Trigonometric Functions

Transcription

1 72 Chapter 4 Trigonometric Functions 4 Trigonometric Functions To efine the raian measurement system, we consier the unit circle in the y-plane: (cos,) A y (,0) B So far we have use only algebraic functions as eamples when fining erivatives, that is, functions that can be built up by the usual algebraic operations of aition, subtraction, multiplication, ivision, an raising to constant powers Both in theory an practice there are other functions, calle transcenental, that are very useful Most important among these are the trigonometric functions, the inverse trigonometric functions, eponential functions, an logarithms In this chapter we investigate the trigonometric functions ÌÖ ÓÒÓÑ ØÖ ÙÒØ ÓÒ º½ When you first encountere the trigonometric functions it was probably in the contet of triangle trigonometry, efining, for eample, the sine of an angle as the sie opposite over the hypotenuse While this will still be useful in an informal way, we nee to use a more epansive efinition of the trigonometric functions First an important note: while egree measure of angles is sometimes convenient because it is so familiar, it turns out to be ill-suite to mathematical calculation, so (almost) everything we o will be in terms of raian measure of angles An angle,, at the center of the circle is associate with an arc of the circle which is sai to subten the angle In the figure, this arc is the portion of the circle from point (,0) to point A The length of this arc is the raian measure of the angle ; the fact that the raian measure is an actual geometric length is largely responsible for the usefulness of raian measure The circumference of the unit circle is 2πr = 2π() = 2π, so the raian measure of the full circular angle (that is, of the 360 egree angle) is 2π While an angle with a particular measure can appear anywhere aroun the circle, we nee a fie, conventional location so that we can use the coorinate system to efine properties of the angle The stanar convention is to place the starting raius for the angle on the positive -ais, an to measure positive angles counterclockwise aroun the circle In the figure, is the stanar location of the angle π/6, that is, the length of the arc from (,0) to A is π/6 The angle y in the picture is π/6, because the istance from (,0) to B along the circle is also π/6, but in a clockwise irection Now the funamental trigonometric efinitions are: the cosine of an the sine of are the first an secon coorinates of the point A, as inicate in the figure The angle shown can be viewe as an angle of a right triangle, meaning the usual triangle efinitions of the sine an cosine also make sense Since the hypotenuse of the triangle is, the sie opposite over hypotenuse efinition of the sine is the secon coorinate of point A over, which is just the secon coorinate; in other wors, both methos give the same value for the sine The simple triangle efinitions work only for angles that can fit in a right triangle, namely, angles between 0 an π/2 The coorinate efinitions, on the other han, apply 7

2 4 Trigonometric Functions Chapter 4 Trigonometric Functions to any angles, as inicate in this figure: Similarly, as angle increases from 0 in the unit circle iagram, the first coorinate of the point A goes from to 0 then to, back to 0 an back to, so the graph of y = cos must look something like this: 2π 3π/2 π π/2 π/2 π 3π/2 2π A (cos,) The angle is subtene by the heavy arc in the figure, that is, = 7π/6 Both coorinates of point A in this figure are negative, so the sine an cosine of 7π/6 are both negative The remaining trigonometric functions can be most easily efine in terms of the sine an cosine, as usual: tan = cos cot = cos sec = cos csc = an they can also be efine as the corresponing ratios of coorinates Although the trigonometric functions are efine in terms of the unit circle, the unit circle iagram is not what we normally consier the graph of a trigonometric function (The unit circle is the graph of, well, the circle) We can easily get a qualitatively correct iea of the graphs of the trigonometric functions from the unit circle iagram Consier the sine function, y = As increases from 0 in the unit circle iagram, the secon coorinate of the point A goes from 0 to a maimum of, then back to 0, then to a minimum of, then back to 0, an then it obviously repeats itself So the graph of y = must look something like this: 2π 3π/2 π π/2 π/2 π 3π/2 2π Eercises 4 Some useful trigonometric ientities are in appeni B Fin all values of θ such that sin(θ) = ; give your answer in raians 2 Fin all values of θ such that cos(2θ) = /2; give your answer in raians 3 Use an angle sum ientity to compute cos(π/2) 4 Use an angle sum ientity to compute tan(5π/2) 5 Verify the ientity cos 2 (t)/( sin(t)) = +sin(t) 6 Verify the ientity 2 csc(2θ) = sec(θ) csc(θ) 7 Verify the ientity sin(3θ) sin(θ) = 2cos(2θ)sin(θ) 8 Sketch y = 2sin() 9 Sketch y = sin(3) 0 Sketch y = sin( ) Fin all of the solutions of 2sin(t) sin 2 (t) = 0 in the interval [0,2π] Ì Ö Ú Ø Ú Ó º¾ What about the erivative of the sine function? The rules for erivatives that we have are no help, since is not an algebraic function We nee to return to the efinition of the erivative, set up a it, an try to compute it Here s the efinition: = sin(+ ) Using some trigonometric ientities, we can make a little progress on the quotient: sin(+ ) = cos +sin cos = cos +cos sin

3 43 A har it Chapter 4 Trigonometric Functions This isolates the ifficult bits in the two its cos an sin Here we get a little lucky: it turns out that once we know the secon it the first is quite easy The secon is quite tricky, however Inee, it is the harest it we will actually compute, an we evote a section to it Ö Ð Ñ Ø We want to compute this it: sin Equivalently, to make the notation a bit simpler, we can compute In the original contet we nee to keep an separate, but here it oesn t hurt to rename to something more convenient To o this we nee to be quite clever, an to employ some inirect reasoning The inirect reasoning is emboie in a theorem, frequently calle the squeeze theorem Figure 43 The squeeze theorem point A are (cos,), an the area of the small triangle is (cos)/2 This triangle is completely containe within the circular wege-shape region borere by two lines an the circle from (,0) to point A Comparing the areas of the triangle an the wege we see (cos)/2 /2, since the area of a circular region with angle θ an raius r is θr 2 /2 With a little algebra this turns into ()/ /cos, giving us the h we seek º THEOREM 43 Squeeze Theorem Suppose that g() f() h() for all close to a but not equal to a If a g() = L = a h(), then a f() = L This theorem can be prove using the official efinition of it We won t prove it here, but point out that it is easy to unerstan an believe graphically The conition says that f() is trappe between g() below an h() above, an that at = a, both g an h approach the same value This means the situation looks something like figure 43 The wiggly curve is 2 sin(π/), the upper an lower curves are 2 an 2 Since the sine function is always between an, 2 2 sin(π/) 2, an it is easy to see that 2 = 0 = 2 It is not so easy to see irectly, that is algebraically, that 2 sin(π/) = 0, because the π/ prevents us from simply plugging in = 0 The squeeze theorem makes this har it as easy as the trivial its involving 2 To o the har it that we want, ()/, we will fin two simpler functions g an h so that g() ()/ h(), an so that g() = h() Not too surprisingly, this will require some trigonometry an geometry Referring to figure 432, is the measure of the angle in raians Since the circle has raius, the coorinates of 0 Figure 432 Visualizing sin / To fin g, we note that the circular wege is completely containe insie the larger triangle The height of the triangle, from (,0) to point B, is tan, so comparing areas we get /2 (tan)/2 = /(2cos) With a little algebra this becomes cos ()/ So now we have cos cos A B

4 43 A har it 77 Finally, the two its cos an /cos are easy, because cos(0) = By the squeeze theorem, ()/ = as well Before we can complete the calculation of the erivative of the sine, we nee one other it: cos This it is just as har as /, but closely relate to it, so that we on t have to o a similar calculation; instea we can o a bit of tricky algebra cos = cos cos+ cos+ = cos2 (cos+) = sin2 (cos+) = cos+ To compute the esire it it is sufficient to compute the its of the two final fractions, as goes to 0 The first of these is the har it we ve just one, namely The secon turns out to be simple, because the enominator presents no problem: Thus, Eercises 43 sin(5) Compute cot(4) 3 Compute csc(3) 5 Compute π/4 cos cos(2) cos+ = sin0 cos0+ = 0 2 = 0 cos = 0 sin(7) 2 Compute sin(2) tan 4 Compute 6 For all 0, 4 9 f() Fin 4 f() 7 For all, 2 g() Fin g() 8 Use the Squeeze Theorem to show that 4 cos(2/) = 0 78 Chapter 4 Trigonometric Functions Ì Ö Ú Ø Ú Ó ÓÒØ ÒÙ º Now we can complete the calculation of the erivative of the sine: sin(+ ) = = cos +cos sin = 0+cos = cos The erivative of a function measures the slope or steepness of the function; if we eamine the graphs of the sine an cosine sie by sie, it shoul be that the latter appears to accurately escribe the slope of the former, an inee this is true: π/2 π 3π/2 2π π/2 π 3π/2 2π Notice that where the cosine is zero the sine oes appear to have a horizontal tangent line, an that the sine appears to be steepest where the cosine takes on its etreme values of an Of course, now that we know the erivative of the sine, we can compute erivatives of more complicate functions involving the sine EXAMPLE 44 Compute the erivative of sin( 2 ) sin(2 ) = cos( 2 ) 2 = 2cos( 2 ) EXAMPLE 442 Compute the erivative of sin 2 ( 3 5) sin2 ( 3 5) = (sin(3 5)) 2 cos = 2(sin( 3 5)) cos( 3 5)(3 2 5) = 2(3 2 5)cos( 3 5)sin( 3 5)

5 Eercises 44 Fin the erivatives of the following functions sin 2 ( ) 2 45 Derivatives of the Trigonometric Functions sin 2 Ö Ú Ø Ú Ó Ø ÌÖ ÓÒÓÑ ØÖ ÙÒØ ÓÒ º All of the other trigonometric functions can be epresse in terms of the sine, an so their erivatives can easily be calculate using the rules we alreay have For the cosine we nee to use two ientities, Now: cos = sin(+ π 2 ), = cos(+ π 2 ) cos = sin(+ π 2 ) = cos(+ π ) = 2 tan = cos = cos2 +sin 2 = cos 2 cos 2 = sec2 sec = (cos) = (cos) 2 ( ) = cos 2 = sectan The erivatives of the cotangent an cosecant are similar an left as eercises Eercises 45 Fin the erivatives of the following functions cos 2 sin(cos) 3 tan 4 tan/(+) 5 cot 6 csc 7 3 sin(23 2 ) 8 sin 2 +cos 2 9 sin(cos(6)) 0 Compute secθ θ +secθ Compute t t5 cos(6t) 2 Compute t 3 sin(3t) t cos(2t) 3 Fin all points on the graph of f() = sin 2 () at which the tangent line is horizontal 80 Chapter 4 Trigonometric Functions 4 Finallpointsonthegraphoff() = 2sin() sin 2 ()atwhichthetangent lineishorizontal 5 Fin an equation for the tangent line to sin 2 () at = π/3 6 Fin an equation for the tangent line to sec 2 at = π/3 7 Fin an equation for the tangent line to cos 2 sin 2 (4) at = π/6 8 Fin the points on the curve y = +2cos that have a horizontal tangent line 9 Let C be a circle of raius r Let A be an arc on C subtening a central angle θ Let B be the chor of C whose enpoints are the enpoints of A (Hence, B also subtens θ) Let s be the length of A an let be the length of B Sketch a iagram of the situation an compute θ 0 +s/ ÁÑÔÐ Ø º Ö ÒØ Ø ÓÒ We have not yet verifie the power rule, a = a a, for non-integer a There is a close relationship between 2 an /2 these functions are inverses of each other, each unoing what the other has one Not surprisingly, this means there is a relationship between their erivatives Let s rewrite y = /2 as y 2 = We say that this equation efines the function y = /2 implicitly because while it is not an eplicit epression y =, it is true that if = y 2 then y is in fact the square root function Now, for the time being, preten that all we know of y is that = y 2 ; what can we say about erivatives? We can take the erivative of both sies of the equation: = y2 Then using the chain rule on the right han sie: Then we can solve for y : ( ) = 2y y = 2yy y = 2y = 2 /2 = 2 /2 This is the power rule for /2 There is one little ifficulty here To use the chain rule to compute /(y 2 ) = 2yy we nee to know that the function y has a erivative All we have shown is that if it has a erivative then that erivative must be /2 /2 When using this metho we will always have to assume that the esire erivative eists, but fortunately this is a safe assumption for most such problems

6 46 Implicit Differentiation 8 Here s another interesting feature of this calculation The equation = y 2 efines more than one function implicitly: y = also makes the equation true Following eactly the calculation above we arrive at y = 2y = 2( /2 ) = 2 /2 So the single calculation leaing to y = /(2y) simultaneously computes the erivatives of both functions We can use the same technique to verify the prouct rule for any rational power Suppose y = m/n Write instea m = y n an take the erivative of each sie to get m m = ny n y Then y = mm mm = nyn n( m/n ) = m n n m m(n )/n = m n m/n This eample involves an inverse function efine implicitly, but other functions can be efine implicitly, an sometimes a single equation can be use to implicitly efine more than one function Here s a familiar eample The equation r 2 = 2 +y 2 escribes a circle of raius r The circle is not a function y = f() because for some values of there are two corresponing values of y If we want to work with a function, we can break the circle into two pieces, the upper an lower semicircles, each of which is a function Let s call these y = U() an y = L(); in fact this is a fairly simple eample, an it s possible to give eplicit epressions for these: U() = r 2 2 an L() = r 2 2 But it s somewhat easier, an quite useful, to view both functions as given implicitly by r 2 = 2 + y 2 : both r 2 = 2 +U() 2 an r 2 = 2 + L() 2 are true, an we can think of r 2 = 2 +y 2 as efining both U() an L() Now we can take the erivative of both sies as before, remembering that y is not simply a variable but a function in this case, y is either U() or L() but we re not yet specifying which one When we take the erivative we just have to remember to apply the chain rule where y appears r2 = (2 +y 2 ) 0 = 2+2yy y = 2 2y = y Now we have an epression for y, but it contains y as well as This means that if we want to compute y for some particular value of we ll have to know or compute y at that value of as well It is at this point that we will nee to know whether y is U() or L() 82 Chapter 4 Trigonometric Functions Occasionally it will turn out that we can avoi eplicit use of U() or L() by the nature of the problem EXAMPLE 46 Fin the slope of the circle 4 = 2 +y 2 at the point (, 3) Since we know both the an y coorinates of the point of interest, we o not nee to eplicitly recognize that this point is on L(), an we o not nee to use L() to compute y but we coul Using the calculation of y from above, y = y = 3 = 3 It is instructive to compare this approach to others We might have recognize at the start that (, 3) is on the function y = L() = 4 2 We coul then take the erivative of L(), using the power rule an the chain rule, to get L () = 2 (4 2 ) /2 ( 2) = 4 2 Then we coul compute L () = / 3 by substituting = Alternately, we coulrealize that the point is onl(), but use the fact that y = /y Since the point is on L() we can replace y by L() to get y = L() =, 4 2 without computing the erivative of L() eplicitly Then we substitute = an get the same answer as before In the case of the circle it is possible to fin the functions U() an L() eplicitly, but there are potential avantages to using implicit ifferentiation anyway In some cases it is more ifficult or impossible to fin an eplicit formula for y an implicit ifferentiation is the only way to fin the erivative EXAMPLE 462 Fintheerivativeofany function efine implicitlyby y 2 +y 2 = We treat y as an unspecifie function an use the chain rule: (y2 +y 2 ) = (y 2+y 2 )+2yy = y 2 +2yy = y 2 y = 2y 2 +2y

7 46 Implicit Differentiation 83 You might think that the step in which we solve for y coul sometimes be ifficult after all, we re using implicit ifferentiation here because we can t solve the equation y 2 +e y = for y, so maybe after taking the erivative we get something that is har to solve for y In fact, this never happens All occurrences y come from applying the chain rule, an whenever the chain rule is use it eposits a single y multiplie by some other epression So it will always be possible to group the terms containing y together an factor out the y, just as in the previous eample If you ever get anything more ifficult you have mae a mistake an shoul fi it before trying to continue It is sometimes the case that a situation leas naturally to an equation that efines a function implicitly EXAMPLE 463 Consier all the points(, y) that have the property that the istance from (,y) to (,y ) plus the istance from (,y) to ( 2,y 2 ) is 2a (a is some constant) These points form an ellipse, which like a circle is not a function but can viewe as two functions paste together Because we know how to write own the istance between two points, we can write own an implicit equation for the ellipse: ( ) 2 +(y y ) 2 + ( 2 ) 2 +(y y 2 ) 2 = 2a Then we can use implicit ifferentiation to fin the slope of the ellipse at any point, though the computation is rather messy Eercises 46 In eercises 8, fin a formula for the erivative y at the point (,y): y 2 = y +y 2 = y 2 = y 3 +y 2 4 4cossiny = 5 + y = 9 6 tan(/y) = +y 7 sin(+y) = y 8 + y = 7 9 A hyperbola passing through (8,6) consists of all points whose istance from the origin is a constant more than its istance from the point (5,2) Fin the slope of the tangent line to the hyperbola at (8,6) 0 Compute y for the ellipse of eample 463 If y = log a then a y = Use implicit ifferentiation to fin y 2 The graph of the equation 2 y+y 2 = 9 is an ellipse Fin the lines tangent to this curve at the two points where it intersects the -ais Show that these lines are parallel 84 Chapter 4 Trigonometric Functions 3 Repeat the previous problem for the points at which the ellipse intersects the y-ais 4 Fin the points on the ellipse from the previous two problems where the slope is horizontal an where it is vertical 5 Fin an equation for the tangent line to 4 = y at (2, 2) (This curve is the kampyle of Euous) 6 Fin an equation for the tangent line to 2/3 +y 2/3 = a 2/3 at a point (,y ) on the curve, with 0 an y 0 (This curve is an astroi) 7 Fin an equation for the tangent line to ( 2 +y 2 ) 2 = 2 y 2 at a point (,y ) on the curve, when y 0 (This curve is a lemniscate) Definition Two curves are orthogonal if at each point of intersection, the angle between their tangent lines is π/2 Two families of curves, A an B, are orthogonal trajectories of each other if given any curve C in A an any curve D in B the curves C an D are orthogonal For eample, the family of horizontal lines in the plane is orthogonal to the family of vertical lines in the plane 8 Show that 2 y 2 = 5 is orthogonal to y 2 = 72 (Hint: You nee to fin the intersection points of the two curves an then show that the prouct of the erivatives at each intersection point is ) 9 Show that 2 +y 2 = r 2 is orthogonal to y = m Conclue that the family of circles centere at the origin is an orthogonal trajectory of the family of lines that pass through the origin Note that there is a technical issue when m = 0 The circles fail to be ifferentiable when they cross the -ais However, the circles are orthogonal to the -ais Eplain why Likewise, the vertical line through the origin requires a separate argument 20 For k 0 an c 0 show that y 2 2 = k is orthogonal to y = c In the case where k an c are both zero, the curves intersect at the origin Are the curves y 2 2 = 0 an y = 0 orthogonal to each other? 2 Suppose that m 0 Show that the family of curves {y = m+b b R} is orthogonal to the family of curves {y = (/m)+c c R} Ä Ñ Ø º Ö Ú Ø We have efine an use the concept of it, primarily in our evelopment of the erivative Recall that f() = L is true if, in a precise sense, f() gets closer an closer to a L as gets closer an closer to a While some its are easy to see, others take some ingenuity; in particular, the its that efine erivatives are always ifficult on their face, since in f(+ ) f() both the numerator an enominator approach zero Typically this ifficulty can be resolve when f is a nice function an we are trying to compute a erivative Occasionally such its are interesting for other reasons, an the it of a fraction in which both numerator an enominator approach zero can be ifficult to analyze Now that we have

8 47 Limits revisite 85 the erivative available, there is another technique that can sometimes be helpful in such circumstances Before we introuce the technique, we will also epan our concept of it, in two ways When the it of f() as approaches a oes not eist, it may be useful to note in what way it oes not eist We have alreay talke about one such case: one-sie its Another case is when f goes to infinity We also will occasionally want to know what happens to f when goes to infinity EXAMPLE 47 What happens to / as goes to 0? From the right, / gets bigger an bigger, or goes to infinity From the left it goes to negative infinity EXAMPLE 472 What happens to the function cos(/) as goes to infinity? It seems clear that as gets larger an larger, / gets closer an closer to zero, so cos(/) shoul be getting closer an closer to cos(0) = As with orinary its, these concepts can be mae precise Roughly, we want f() = to mean that we can make f() arbitrarily large by making close enough a to a, an f() = L shoul mean we can make f() as close as we want to L by making large enough Compare this efinition to the efinition of it in section 23, efinition 232 DEFINITION 473 If f is a function, we say that f() = if for every N > 0 a there is a δ > 0 such that whenever a < δ, f() > N We can eten this in the obvious ways to efine f() =, f() = ±, an f() = ± a a a + DEFINITION 474 Limit at infinity If f is a function, we say that f() = L if for every ǫ > 0 there is an N > 0 so that whenever > N, f() L < ǫ We may similarly efine f() = L, an using the iea of the previous efinition, we may efine f() = ± ± We inclue these efinitions for completeness, but we will not eplore them in etail Suffice it to say that such its behave in much the same way that orinary its o; in particular there are some analogs of theorem 236 Now consier this it: 2 π 2 π 86 Chapter 4 Trigonometric Functions As approaches π, both the numerator an enominator approach zero, so it is not obvious what, if anything, the quotient approaches We can often compute such its by application of the following theorem THEOREM 475 L Hôpital s Rule For sufficiently nice functions f() an g(), if f() = 0 = g() or both f() = ± an a g() = ±, an if a a a f () f() eists, then a g () a g() = f () This remains true if a is replace by a g () or This theorem is somewhat ifficult to prove, in part because it incorporates so many ifferent possibilities, so we will not prove it here We also will not nee to worry about the precise efinition of sufficiently nice, as the functions we encounter will be suitable 2 π 2 EXAMPLE 476 Compute in two ways π First we use L Hôpital s Rule: Since the numerator an enominator both approach zero, 2 π 2 π = 2 π cos, provie the latter eists But in fact this is an easy it, since the enominator now approaches, so 2 π 2 π = 2π = 2π We on t really nee L Hôpital s Rule to o this it Rewrite it as an note that (+π) π π π π = π π sin( π) = since π approaches zero as approaches π Now as before (+π) π π = (+π) = 2π( ) = 2π π EXAMPLE 477 Compute in two ways

9 47 Limits revisite 87 As goes to infinity both the numerator an enominator go to infinity, so we may apply L Hôpital s Rule: = In the secon quotient, it is still the case that the numerator an enominator both go to infinity, so we are allowe to use L Hôpital s Rule again: = 4 2 = 2 So the original it is 2 as well Again, we on t really nee L Hôpital s Rule, an in fact a more elementary approach is easier we ivie the numerator an enominator by 2 : = = Now as approaches infinity, all the quotients with some power of in the enominator approach zero, leaving 2 in the numerator an in the enominator, so the it again is 2 EXAMPLE 478 Compute sec Both the numerator an enominator approach zero, so applying L Hôpital s Rule: Eercises 47 sec sectan = = 0 = 0 cos Compute the its cos (/t) 3 4 t + t 2 2t t+5 2/t /t t 3t+2 /t 2 y ++ y 7 8 y y 3 ( ( ) /4 9 0 t+ ) ((4 t) 3/2 8) t 0 t 88 Chapter 4 Trigonometric Functions ( t 0 + t + ) ( 2 t+ ) 2 t 2+ 2+(/) (u ) 3 3 u (/u) u 2 +3u (2/) +5/ / 3+ / /2 + /2 + / /3 + /4 t t t 9 t t cos 2 π/2 (π/2) t 20 t t+ 4t+ t+2 / / / (+5) ( 2 + ) ( 34 (+5) ( 36 (+5) ) ) 40 The function f() = has two horizontal asymptotes Fin them an give a rough 2 + sketch of f with its horizontal asymptotes