiun aruund a eireiei the anguiartredueney 7r radians per seeund ii is Equivaient tn unehait uf a eyeie per seeund Nutice that uur graph in Figure 5 shuws a tunetiun that eurnpietes unehait uf a Enud in one unit uf H Haberman MTH 112 Section 1 Periodic Functions and Trigonometry Module 3 The Sine and Cosine Functions As we noticed in the Ferris wheel example in Module 2 that a circle rotating about its center lends itself naturally to the study of periodic functions The sine and cosine functions are fundamental periodic functions defined in terms of the unit circle DEFINITION The sine function denoted s0 sin0 associates each angle 0 with the vertical coordinate ie the ycoordinate of the point P specified by the angle 0 on the unit circle The cosine function denoted ct9 cos0 associates each angle 0 with the horizontal coordinate ie the xcoordinate of the point P specified by the angle 0 on the unit circle 80 the point P in Figure 1 has coordinates x y 0050 sin0 Figure1 In order to enhance our understanding of the sine and cosine functions we should determine some particular values for the functions and sketch their graphs But before we confront these details let s determine the signs positive or negative of the sine and cosine functions in the four different quadrants of the coordinate plane To review the quadrants of the coordinate plane see Section I Module 1 Figure 6 When the terminal side of angle 0 is in Quadrant I both the x and ycoordinates of point P are positive so 0056 gt 0 Sinw gt 0 In Quadrant I When the termthat stde etahgte a tsth Quadrant the yrcuurdmate etpmht P S pesttwe butthe xrcuurdmate S negatwe su 2055 lt 0 5 gt 0 m Quadrant When the tetmthat see at ahgte a tsth Quadrantl buth the x7 and yrcuurdmates etpmht P are negatwe su 5055 lt 0 ma 0 nQuadrantDI When the termthat stue etahgte 5 S W QuadrantIV thE xrcuurdmate etpmht P Spusmve butthe yrcuurdtnate S negatwe su cos gt 0 my 0 m Ouadrantl v Let s summanze what we ve determmed abuut the stghs at the SWE a d EDSWE mhmehs m the mttereht quadrants cus gt0 slto QIv Figurez The stghs et SWE a d EDSWE m the fuur quadrants Nuwtet s dEtErWWE the SWE and EDSWE at a few pamcmar SrvaMEs the Euurdmate axes h thure 3 the nurdmates uf these fuur mm are gWEn Keep m mmd that EDSWE represents the xrcuurdtnate and SWE represents the yrcuurdmate EH Figure The pmnt 1 0 eh the UM eheTe ts spammed by the ahgTe a 0 tadtahs 1e a 0 Sn cos0 1 arm sm0 0 The pmnt o 1 eh the UM eheTe ts spammed by the ahgTe a 90 1e vad12ns su mew and 5m 7 The pu1n1710unthe umt eheTe ts spammed by the ahgTe 5 180 12 5 7r Yad ans Sn cos7r and sm7r The pmnt 0 e1 eh the UM eheTe ts spammed by 5 270 1e 5 37quotvad12ns su Nettee that ahgTes uf measure 27rvad12ns t e 5360 a d stamens spemiy the same pumt 1 0 Thus the SWE a d EDSWE vames fur 27Hadmvs arm 0 tadtahs are the same shee ANY ahgTe 5 arm 5 27r spemiy the same pmnt eh the UM arch the SWE a d EDSWE vames uf 9 arm a 27r are the same therefure the pEHEId uf the SWE a d EDSWE mhettehs ts 27rvad12ns Para 5 51115 sm5 27r a d 12055 cos5 270 Sn the period etheth 59 my arm 55 12055 ts 2madtahs 1e 360 Now we ll sketch graphs of the sine and cosine functions Let s start by organizing the function values we determined above in a table and then plot this information on the coordinate planes in Figure 4 below 399 degrees 0 90 180 270 360 450 540 630 720 6 radians 0 7239 37 27239 57 37239 77 47239 y 0056 1 0 71 0 1 0 71 0 1 y 5in6 0 1 0 71 0 1 0 71 0 10 y o o 6 m 2 52 a 35x2 2 55x2 3 752 41 1 o 0 Figure 4a Some points on y 0056 1 y o o 9 1rr 2 m 139 3m 2391 5m 3391 7m 4391 1 o 0 Figure 4b Some points on y 5in6 In the next modules we ll find many more values of sine and cosine but we can go ahead and connect the dots in these graphs to obtain reasonable sketches of the graphs of the sine and cosine functions in Figures 5a and 5b below We can use what we observed in Example 4 from Module 2 when we studied the Ferris wheel 5 y f3 Wig 1 asz 215sz 4 Figure 5a The graph of y 0056 k 1 3 v I 1w 2 w W 5m 3W 1 Figure 5b The graph of y 5in6 Notice that our analysis of the signs of the sine and cosine functions see Figure 2 above agree with the graphs of the sine and cosine amp KEY POINT To get these graphs on your calculator be sure to change the mode of to the setting radian under the heading angle If you want to get graphs of sine and cosine in degree mode be sure the window has a horizontal interval like 300 1440 since the period of the function is 360 Notice that the graphs of y 0056 and y 5in6 are very similar In fact if we shift y 5in6 to the left units we ll obtain the graph of y 0056 Using what we learned about graph transformation in MTH 111 we know that 00565in6 This is equation is known as an identity since the left and right sides of the equation are always identical no matter what value of 6 is used DEFINITION An identity is an equation that is true for all values in the domains of he involved expressions Earlier in this module we observed a couple of identities but didn t call them identities The equa ons 5in6 5in6 27239 and 0056 0056 27239 are identities since they are true for all values of 6 We can use the graphs of sine and cosine to determine a few other important identities Do your best to convince yourself that each of following identities are true I strongly encourage you to graph on your graphing calculator both sides of the identities and notice that the graphs are identical Also use what you learned in MTH 111 about graph transformations and symmetry to make sense of WHY these identities are true In Module 5 we will review graph transformations SOME IDENTITIES 5in6 5in6 27239 0056 0056 27239 0056 5in6 5in6 0056 005 6 0056 5in 6 5in6 5in6 5in7239 6 The last identity on this list is hardest one to make sense of but it is worth taking time to understand it since it is a useful identity We will use it in Module 7 when we solve equations involving the sine function The easiest way to understand it is to focus on the 0 values between 0 and lf 0 is in this interval then it should be clear if you study the graph of y sint9 that sin7r 9 is the same as sint9 due to the symmetry of the sine function between 0 and 7239 Once you see why sint9 sin7r 9 for 0 values between 0 and it will be easier to convince yourself that the identity holds for all values of 0 We ll study a few more identities in Modules 4 and 6 and focus on identities in Section II Although the sine and cosine functions are defined via the unit circle we can use sine and cosine to find the coordinates of a point on the circumference of any circle First let s notice a few things about the unit circle Figure 6 The unit circle with a point P specified by the angle 0 As shown in Figure 6 we can construct a righttriangle using the terminal side of angle 0 and the horizontal and vertical components of the point P By construction this right triangle has a hypotenuse of length 1 unit since this is the radius a horizontal component of length 0056 and a vertical component of length sint9 See Figure 7 below Figure 7 The right triangle induced by point P on the unit circle uWEt s eehstuer a ehete WW1 a mtterehtramus See gure a hem keep m mmd thatthe ahgte 5 ts the same m gure a as twas m gures 6 and 7 Figures The hght thahgte thuueeu by pmnt T eh a mrde etramus 39 Wu thahgtes are similar 1 smltagt y A I A l cos5 x Figures Stmttarhghtthahgteswth ahgte a shee the Wu thahgtes m gures 7 and a are heth hghtthahgtes t e they heth have a 90 5 A va LknuWn tact abuut shhttar thahgte EXamp E the ratm at the hetg t S m ar y the ram at the hehz nns tant WE can u s ts that the ram at StdEJEngt s e se these t equauDhs h ts euhstaht Fur tn the hyputenuse uft e VESpEEUVE thahgtes ts euhstah htat Ength tn the hyputenuse uf the respeetwe thahgtes ts aets and the thahgtes W thure a El Dbtam the quuvvm cos5 1 and sm5 L 1 r 1 r SEIMHQ these Equatmnsfur x a d y respemve y vva Dbtam x rcos and y rsm5 Luukmg back at ngure 8 We see thatvvhat we ve fuund are the cuurdmates ef the pawl T spem ed by the ang e a an we enemerenee elf a eme elf radms 39 SEE the bux be uvv Suppuse thatthe pumt T xy sspem ed bythe ang e a an the ewemerenee elf a mrde elf radms 39 Then x 2055 and ysm Figure 10 erde ufradms r

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