4 Trigonometric and Inverse Trigonometric Functions

Transcription

1 MATH983/954 Mathematics 0C/C. Geneal infomation fo the academic yea 0-03: Lectue: D Theodoe Voonov, Room.09, School of Mathematics, Alan Tuing Building, theodoe.voonov@mancheste.ac.uk. Lectues: Tuesdays am in Renold/D7 and Wednesdays 9am in Sackville/C4. Tutoials: Thusdays am : Geoge Begg/C3 (optomety only) and Fidays 9am : ooms announced on the webpage (foundation). Assessment: cousewok (deadline in week 4): 0%, cousewok (deadline in week 0): 0%, hou end of semeste eam: 80%. Couse webpage: 4 Tigonometic and Invese Tigonometic Functions 4. Recollection of tigonometic functions 4.. Angula measue Conside a cicle of adius. It is known that its cicumfeence equals π (in fact, this the definition of the numbe π) and aea of the disc equals π. θ Aea = π cicumf. = π ad Angles can be measued in degees and in adians. Notation: a o b ad. The notation fo the unit ad is commonly omitted. (So if no units ae indicated, that means adians.) A full cicle is defined to be 360. In paticula, it follows that a ight angle (a quate of a full cicle) is 90. Radians ae less abitay units of angle because they ae defined in tems of ac length. An angle of adian is defined to be the angle which makes an ac of length on a cicle of adius. Since the total ac length of a cicle is π, thee ae π adians in a cicle. So π ad = 360. Angles ae nomally measued anti-clockwise fom the -ais as indicated.

2 4.. Definitions of tigonometic functions Given a ight angled tiangle as in the diagam: y θ The side labelled is called the hypotenuse, the side labelled the adjacent and the side labelled y the opposite. The following functions ae defined fo the vaiable θ: sin θ = y cos θ = tan θ = y = sin θ cos θ cosec θ = y = sin θ sec θ = = cos θ cot θ = y = tan θ Note that cos θ = sin ( π θ), sin θ = cos ( π θ). The angles θ and π θ ae called complementay. (Hence the names: cosine, i.e., cosinus means complementi sinus.) These functions ae called tigonometic o cicula. Altenative notations: tan θ = tg θ, cot θ = ctg θ, sec θ = sc θ, and cosec θ = csc θ. The main tigonometic functions, which ae sine and cosine, ae define as above fo an acute angle θ, i.e., fo 0 θ π. Howeve, we may notice that cos θ and sin θ ae espectively the - and the y-coodinates of a point on the unit cicle. That immediately allows to etend

3 them to othe values of the angles and in paticula note thei peiodicity: cos(θ + π) = cos θ, sin(θ + π) = sin θ, tan(θ + π) = tan θ Main identities You should be familia with the following esult. Theoem (Pythagoas Theoem). In a ight-angled tiangle, the sum of the squae on the hypotenuse is equal to the sum of the squaes on the othe two sides, i.e., = + y whee is the length of the hypotenuse,, y, the lengths of the othe two sided. Poof. Stat with one tiangle: y and place thee moe identical ones aound it 3

4 y y The aea of the oute squae can be epessed as the aea of the inne squae plus the aeas of the fou tiangles: fom which we obtain ( + y) = + 4 y + y + y = + y, + y =, which is the statement of the Pythagoas Theoem. We can use the definitions of the tigonometic functions, togethe with the Pythagoas Theoem to obtain the following main identities satisfied by tigonometic functions. Theoem. Fo all values of the agument θ, cos θ + sin θ =. Poof. Fom the Pythagoas theoem we have ( ) ( y ) cos θ + sin θ = + = + y = + y = = 4

5 Coollay. Fo all values of θ fo which the functions ae defined: + tan θ = sec θ cot θ + = cosec θ Poof. These fomulas ae obtained by dividing thoughout by cos θ and sin θ espectively. Eample. Thee common ight angled tiangles ae: = π 45 = π 60 = π sin π = sin π = 6 4 sin π = 3 3 cos π = 3 cos π = 6 4 cos π = 3 tan π = 6 3 tan π = tan π = They ae obtained by the Pythagoas theoem. In paticula, one takes an equilateal tiangle and divide it into two to get the values fo tigonometic functions of π/3 and π/ Gaphs of tigonometic functions f(θ) = sin θ 0 π 5

6 f(θ) = cos θ 0 π 4 f(θ) = tan θ 0 π 4 6

7 Note some useful elations: sin( π cos( π sin( π θ) = cos θ θ) = sin θ + θ) = cos θ cos( π + θ) = sin θ sin(π + θ) = sin θ sin(π θ) = sin θ cos(π ± θ) = cos θ All of them can be seen fom the diagam of a unit cicle at the y plane. 7

8 4. Addition fomulas and futhe identities 4.. Addition fomulas Theoem 3 (Addition fomulas). Fo any angles A and B, sin (A + B) = sin A cos B + cos A sin B sin (A B) = sin A cos B cos A sin B cos (A + B) = cos A cos B sin A sin B cos (A B) = cos A cos B + sin A sin B Poof. To pove sin (A + B) = sin A cos B + cos A sin B conside the following diagams. cos B sin A cos B A cos A cos B B cos A sin B cos A B cos B p q A+B 90 In both diagams the lowe angle is 90 B, which poves that the angle at the top of the p ight hand diagam eally is B. Now we have cos B = sin (A + B) and p = cos B. Togethe q this gives sin (A + B) = p cos B = q cos B cos B = q and so: sin (A + B) = sin A cos B + cos A sin B as equied. To pove cos (A + B) = cos A cos B sin A sin B, note that sin (θ + π ) = cos θ and cos (θ + π ) = sin θ. (This can be seen fom the gaphs.) So now cos (A + B) = sin (A + B + π ) = sin A cos (B + π ) + cos A sin (B + π ) Thus cos (A + B) = sin (A + B + π ) = sin A sin B + cos A cos B giving the esult. 8

9 4.. Futhe tigonometic identities Othe identities may be obtained fom the fomulas sin (A + B) and cos (A + B). Theoem 4 (Addition fomulas fo tangent). tan(a + B) = tan(a B) = Poof. Use the addition fomula fo sine and cosine: tana + tanb tana tan B tan A tan B + tana tan B tan(a + B) = sin(a + B) cos(a + B) = sin A cos B + sin B cos A cos A cos B sin A sin B = The second fomula follows in the same way. Theoem 5 (Double angle fomulas). The latte identity may also be witten sin A cos B sin B cos A + cos A cos B cos A cos B sin A sin B cos A cos B sin θ = sin θ cos θ cos θ = cos θ sin θ = sin A + sin B cos A cos B sin B cos A cos B sin A = tana + tanb tana tan B cos θ = cos θ = sin θ. Poof. Let θ = A = B in the sum identities sin θ = sin (θ + θ) = sin θ cos θ + cos θ sin θ = sin θcos θ cos θ = cos (θ + θ) = cos θ cos θ sin θ sin θ = cos θ sin θ Using the identity cos θ + sin θ = to eliminate eithe cos θ o sin θ fom the identity fo cos θ completes the poof. It is possible to deduce geneal fomulas fo cos n and sin n. We shall not do that, but conside paticula eamples instead. Eample. cos 3 = cos( + ) = cos cos sin sin = (cos sin ) cos sin cos sin = ( cos ) cos cos ( cos ) = 4 cos 3 3 cos. Theefoe cos 3 = 4 cos 3 3 cos. 9

10 Eample 3. cos 4 = cos(3 + ) = (4 cos 3 3 cos ) cos (3 sin 4 sin 3 ) sin = 4 cos 4 3 cos 3 sin + 4 sin 4 = 4 cos sin 4 3 = 4 cos 4 + 4( cos ) 3 = 4 cos 4 + 4( cos + cos 4 ) 3 = 8 cos 4 8 cos +. Theefoe cos 4 = 8 cos 4 8 cos +. Fom the fomulas fo multiple angles, we can deduce fomulas fo powes (useful fo integation). Eample 4. Epess cos 3 in tems of multiple angles. Solution. We have cos 3 = 4 cos 3 3 cos, hence 4 cos 3 = cos cos and finally Theoem 6. sin θ = cos 3 = (cos cos ). 4 tan θ + tan θ, cos θ = tan θ + tan θ. Poof. Conside the standad fomulas fo double angle and divide by = cos θ + sin θ: sin θ = sin θ cos θ = Similaly fo cos θ: sin θ cos θ cos θ + sin θ = sin θ cos θ cos θ + sin θ cos θ = sin θ cos θ + sin θ cos θ = tan θ + tan θ. cos θ = cos θ sin θ = cos θ sin θ cos θ + sin θ = tan θ + tan θ. Remak. The above goup of fomulas is often e-witten in tems of = θ and efeed to as tangent half-angle fomulas : sin = tan + tan, cos = tan + tan. Theoem 7. ( X + Y sin X + sin Y = sin ( X + Y sin X sin Y = cos ( X + Y cos X + cos Y = cos ( X + Y cos X cos Y = sin ) ( X Y cos ) sin ) cos ) sin ) ( ) X Y ( ) X Y ( ) X Y 0

11 Poof. sin (A + B) = sin A cos B + cos A sin B sin (A B) = sin A cos B cos A sin B Adding, sin (A + B) + sin (A B) = sin A cos B Let X = A + B, Y = A B then A = (X + Y ), B = (X Y ) and ( ) ( ) X + Y X Y sin X + sin Y = sin cos Subtacting, i.e. sin (A + B) sin (A B) = cos A sin B ( ) ( ) X + Y X Y sin X sin Y = cos sin cos (A + B) = cos A cos B sin A sinb cos (A B) = cos A cos B + sin A sinb Adding, i.e. Subtacting, cos (A + B) + cos (A B) = cos A cos B ( ) ( ) X + Y X Y cos X + cos Y = cos cos cos (A + B) cos (A B) = sin A sin B i.e. ( ) X + Y cos X cos Y = sin ( ) X Y sin One can use the above fomulas to do the convese: to epess the poduct of sines and cosines via the sum. Eample 5. Conside the fomula ( X + Y cos X + cos Y = cos ) ( ) X Y cos.

12 Intoduce α = X+Y and β = X Y. We have α + β = X and α β = Y. Theefoe in the new notation the above fomula becomes cos α cos β = This fomula is useful fo integation. Let us conside some moe eamples. Eample 6. Calculate the poduct: ( cos(α + β) + cos(α β) ). cos 0 cos 40 cos 80. Solution: denote the poduct by ; multiply though by sin 0 ; we get sin 0 = sin 0 cos 0 cos 40 cos 80 = sin 40 cos 40 cos 80 = sin 80 cos 80 = sin 60 = 8 sin(80 0 ) = 8 sin 0, by a epeated application of the double angle fomula. Theefoe cos 0 cos 40 cos 80 = 8.

13 4.3 Invese tigonometic functions Fistly, ecall the idea of an invese function. Eample 7. The functions y = ep and y = ln ae mutually invese functions. That means the following. Suppose > 0. If we take ln and then apply the eponential, the we etun to the oiginal : ep(ln ) =. Similaly, if we take fist ep and then apply logaithm, we etun to the oiginal : ln(ep ) =. The effect of one function is undone by its invese. The situation is not always that simple. Eample 8. Conside the functions y = and y =. It is natual to think that they ae mutually invese functions. Indeed, ( ) =. Note that this makes sense fo > 0 only. Conside also. Hee can be both positive and negative. Is it tue that =? The symbol has two intepetations. If we agee that 0 always (the aithmetic squae oot ), then fo 0 = =. fo 0 Altenatively, we may agee that takes two values (positive and negative), e.g., 4 = ±. Then = ±. In the sequel we stick to the fist intepetation; so a fo any a 0 will always mean fo us the aithmetic squae oot, a 0, and we wite ± a fo the geneal squae oot. 3

14 To summaize, when we solve an equation f() = y () fo (so y is given), two things may happen:. Solution eists not fo all values of y ;. Fo a given y, thee may be many solutions. In the latte case it is often said that the invese function f, which sends y to if f() = y (the invese function fo a function f) is multi-valued. Typically, one paticula value (paticula solution) is chosen as the pincipal value (o pincipal banch). Othe solutions of the equation () ae epessed in tems of the pincipal solution given by the pincipal value. Eample 9. The aithmetic squae oot y (defined fo y 0) is the pincipal value of the invese function fo the function f() =. All solutions of the equation = y ae epessed as (whee y is the aithmetic squae oot). = ± y Now we poceed to the invese functions of the tigonometic functions. Definition. Let sin = y. Then wite = Acsin y (ponounced ac sine ; capital A) fo any angle such that sin = y. Similaly fo cos and tan. Note the capital lettes used hee fo Acsin, Accos and Actan. We ae not obliged to use y fo the agument of the invese functions. In the sequel we usually wite Acsin, Accos and Actan o use any othe lette fo the agument which is suitable. Clealy, the so defined functions (called invese tigonometic functions o cicula functions) ae multi-valued because, in paticula, the functions sin, cos and tan ae peiodic. The gaphs of y = Acsin and y = Accos ae shown on the net page. 4

15 π π π π 0 y = acsin Pincipal Values y = accos 0 π y = Acsin To get a unique value, we specify the ange. y = Accos Definition. y = acsin is the value of y = Acsin such that π y π. y = accos is the value of y = Accos such that 0 y π y = actan is the value of y = Actan such that π y π. Remak. Invese tigonometic functions with lowe case such as acsin, accos and actan ae single-valued functions egaded as the pincipal values of the multi-valued Acsin, Accos and Actan. (Thei position is simila to that of the aithmetic squae oot a, a 0, in elation to the double-valued squae oot ± a.) Thee is an altenative notation fo them, namely, sin, cos and tan. Because of the dange of confusion with powes such as sin = (sin ), it is pefeable to use the ac notation. (Its oigin is in Latin acus, ac, so that, e.g., ac sine means the ac length of a unit cicle coesponding to a given sine, since ac length fo a unit cicle coincides with angle measued in adians.) The geneal values of the invese tigonometic functions ae epessed via thei pincipal 5

16 values as follows. Acsin = acsin + kπ o π acsin + kπ = ( ) n acsin + nπ (k, n = 0, ±, ±, ±3,...), Accos = ± accos + kπ (k = 0, ±, ±, ±3,...), Actan = actan + kπ (k = 0, ±, ±, ±3,...). The agument fo Acsin and Accos should be between and ; the agument fo Actan can be any numbe. Eample 0. Some fequently met values of the ac functions: acsin 0 = 0, accos 0 = π, actan 0 = 0, acsin = accos = π 4, acsin 3 = π 3, acsin = π 6, actan = π 4. Theoem 8 (Popeties of invese tigonometic functions). acsin( ) = acsin, accos( ) = π accos, actan( ) = actan. Poof. Note that both sides of the above identities take values in the same ange. To pove them, it is sufficient to apply sin, cos and tan espectively to to both sides. Theoem 9 (Relations between invese tigonometic functions). acsin = π accos. acsin = actan. Poof. Conside the fist identity. Since accos takes values between 0 and π, the epession π accos takes values between π and π as is equied fo acsin. Theefoe it is sufficient to check that sine applied to the.h.s. gives. Indeed, ( π ) sin accos = cos(accos ) =. So the l.h.s. is indeed equal to the.h.s. Fo the second identity, note that the l.h.s. and.h.s. take values in the same ange by the definition of acsin and actan. Theefoe it is sufficient to check that tan applied to both sides gives the same numbe. Indeed, if α = acsin, then sin α = and π α π. In this ange, cos α = sin α =. Hence which completes the poof. tan acsin = tan α = sin α cos α =, 6

17 Conside eamples. Eample. Let cos α =. Find all values of α. Solution: We have α = Accos = ± accos + kπ = ± π + kπ. Note that π + π = 5π, so the solution can be witten in an altenative fom α = π 3 + kπ o α = 5π 3 + kπ. Eample. Find θ in the ange 0 θ < π such that tan θ = 5. Solution: The unique solution in the ange between π and π is θ = actan 5.37 (using a calculato o tables). Geneal solution of the equation tan θ = 5 is obtained by adding integal multiples of π. Howeve, adding any multiple of π to to θ = actan 5 takes it out of the ange 0 θ < π. Hence the answe is: θ = actan 5.37 Eample 3. Solve fo all values: cos = 7. Solution. We have cos = 7 o cos = 7. Theefoe = ± accos 7 + kπ o = π ± accos 7 + kπ. This combines into = ± accos 7 + kπ. Eample 4. Solve the equation: sin + 5 sin 3 = 0. Solution. We factoize: ( sin )(sin + 3) = 0, so sin = o sin = 3. The second equation has no solutions, so we have sin = = ( )k acsin + kπ = ( )k π + kπ. 6 Eample 5. Epess cos + 3 sin in the fom A sin( + 0 ) whee A and 0 ae to be detemined. Solution. We can wite cos + 3 sin = ( ) 3 cos + sin = ( 3 cos + 3 ) sin Now we look fo 0 such that cos 0 = 3 3 and sin 0 = 3. We can take 0 = acsin 3 = accos 3 3. So cos + 3 sin = 3 sin( + 0 ) whee 0 = acsin 3. 7