Use of Trigonometric Functions

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1 Unit 03 Use of Trigonometric Functions 1. Introduction Lerning Ojectives of tis UNIT 1. Lern ow te trigonometric functions re relted to te rtios of sides of rigt ngle tringle. 2. Be le to determine te unknown side(s) nd/or ngles of tringle using trigonometric functions nd identities. 3. Solve rel world prolems involving trigonometric functions. Te term trigonometry mens mesurement (-metry) involving tringles (trigon). In plne trigonometry our suject of interest consists of plne ngles nd tringles. However, in trigonometric prolems tere my not lwys e n explicit mention of tringles. Solving trigonometric (trig.) prolems will involve determining te mesure of unknown ngles or sides of tringles from given informtion. You my wonder wy develop wole suject out te property of tringles. To pprecite te importnce of trigonometry, consider tis: ny plne geometricl spe cn e ultimtely roken up into tringulr prts. Furtermore, ny tringle cn e tougt of s consisting of no more tn two rigt-ngle tringles. Tus, trigonometry relly consists of properties of rigt- ngle tringle. Tese properties cn ten e employed to study ny figure in plne geometry. In tringle, s you ve lredy lerned, te sum of te internl ngles equls Tus, knowing ny two ngles of tringle, you know te spe of tt tringle. In te cse of rigt-ngle tringle, you only need one dditionl ngle to determine te spe of te tringle. Tis mens tt ll rigt-ngle tringles, wic ve te sme one cute ngle re similr nd terefore te rtios of teir corresponding sides will e only function of tis specified cute ngle. Tese rtios nd te reltions mong tem is te suject of trigonometry. 1

2 Te side opposite te specified cute ngle is clled te ltitude, wic is perpendiculr to te se. Te remining side (te longest side of te tringle) is clled te ypotenuse. In te figure elow te ltitude, te se, nd te ypotenuse re, respectively, represented y,, nd. 2. Trigonometric Functions Tere re only two independent trig. functions: sineθ = sinθ = / cosineθ = cosθ = /. All oter trig functions cn e otined from tese two functions. For exmple: tngentθ = tnθ = / = (/)/(/) = sinθ/ cosθ cotngentθ = cotθ = 1/ tnθ = / cosecntθ = cosecθ = 1/ sinθ = / secntθ = secθ = 1/ cosθ = / Te cosine, cotngent, nd cosecnt functions re clled te co-functions, respectively, of te sine, tngent nd te secnt. 3. Te rdin mesure of n ngle B C o r q l A 2

3 A plnr ngle θ is defined s rtio of n rc lengt to its rdius. For exmple if te rc of rdius r s lengt AB = l, ten AOB = = l / r. Te unit of mesurement of n ngle defined in tis mnner is rdin. Since te rc lengt ABC = πr, AOC = r / r = rdins = Smll Angle Approximtion In te tringle (see figure) sinθ = / = /( ) 1/2 wen θ is very smll, <<, nd terefore cn e neglected in comprison to in te denomintor. Tus ( ) 1/2 ~. Terefore, sinθ = 1/(1+(/) 2 ) 1/2 = / = tnθ. Furtermore, for smll ngles, ~ nd terefore / = θ. Tus, wen ngles re smll, sinθ ~ tnθ θ ( mesured in units of rdin). You my e curious out exctly ow smll is smll ngle. Te term smll is oviously reltive term. An ngle cn e considered smll, for te purpose of te ove pproximtion, s long s it is smll frction of rdin. 5. Signs of Trig. Functions y-xis In te xy-coordinte system, te position of ny point cn e expressed in terms of two numers clled te x- nd P(x,y) y-coordintes. In tis co-ordinte system, te x- nd y-xes re perpendiculr (just like te ltitude nd O r x y x-xis te se in rigt-ngle tringle) to ec oter. Suc co-ordinte system is lso clled n ortogonl co-ordinte 3

4 system. Te ortogonlity of te xy-coordinte system offers convenient wy to express te trig. functions in terms of te coordintes of point on circle. sinθ = ordinte/rdius = y/r cosθ = sciss/rdius = x/r tnθ = ordinte /sciss = y/x Since scisss (x-coordintes) nd te ordinte (y-coordintes) cn e positive or negtive depending upon wic qudrnt te point on te circle lies in, te numericl vlues of te vrious trig, functions cn e positive (+) or negtive ( ). For exmple, in te second qudrnt (ngles greter tn 90 o ut less tn 180 o ), te x-coordinte of point will e negtive ut te y-coordinte will e positive, terefore sine of ngles in te second qudrnt will e positive, ut te cosine nd te tngent functions will e negtive. You cn exmine to convince yourself tt only te tngent function is positive in te tird qudrnt nd only te cosine function will e positive in te fourt qudrnt. 5. Solved Exmples Exmple 1 Prove te identity sin 2 θ +cos 2 θ = 1 In te rigt-ngel tringle of sides,, nd (see Figure) sinθ = /, nd cosθ =/ Terefore, sin 2 θ +cos 2 θ = (/) 2 + (/) 2 = (( )/ 2 = 1 ( since = 2 from te Pytgoren teorem) Exmple 2 Given, sinθ = 0.6, wt is te vlue of tnθ? 4

5 cos 2 θ = 1 - sin 2 θ cosθ = [1 - sin 2 θ] 1/2 =[ ] 1/2 = 0.8 Terefore, tnθ = 0.6/0.8 = 0.75 Exmple 3. Te Lw of Cosines Consider tringle wose tree sides re,, nd c s sown. Te ngle etween nd is α. Sow tt: c = ( cos α) 1/2 c d From te Pytgoren teorem: c 2 = 2 + (+d) 2 = d 2 + 2d sustitute, d= cosβ nd 2 +d 2 = 2 in te ove eqution to get c 2 = cosβ = cos (180-)= cosα or, c = ( cos α) 1/2 (Note: te ove eqution is lso known s te Lw of Cosines. Te Lw of Cosines sttes tt te squre of ny one side of tringle is equl to te sum of te squres of te oter two sides minus twice te product of tese sides nd te cosine of te included ngle.) Exmple 4 In te tringle sown, find te unknown side. Using te Lw of Cosines: 37 o

6 = cos 37 0 = = 85 = 9.2 Exmple 5. Te Lw of Sines: In te tringle sown prove tt : c C 1 sinα/ = sinβ/ = sinγ/c A D B (Te eqution ove is lso known s te Lw of Lines wic sttes tt in ny tringle te lengts of te sides re proportionl to te sines of te opposite sides. Drw CD perpendiculr to AB. Ten, note tt: = sinγ ( from te rigt-ngle tringle CDB) = c sinα (from te rigt-ngle tringle CAB) Te equlity cn e rerrnged to red: sinα/ = sinγ/c.[1] Similrly, y drwing perpendiculr from B on to te extension of side AC, one cn write: 1 = sinδ = sinα sinα/ = sinδ/ = sin(180 β)/ = sinβ/.[2] Tus, comining eqn.[1] nd [2], sinα/ = sinβ/ = sinγ/c Exmple 6 In te tringle sown, find te unknown side c nd te ngle γ. Using te Lw of Sines: 8 sin = 10 sin 30 = c 30 o 8 6

7 Terefore, 8 = sin 1 & % 20' ( = γ = = Applying te Lw of Sines once more: c sin126.4 = 10 0 sin 30 = 20 0 c = 20sin = 16 Exmple 7 An ellipsoidl stronomicl oject, X of semi-mjor xis R is oserved from two strs S 1 nd S 2, 1.5 ligt-yers prt. Te ngles sutended y X t te two strs re s sown wit α = 3.0 sec nd β = 4.0 sec. Clculte R nd te distnce OS 2. Express your nswer in meters. (Speed of ligt, c =3x10 8 m/s.) S 1 S 2 R o X One ligt yer equls te distnce ligt trvels in one yer. Let d represent te distnce etween S 1 nd S 2. d = 1.5*365*24*60*60*3*10 8 = 1.42*10 16 m Note: 1 deg = 60 min; 1min = 60 sec; 180 deg = π rd. Tus, 1sec = π/60*60*180 rd. Using te smll ngle pproximtion: R/OS 1 = R/(d+OS 2 ) = 1.5 π/60*60*180 rd [1] R/OS 2 = 2 π/60*60*180 rd [2] Divide eqn.[2] y eqn.[1] to get: (d+os 2 )/ OS 2 = 4/3, or 3d+3OS 2 = 4OS 2 Tus, OS 2 = 3d = 3*1.42*10 16 = 4.26*10 16 m. Sustitute tis vlue of OS 2 in eqn.[2] to get : R = OS 2 * 2.0*4.85*10-6 = 4.25*10 16 *2.0*4.85*10-6 = 4.12*10 11 m = 4.12*10 8 km. 7

8 Exmple 8. An ntenn of eigt, on te top of 20 ft tll uilding is viewed from point A on te ground nd noter point B on te roof of te uilding. Te ngles sutended y te ntenn nd te distnces involved re sown in te digrm. Determine: [] te eigt of te ntenn. [] te widt, W of te uilding. From te figure: 63.5 o B /w = tn 63.5 o = 2, terefore = 2w, lso (+20)/(160-w) = tn37 o = 0.75, A 37 o 160 ft w (+20) = 0.75*(160-w).[1] Also, sustitute = 2w in eqn.[1] to get 2w + 20 = w, wic cn e solved to get, w = ft =36.4 ft, nd finlly = 2w = 2*36.36 = 72.7 ft Exmple 9 E Sow tt: sin(α+β) = sinαcosβ + cosαsinβ ' C Consider tringle ABC s sown. A D m n B m + n = c Drw perpendiculr from C to te se AB. Similrly drw perpendiculr from A to BC extended. Since δ+γ = 180 = α+β+γ, ngle δ = α+β 8

9 In te rigt-ngle tringle AEC, / = sin(α+β), or = *sin(α+β) nd in te rigt-ngle tringle AEB, /(m+n) = sinβ, = (m+n)sinβ = (cosα + cosβ) sinβ = sin(α+β) sin(α+β) = cosα sinβ + cosβ sinβ sin(α+β) = (/)cosβ sinβ + cosα sinβ...[1] Applying lw of sines to tringle ABC, gives, sinβ/ = sinα/, wic wen sustituted in te eqution ove gives: sin(α+β) = sinα cosβ + cosα sinβ. (For noter wy to prove tis identity, see Exmple 12 elow.) Exmple 10 Given, sin (α+β) = 0.6 cosβ+ 0.5 cosα, wt is te vlue of cos(α+β)? sin (α+β) = sinα cosβ+ cosα sinβ = 0.6 cosβ+ 0.5 cosα Compring te two sides gives: sinα = 0.6 or α = sin -1 (0.6) = 37 o sinβ = 0.5 or β = sin -1 (0.5) = 30 o Terefore, cos(α+β) = cos67 o =

10 sin( + ) cos sin cos sin Exmple 11 - Te Slider-Crnk Mecnism Te slider-crnk is n importnt mecnism for converting rottionl motion into reciprocting liner motion. Te figure depicts tree components te crnk rm, te connecting rm, nd te piston - of slider-crnk mecnism. o L Crnk Arm q Connector Arm R x PISTON P Slider-Crnk Te crnk rm cn rotte wit n xis of rottion out O. Te piston is constrined to move linerly ck nd fort (te reciprocting motion). Te slider-crnk eqution expresses te distnce of te piston from O in terms of R, L, nd θ. x = OP = O + P O = Lcosθ P = [R 2 2 ] 1/2 = [R 2 L 2 sin 2 θ ] 1/2 terefore, x = Lcosθ + [R 2 L 2 sin 2 θ ] 1/2 Exmple 12 Anoter wy to prove: sin(α+β) = sinαcosβ + cosαsinβ Explntion: In te digrm te ypotenuse AD s een ritrrily ssigned vlue of 1. Te digrm is supposed to prove te identity: sin(α+β) = sinαcosβ + cosαsinβ 1 D sin E C Exmine te digrm nd see if you re convinced. You cn use te sme drwing to sow tt: cos cos(α+β) = cosαcosβ - sinαsinβ A F B 10

11 Some Common Trigonometric Identities sin = / c cos = / c tn = sin cos = / csc = 1/ sin = c / sec = 1/ cos = c / cot = 1/ tn = / q c sin 2 θ + cos 2 θ = 1 tn 2 θ + 1 = sec 2 θ cot 2 θ + 1 = csc 2 θ In te following x nd y re ngles in rigt ngle tringle. sin(-x) = -sin(x) csc(-x) = -csc(x) cos(-x) = cos(x) sec(-x) = sec(x) tn(-x) = -tn(x) cot(-x) = -cot(x) tn(x ± y) = tn x ± tn y 1 tn x tn y sin2x = 2sin x cos x cos2x = cos 2 x sin 2 x = 2cos 2 x 1 = 1 2sin 2 x tn2x = 2 tn x 1 tn 2 x sin x + sin y = 2sin x + y 2 % & cos x ' y 2 % & cos x + cos y = 2cos x + y 2 % & cos x ' y 2 % & sin x sin y = 2sin cos x cos y = 2sin x y 2 % & ' cos x + y % 2 & ' x y 2 % & ' sin x + y 2 % & ' 11

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