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Activity 1 Identify whether the given equation is an identity or a conditional equation. For each conditional equation, provide a value of the variable in the domain that does not satisfy the equation. i. and 2:2 etric expre (1) 1+2+ 1 - 2 1- 2 PAY RENG (2) cos' 0 - sin2 A = cos 0 - sin 8 cos # + sin A B Chane (3) tand = cot 0 Activity 2 line: Watch who CON DIS A. Use the identities presented in this lesson to simplify each trigonometric expression. Bart call, or video charl 1. cot 0 cos 0Periodic motions are usually modeled by either sine on cosine function, and are called simple harmonic motions. Unimpeded movements ofjobjects like oscillation, vibration, rotation, and motion due to water waves are real-life occurrences that behave in simple harmonic motions Equations of Simple Harmonic Motion The displacement y (directed height or length) of an object, behaving in a simple harmonic motion with respect to time t is given by one of the following equations.A pull upsidad THALyasinb(t-offd "when or y = a cosb(t - c) + d. In both equations, we have the following information: . amplitude = lal = =(M - m) - the maximum displacement above and below the rest position or central position or equilibrium, where M is the maximum height and m is the minimum height; "s.\\ 51052 . period = - the time required to complete one cycle (from one highest or lowest point to the next); noitom amormail olgatie oltival . frequency = 2 - the number of cycles per unit of timeline ovlee. [ . c - responsible for the horizontal shift in time; and onsups Isnowlibno . d - responsible for the vertical shift in displacement. "onogiil vilqmi Example 1. A weight is suspended from a spring and is moving up and down in a simple harmonic motion. At start, the weight is pulled down 5 cm below the resting position, and then released. After 8 seconds, the weight reaches its highest location for the first time. Find the equation of the motion. " ical Solution. We are given that the weight is located at its lowest position at t - 0; that is, y = -5 when t = 0. Therefore, the equation is y = --5 cos bt. Because it took the weight 8 seconds from the lowest point to its immediate highest point, half the period is 8 seconds. anilno 8 -bonsq-doing 1 27 =8 6=21 By =1-5 cos O Ad-snuge-bald 2 6-to-momswold moo oduin Example 1. Suppose you ride a Ferris wheel. The lowest point of the wheel is 3 meters of the ground, and its diameter is 20 m. After it started, the Ferris wheel revolves at a constant speed, and it takes 32 seconds to bring you back again to the riding point. After riding for 150 seconds, find your approximate height above the ground. nil wov II (quinns i boon 1-(197 wil) OR~ Dale MINon . sanseesin signoult isdonei bosidna sill sguzzsin To thienos of soil lost senalq ,hobizorg musings silt Solution. We ignore first the fixed value of 3 m of the ground, and assume that the central positions bys passes through the center of the wheel and is parallel to the ground. Let t be the time (in seconds) elapsed that you have been riding the Ferris wheel, and y is the directed distance of your location with respect to the assumed central position at time t. Because y = -10 when t = 0, the appropriate model is y = -10 cos bt for t 2 0. Given that the Ferris wheel takes 32 seconds to move from the lowest point to the next, the 27 b = 32 = b= - 16 = y= -10 cos 16 When t = 150, we get y = 10 cos = ~ 3.83.Bringing back the original condition given in the problem that the riding point is 3 m off the ground, after riding for 150 seconds, you are approximately located 3.83 + 13 = 16.83 m off the ground. Example 2. A signal buoy in Laguna Bay bobs up and down with the height h of its transmitter (in feet) above sea level modeled by h(1) - a sin bird at time ) (in seconds). During a small squall, it's height varies from 1 ft to 9 fit above sea level, and it takes 3.5 seconds from one 9-ft height to the next. Find the values of the constants a, b, and de bus Idan all of win des " svom of zer b f durian Solution. We solve the constants step by step. sado ow ,a lo sulav all of sA Jnomugs and of malimid .( 1- = w) sulev seewol all of brawnwob og bun (d == y) of . The minimum and maximum values of h(t) are left and 9 ft, respectively.au) Thus, the amplitude is a = ;(M - m) = (9+1) = 4:danielemi idgi ads . Because it takes 3.5 seconds from one 9-ft height to the next, the period is 3.5. Thus, we have ? - 3,5, which gives 6 757. () ned =. . Because the lowest point is 1 ft above the sea level and the amplitude is 4w it follows that d = 5. aVS- SV8+ 5-N- (8 - 81) a180 = 3. Example 3. A variable star is a star whose brightness fluctuates as observed from Earth. The magnitude of visual brightness of one variable star ranges from 2.0 to 10.1, and it takes 332 days to observe one maximum brightness to the next. Assuming that the visual brightness of the star can be modeled by the equation y = a sin b(t - c) + dot in days, and putting t = 0 at a time when the star is at its maximum T brightness, find the constants a, b, c, and d, where a, b > 0 and c the least nonnegative number, possible. Solution. M - m 10.1 - 2.0 .nobigups land = 2 27 = 332 b= "noisupd (snoisibno' ban vingbl nousups sell to mismob ord ni oldshor sidi To zoulay (is 101 ound66eds nousups an at quitshi nA bill lo zsulev smoe li .eblow mario d = a t m = 4.05 + 2.0 = 6.05 viunsbi na ton zi tardi nousuponA Ianoinbrog sel notlupe ord mort noitsups ord viause ton of nousups and to mamob ords mu oldsisy For the (ordinary) sinc function to start at the highest point at t = 0, the least possible horizontal movement to the right (positive value) is , units. Solu foilsups silt vlzine ton bond 37 249 bivorq . noilsups Isnoilibnoo 2 26 21 166 4 8 -1) =6- 2 (1) I + 0 Boo = 0'nis (S) Example 4. The path of a fast-moving particle traces a circle with equation (x + 7) + (y - 5) =36nie (8) It starts at point (-1, 5), moves clockwise, and passes the point (-7, 11) for the first time after traveling 6 microseconds. Where is the particle after traveling 15 microseconds? 3V + 1 Solution. We may choose sine or cosine function. Here, we choose the sine function to describe both . and y in terms of time t in microseconds; that is, we let noisup2 p or loxasin b(1-c) + dandy =esin f(-8) -h, ibnooziei T (S) where we appropriately choose the positive values for a, b, e, and f, and the least nonnegative values for c and g. noitsups odd to cobia dod gods ,0 = 0HI .noitsups Isnoidibucs a bala ai aiifT (8) The given circle has radius 6 and center (-7, 5). Defining the central position of the values of x as the line x = -7 and that of the values of y as the line y = 5, we get a me $6, d :7, and h #15, oliniw From the point (-1, 5) to the point (-7; 11) (moving clockwise), the particle has traveled three- fourths of the complete cycle; that is, three-fourths of the period must be 2."( un a am able basil-flot oil to Toinsimonsb on anisilagora3 27 3 27 As the particle starts at (-1, 5) and moves clockwise, the values of z start at its highest value (a = 1) and move downwar iDownward toward its central position wo ("() and continue to its lowest value (2 = 13), Therefore, the graph of a sin it + d has to move s = 6 units to the right, and so we get C 1 9.no sdi to aoutEv As to the value of 9, we observe the values of y start at its central position (y = 5) and go downward to its lowest value (y = -1). Similar to the argument used in determining c, the graph of y = esin ft th has to move Ti= 4 units to the right, implying that g # 4. 0) = (m - \\)janei obutilquis ads ,and'T Hence, We have the following equations of a and y in terms of tanga . * = 6sin *(t -6) -7 and y = 6sin # (124) 45. T E.8 When ? = 15, we get bun lovol son ads avoda it I ai iniog taowol offi seuspoll s .d = b fordd awollot di = 6sin # (15 - 6) -7=-7+3V/2~-2.76 sbutingem silT dhud moil boviedo as asisuloull eboniignd sodwine s ai inte sidenav A & olquinzel and bvisedo of zenb Lef auto offl Yd bolabom ad na , # 6sin (15 -4) +5- 5 + 3V2 ~9.24 2zoningid mixer That is, after traveling for 15 microseconds, the particle is located near the pointsup? olt (--2.76, 9.24). vitegannon tanol ardi 3 bre 0 unit atit (4) This is an identity because the right-hand side of the equation is obtained by rationalizing the denominator of the left-hand side.LESSON 3 Simplify the trigonometric expressions using fundamental trigonometric identities. The Fundamental Trigonometric Identities Reciprocal Identities SAINT MARISWITHCHASE IPR TO sin 0 = - cos 0 = tan 0 =- csc 0 sec 6 cot . 0 CSC 0 = sec 0 = cot 0 IS e nie sin 6 cos 8 tan e Quotient Identities IN DANNA MALLS cutty 0 209 tan 0 = sin 0 cos e cot 8 = cos sin e Inclusive Jives ISVS Pythagorean Identities IS sin2 0 + cos2 0 = 1 1 + tan2 0 = sec2 0 1 + cot2 0 = csc2 0 Even-Odd Identities an witity or a conditional equation. For sin (-0) = -sin 0 cos(-0) = cos 0 the variable m the domain that does tan(-0) = - tan 0 Example. Simplify: 1. sind cot 0 Solution: sin 0 cot 0 = sin 0 (cose sin 0 = cos 2. tan 0 cos 0 sec 0 Solution: cose (cos 0)(sec 0) = 1 3. cos' 0 + cos2 0 tan? A Solution: cos' 0 + cos' @ tan' 0 = (cos? @) (1 + tan? () = cos2 0 sec2 0 = 1 3. If sino= = and cos 0 > 0. Find cos 0. Solution. Using the identity sin? ( + cos2 0 = 1 with cos > 0, we have cos 0 = V1 - sin20 = /1 - (-3) 4. If sec 0 = = and tan 0