STUDENT S COMPANIONS IN BASIC MATH: THE FOURTH. Trigonometric Functions

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1 STUDENT S COMPANIONS IN BASIC MATH: THE FOURTH Trigonometric Functions Let me quote a few sentences at the beginning of the preface to a book by Davi Kammler entitle A First Course in Fourier Analysis to emphasize the importance of trigonometric functions (or, in engineering literature, sinusoial functions): You can synthesize a variety of complicate functions from pure sinusois in much the same way that you prouce a major chor by striking nearby C, E, G keys on a piano.... It was Joseph Fourier, however, who evelope moern methos of using trigonometric series an integrals as he stuie the flow of heat in solis. Toay, Fourier analysis is a highly evolve branch of mathematics with an incomparable range of applications an with an impact that is secon to none. Due to this clear message, in selecting materials in elementary aspects of trigonometric functions here, your FOURTH COMPANION keeps Fourier analysis in min. Trigonometric functions are sometimes calle circular functions, because they have something to o with a circle (just as hyperbolic functions have something to o with a hyperbola). As you know, the ratio between the circumference C an the iameter D of a circle is π: C/D = π. If R is the raius of the circle, then D = R an hence C/D = π gives C = πd = π (R), or C = πr. From now on we assume that R = 1, that is, the circle we are looking at is a unit circle. Through the center O of this unit circle we raw a horizontal line an a vertical line, giving us the x axis an the y axis of a cartesian coorinate system. It follows from Pythagoras theorem that the unit circle is escribe by the equation x + y = 1 in this cartesian plane. QUESTION 1. How o we get this equation? Explain. 1

2 Take an arbitrary point P (x, y) on the circle. The raial line OP an the x axis etermines an angle θ, measure by raians, that is, the size of this angle is the arc length of the portion from the circle facing the angle. The cosine function an the sine function at θ is just the x coorinate an the y coorinate respectively of point P : cos θ = x, sin θ = y. Notice the basic relation: cos θ + sin θ = 1. (It is a bit unfortunate that cosine shoul lea the way instea of sine, but people ten to let sine lea.) QUESTION. Why o we have cos θ + sin θ = 1? EXAMPLE 3. (In each of the following cases, raw a figure to visualize the situation.) When P = (1, 0), we have θ = 0, x = 1 an y = 0. Hence cos 0 = 1 an sin 0 = 0. When P = (0, 1), we have x = 0 an y = 1. In this case θ (in egrees, is 90 o ) correspons to a quarter of a circle an hence is π/4 = π/ in raians. Thus we have cos π/ = 0 an sin π/ = 1. Similarly, when P = ( 1, 0) we have θ = π from which we get cos π = 1 an sin π = 0. When P = (0, 1) we can take either θ = 3π/ or θ = π/ an hence cos 3π/ = cos( π/) = 0 an sin 3π/ = sin( π/) = 1. EXERCISE 4. (a) Check that if θ correspons to one-eighth of the circle, we have θ = π/4 an x = y. Deuce that x = y = / an consequently cos π/4 = sin π/4 = 1/. (b) Check that if θ is one sixth of the circle, we have θ = π/3 an x = 1. (Hint: raw an appropriate equilateral triangle.) Deuce that cos π/3 = 1/ an sin π/3 = 3/. (c) Verify that cos π/6 = 3/ an sin π/6 = 1/. The cosine function is an even function an the sine function is an o function. In other wors, cos( x) = cos x sin( x) = sin x for all x. Recall that, a function f efine on the real line is an even function if f( x) = f(x) for all x an f is an o function if f( x) = f(x) for all x. QUESTION 5. Which of the following functions is an even function, an o function, or, neither even nor o? (a) cos x + sin x, (b) cos x sin x, (c) sin x, () cos 3 x sin 3 x.

3 Both sine function an cosine function are perioic functions of perio π: cos(x + πn) = cos x, sin(x + πn) = sin x; n = 0, ±1, ±,... for all x. Also, their averages over one perio are zero: 1 π π 0 cos x x = 0, 1 π π 0 sin x x = 0. (These ientities can be unerstoo easily from the graphs of sine an cosine functions.) QUESTION 6. In each of the following cases, what is the perio of the given function? (a) cos 4x, (b) sin 6x, (c) cos 4x + sin 6x, () cos x 4, (e) sin x 6, (f) cos x 4 + sin x 6. For the next four exercises (EXERCISES 7, 8, 9, 10), you nee to know a little bit of integration, especially cos x x = sin x + C an sin x x = cos x + C. EXERCISE 7. Verify that, if n is any nonzero integer, π 0 sin nx x = 0, π 0 cos nx x = 0. Why o we require n to be nonzero? The most basic ientity in trigonometry is the following aition formulas: cos(α + β) = cos α cos β sin α sin β sin(α + β) = sin α cos β + cos α sin β. (1) Other trigonometric ientities are more or less consequences of these two ientities. Let us consier a special case of (1): β = α = θ. They become cos θ = cos θ sin θ an sin θ = sin θ cos θ. The first ientity shoul be put together with cos θ + sin θ = 1: cos θ + sin θ = 1 cos θ sin θ = cos θ 3

4 If we take the sum an the ifference of these two ientities an then ivie by, we get cos θ = 1 + cos θ sin θ = 1 cos θ. () What is the big eal of the above ientities? Well, they tell you how to fin the inefinite integrals cos x x an sin x x from the known ientities cos x x = sin x + C an sin x x = cos x + C. EXERCISE 8. Fin cos x x an sin x x. EXERCISE 9. Fin cos 4 x x an sin 4 x x. (Hint: cos 4 x = (cos x) = ) Replacing β by β in (1), an noticing cos( β) = cos β, sin( β) = sin β, we obtain cos(α β) = cos α cos β + sin α sin β sin(α β) = sin α cos β cos α sin β (3) Aing or subtracting various ientities from (1) an (3) we can convert proucts in cosines an sines to sums or ifferences in cosines an sines. The etail is left to you as the following exercise. EXERCISE 10. Deuce from (1) an (3) the following cos α cos β = 1 (cos(α β) + cos(α + β)) sin α sin β = 1 (cos(α β) cos(α + β)) sin α cos β = 1 (sin(α + β) + sin(α β)) From these ientities an EXERCISE 7 euce that, for all integers m, n π 0 sin mx cos nx x = 0 (4) an for all integers m, n with m n, π 0 cos mx cos nx x = 0, π 0 sin mx sin nx x = 0. (5) Why o we nee to specify m n in the last two ientities? 4

5 Ientities (4) an (5) constitute the orthogonality relations for trigonometric functions, which are funamental in Fourier analysis. The most basic fact about trigonometric functions for ifferential calculus is, the limit of sin h/h is 1 as h tens to zero: lim h 0 sin h h = 1. (6) Here, h is ummy in the sense that nothing is change if we replace h by another letter in all of its occurences. For example, we can rewrite (6) as lim x 0 sin x/x = 1. EXERCISE 11. Use (6) to erive x sin x = cos x an x (Note that, accoring to the efinition of erivative, x Now you realize that (1) is neee. sin x = lim h 0 cos x = sin x. sin(x + h) sin x. h PROBLEM 1. What makes you think that (6) is correct? (Hint: The total length of all sies of a regular n gon inscribe in a unit circle tens to π as n.) EXERCISE 13. Fin the limit of 1 cos h h as h 0. (Hint: use the following trick: 1 cos h h = (1 cos h)(1 + cos h) h (1 + cos h) = sin h h cos h. If you know L Hospital s rule, you can avoi this trick.) EXERCISE 14. Accoring to Newton s secon law an Hooke s law, the motion of a mass spring system is escribe by the ifferential equation m x + kx = 0 (7) t where m is the mass an k is Hooke s constant. Both m an k are positive numbers. Check that x = A sin(at) + B cos(at) 5

6 k (representing oscillatory motion) is a solution to (7), where a = is the frequency an m A, B are arbitrary constants. Verify that this solution can be rewritten as M sin(at + b), where M = A + B is the amplitue an b is an appropriate constant. In Fourier analysis, one crucial step is to fin a close form for the sum S n = 1 + (cos x + cos x + cos 3x + + cos nx) (8) where n is an arbitrary positive integer. A VERY SLICK way to eal with it is to see what happens if both sies are multiplie by sin(x/): ( sin x ) S n = sin x + sin x cos x + sin x cos x + sin x cos 3x + + sin x cos nx. Starting with the secon term, apply the recipe sin α cos β = sin(α + β) + sin(α β) (see EXERCISE 10) to each term. Then we realize that many terms can be cancele. After some simplification an rearranging, we arrive at S n 1 + (cos x + cos x + cos 3x + + cos nx) = sin (n + 1)x sin x. PROBLEM 15. Follow the suggestion here to erive the last ientity. (A more natural way to erive this ientity is to regar cos kx as the real part of e ikx so that S n given by (8) becomes a geometric progression. See EXAMPLE 1 from the SIXTH COMPANION.) PROBLEM 16. By applying sin θ = sin θ cos θ repeately, we have 1 = sin π = cos π 4 sin π 4 = cos π 4 cos π 8 sin π 8 = By noticing that lim n n sin π/ n = π, erive π = cos π 4 cos π 8 cos π (an infinite prouct) n Rewrite the ouble angle formula cos 1 + cos θ + cos θ θ = as cos θ =. Starting with cos π/4 = /, use the last ientity to churn out: cos π 4 =, cos π 8 = +, cos π 16 = 6 + +,...

7 Hence erive the following π = which is calle Vieta s formula for π/. The tangent function is the quotient sine over cosine (in some books tg x is use instea of tan x) an the secant function is the reciprocal of cosine tan x = sin x cos x sec x = 1 cos x. Notice that the cosine function vanishes at half-integral multiples of π; in other wors, cos(n + 1 )π = 0 for all integers n. So both of tan x an sec x are unefine for x = (n + 1 )π. QUESTION 17. Why o we have 1 + tan x = sec x? EXERCISE 18. Use (1) to verify the aition formula for the tangent function: tan(α + β) = tan α + tan β 1 tan α tan β. Use this ientity an the fact that tan h lim h 0 h = lim h 0 sin h h cos h = 1 to euce that x tan x = 1 + tan x sec x. sin h 1 cos h Also, use lim = 1, lim h 0 h h 0 h = 1 to euce sec x = sec x tan x. (Certainly, these two ientities can also euce from (/x) sin x = cos x an (/x) cos x x = sin x by means of the quotient rule.) A rather amazing fact about the tangent function is that all trigonometric functions sin t, cos t, tan t, sec t etc. can be expresse as a rational function of tan(t/). Inee, sin t = sin t cos t = sin t cos t cos t = tan t sec t = tan t 1 + tan t an cos t = cos t sin t = cos t (1 tan t ) = 1 tan t sec t 7 = 1 tan t 1 + tan t.

8 EXERCISE 19. Verify that, setting u = tan t, we have sin t = u 1 u u 1 + u, cos t =, tan t =, sec t = 1 + u 1 + u 1 u 1 u, t = 1 + u u. These ientities can be use to integrate all rational functions of trigonometric functions. But this topic, interesting though, is normally not covere in a calculus course toay. The ientities in the above exercise seem to be rather mysterious. Actually, they can be visualize by means of so calle stereographic projection. Again, consier the unit circle x + y = 1 in the cartesian plane. The topmost point on the circle is N = (0, 1) an let us call it the North Pole. Take any point P (x, y) on the circle an let Q(u, 0) be the intersection of the line N P (Draw a picture!) u = an the x axis. Then we have x 1 y, x = u u + 1, y = u 1 u + 1. (9) EXERCISE 0. Verify the above ientities. (Hint: To verify u = x/(1 y), use the fact that N, P an Q are colinear. Then compute 1 + u = ((1 y) + x )/(1 y) = an use the fact that x + y = 1.) Let t be the angle between OP Then the angle between N P an OS, where S is the is the South Pole (0, 1). an N S is t/. EXERCISE 1. Check that u = tan t/, x = sin t an y = cos t. Then use (9) to verify the ientities in EXERCISE 19. The ientities x = u u + 1, y = u 1 u + 1 gives us a rational parameterization of the unit circle. A irect computation shows x + y = ( ) u 1 + u + ( u ) u = 1. Notice that, if u is a rational number, so are x an y, an henceforth a rational point (x, y) on the unit circle is prouce. All rational points on the unit circle can be obtaine in this way. Inee, if x an y are rational numbers, then so is u = x/(1 y). 8

9 A rational point gives us a Pythagoras triple an vice versa. (A triple (a, b, c) of integers is calle a Pythagoras triple if a + b = c, e.g. (3, 4, 5).) For example, letting u = 4, we have x = u/(u + 1) = 8/17, y = (u 1)/(u + 1) = 15/17 an hence we obtain the Pythagoras triple (8, 15, 17). You can easily check that = 17 (= 89). You may secretly take a large u to prouce a huge triple to perplex your friens. One amazing ientity in calculus concerning trigonometric functions is the ientity x ln sec x + tan x = sec x. (10) It is amazing because normally we expect that the erivative of a function looks more complicate, but in this case it is simpler! We nee this in calculus because it tells us what the integral of sec x is: sec x x = ln sec x tan x + C. Ientity (10) is quite straightforwar to check by means of the chain rule an the ientities x tan x = sec x an x sec x = tan x sec x. The expression ln sec x + tan x looks mysterious. Actually it occurs in the parametric equations of a curve calle tractrix. If you tow an object on the y axis by a string an walk along the x axis, the curve trace out by the object is a tractrix. The surface of revolution obtaine by rotating this curve about the x axis is calle a pseuo sphere. It has a constant negative curvature. (A usual sphere is a surface of a constant positive curvature.) EXERCISE. Check (10). QUESTION 3. What are cos x x, sin x x, tan x x an sec x x? Finally, we consier inverse trigonometric functions arcsine, arccosine an arctangent. Strictly speaking, none of the trigonometric functions is invertible. But this oes not bother us, if we are willing to narrow own their omains with care. For example, the function x = u is not invertible. This is because ifferent u values may give the same x, e.g. both u = an u = give x = 4. By if we impose the restriction that u is nonnegative, then u is uniquely etermine by x an we can write u = x. To express in a proper way, we say that u = x is the inverse function of x = u restricte to [0, ), the set of nonnegative numbers. Now we efine 9

10 u = Arcsin x = the inverse function of x = sin u restricte to [ π, π ]. u = Arccos x = the inverse function of x = cos u restricte to [0, π]. u = Arctan x = the inverse function of x = tan u restricte to ( π, π ). You shoul raw a picture in each case to see what is behin each restriction. To inicate the omain an the range of each of these inverse functions, we write: Arcsin : [ 1, 1] [ π, π ] (, Arccos : [ 1, 1] [0, π], Arctan : R π, π ). The erivatives of these inverse functions are x Arcsin x = 1, 1 x x Arccos x = 1, 1 x x Arctan x = x. EXERCISE 4. Verify the above ientities. The most fascinating inverse trig function is arctangent. Its erivative 1/(1 + x ) looks like a bell curve but it is not (because of its big tails on both sies). The total area of the region unerneath this curve an above the x axis is x M 1 + x lim M M x 1 + x = lim Arctan x M = π ( M M π ) = π. We normalize 1/(1 + x ) by iviing π. The function f(x) = 1 π x obtaine in this way is calle the Cauchy istribution in probability theory. There are some intriguing ientities giving us π in terms of arctangents: π = 16Arctan 1 5 4Arctan 1 (Machin s formula, 1706) 39 π = 1Arctan Arctan Arctan 1. (Ferguson s ientity) 1985 PROBLEM 5. Verify the above ientities. 10