Supporting Australian Mathematics Project. A guide for teachers Years 11 and 12. Calculus: Module 15. The calculus of trigonometric functions

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1 Supporting Australian Mathematics Project A guie for teachers Years an Calculus: Moule 5 The calculus of trigonometric functions

2 The calculus of trigonometric functions A guie for teachers (Years Principal author: Peter Brown, University of NSW Dr Michael Evans, AMSI Associate Professor Davi Hunt, University of NSW Dr Daniel Mathews, Monash University Eitor: Dr Jane Pitkethly, La Trobe University Illustrations an web esign: Catherine Tan, Michael Shaw Full bibliographic etails are available from Eucation Services Australia. Publishe by Eucation Services Australia PO Bo 77 Carlton South Vic 3053 Australia Tel: ( Fa: ( info@esa.eu.au Website: 03 Eucation Services Australia Lt, ecept where inicate otherwise. You may copy, istribute an aapt this material free of charge for non-commercial eucational purposes, provie you retain all copyright notices an acknowlegements. This publication is fune by the Australian Government Department of Eucation, Employment an Workplace Relations. Supporting Australian Mathematics Project Australian Mathematical Sciences Institute Builing 6 The University of Melbourne VIC enquiries@amsi.org.au Website:

3 Assume knowlege Motivation Content Review of raian measure An important limit Differentiating trigonometric functions Applications of the erivatives Integrating trigonometric functions Simple harmonic motion Links forwar Inverse trigonometric functions The hyperbolic functions History an applications Answers to eercises

4 The calculus of trigonometric functions Assume knowlege The content of the moules: Trigonometric functions an circular measure Introuction to ifferential calculus Integration. Motivation It is an interesting eercise to sit back an think about how we have evelope the topic of trigonometry uring the earlier seconary school years. We commence by looking at ratios of sies in a right-angle triangle. This enable us to fin unknown sies an angles. We etene this to inclue non-right-angle triangles using the sine an cosine rules. In the moule Trigonometric functions an circular measure, we reefine the sine an cosine functions in terms of the coorinates of points on the unit circle. This enable us to efine the sine an cosine of angles greater than 90 an to plot the graphs of the trigonometric functions an iscover their perioic nature. The sine an cosine functions are use to moel perioic phenomena in nature, such as waves, ties an signals. Inee, these functions are use to moel all sorts of oscillatory motion arising in a range of subjects, incluing economics an ecology. In this moule, we continue this evelopment by applying the ieas an techniques of calculus to the trigonometric functions. For eample, if we wish to analyse the motion of a particle moelle by a trigonometric function, we can use calculus to fin its velocity an acceleration. The simplicity of the results obtaine by oing this is amazing an has wie-ranging impact in physics an electrical engineering, an inee in any area in which perioic motion is being moelle.

5 A guie for teachers Years an {5} Content Review of raian measure We saw in the moule Trigonometric functions an circular measure that angles can be naturally efine using arc length. We efine raian (written as c to be the angle subtene in the unit circle by an arc length of one unit. y (0, (,0 O (,0 (0, Since the circumference of the unit circle is π an the angle in one revolution is 360, we can relate the two units by π c = 360 or π = 80. (As usual, we will rop the superscript c when it is clear that the angle uner iscussion is in raians. Many commonly occurring angles can be epresse in raians as fractions of π. For eample, 60 = π 3 an 330 = π 6. We have also seen that the arc length l of a sector of a circle of raius r, containing an angle θ (in raians, is given by l l = r θ, θ r while the area A of the sector is given by O A = r θ.

6 {6} The calculus of trigonometric functions An important limit In orer to apply calculus to the trigonometric functions, we will nee to evaluate the funamental limit sin lim 0, which arises when we apply the efinition of the erivative of f ( = sin. It must be stresse that, from here on in this moule, we are measuring in raians. You can see that attempting to substitute = 0 into sin is fruitless, since we obtain the ineterminate form zero over zero. On the other han, if we try substituting say = 0. (in raians, then the calculator gives , an if we try = 0.0, we get We guess that, as approaches 0, the value of sin approaches. We will now give a nice geometric proof of this fact. Consier a sector OAB of the unit circle containing an acute angle, as shown in the following iagram. Drop the perpenicular BD to OA an raise a perpenicular at A to meet OB prouce at C. Since OA = OB =, we have BD = sin an AC = tan. O B D C A The area of the sector OAB is. The sector OAB clearly contains the triangle OAB an is containe insie the triangle OAC. Hence, comparing their areas, we have Area OAB Area OAC = OA BD OA AC = sin tan = sin tan. Now, since tan = sin cos an since sin > 0 for 0 < < π, we can ivie both inequalities by sin to obtain sin cos.

7 A guie for teachers Years an {7} Since cos approaches as approaches 0, we see that sin as 0. Taking the reciprocal, it follows that sin as 0. (Here we have use the pinching theorem an the algebra of limits, as iscusse in the moule Limits an continuity. Thus we can write sin lim =. 0 Note that this means that, if the angle is small, then sin. This fact is often use by physicists when analysing such things as a simple penulum with small angle. Eercise Using the limit above, prove that tan lim =. 0 Other relate limits can be foun by manipulating this basic limit. Eample Fin sin lim 0 3. Solution We can write sin 3 = sin 3. If we let u =, then as 0, we have u 0. Hence sin lim = lim sinu u 0 u = 3.

8 {8} The calculus of trigonometric functions Eercise Fin lim 0 sin3 + sin7. 5 Eercise 3 a Show that lim 0 cos =. b Deuce that cos, for small. Differentiating trigonometric functions The erivative of sine Since the graph of y = sin is a smooth curve, we woul like to fin the graient of the tangent to the curve at any point on it. Before oing this, we erive a useful trigonometric ientity that will assist us. Using the compoun-angle formulas, we have sin(a + B sin(a B = sin A cosb + cos A sinb (sin A cosb cos A sinb = cos A sinb. If we put C = A + B an D = A B, we can a these equations to obtain A = (C + D an subtract them to obtain B = (C D. Substituting these back, we obtain the sine ifference formula: ( C + D sinc sind = cos sin ( C D To fin the erivative of sin, we return to the first principles efinition of the erivative of y = f (: y = lim h 0 f ( + h f (. h Substituting y = sin, we have y = lim h 0 sin( + h sin. h.

9 A guie for teachers Years an {9} Applying the sine ifference formula, we have ( ( sin( + h sin cos + h sin h lim = lim. h 0 h h 0 h We can put u = h. Then u 0 as h 0. So the limit becomes y = lim cos( + u sinu u 0 u ( ( = lim cos( + u sinu lim u 0 u 0 u = cos = cos. We can thus conclue that (sin = cos. This result is both simple an surprising, an stuents nee to commit it to memory. The erivation above involve a number of ingreients an is often ifficult for stuents the first time through. Derivatives of other trigonometric functions Now that the erivative of sine is establishe, we can use the stanar rules of calculus the chain, prouct an quotient rules to procee. Since cos = sin ( + π, we can apply the chain rule to see that Thus ( (cos = sin ( + π = cos ( + π = sin. (cos = sin. This is also a simple an surprising result that nees to be committe to memory.

10 {0} The calculus of trigonometric functions The following graphs illustrate what is happening geometrically. If we raw the tangent to the curve y = sin at a point with 0 < < π, then the tangent clearly has positive graient, while a tangent to y = cos, in the same range, clearly has negative graient. y y = sin 0 y 4 y = cos 0 4 The erivative of cos can also be foun by using first principles. Eercise 4 a b Show that ( C + D cosc cosd = sin Show by first principles that sin ( C D. Eercise 5 (cos = sin. By writing tan = sin an applying the quotient rule, prove that cos (tan = sec.

11 A guie for teachers Years an {} Stuents nee to remember the erivatives of sin, cos an tan. The rules of calculus can also be use to fin the erivatives of the reciprocal functions. Eercise 6 Show that a b c (cosec = cosec cot (sec = sec tan (cot = cosec. These three erivatives nee not be committe to memory. Further eamples Eample Use the rules of calculus to ifferentiate each of the following functions with respect to : 4 sin( cos( 3 e 3 tan(4. Solution 3 ( 4 sin( = 6 cos( ( cos( = sin( + cos( ( e 3 tan(4 = 4e 3 sec (4 + 3e 3 tan(4. Eercise 7 Show that log e ( + sin cos = sec.

12 {} The calculus of trigonometric functions Applications of the erivatives Arme with the ability to ifferentiate trigonometric functions, we can now fin the equations of tangents to trigonometric functions an fin local maima an minima. Eample Fin the equation of the tangent to the curve y = sin + cos at the point = π. Fin the minimum value of y = sin + cos in the interval 0 π. Solution The graient of the tangent is given by y = cos sin = at = π. The y-value at this point is. Hence, the equation of the tangent is y = ( π or, equivalently, y + = + π. Since the function is continuous, the minimum will occur either at an en point of the interval 0 π or at a stationary point. The y-value at each enpoint is. To fin the stationary points, we solve y = 0. This gives cos = sin. To procee, we use a ouble-angle formula: cos = sin = cos = sin cos = ( sin cos = 0. Hence cos = 0 or sin =. The solutions in the range 0 π are = π, 3π, π 6, 5π 6. The smallest y-value, which is 3, occurs at = 3π. Hence the minimum value is 3. is increasing in each interval in which the enom- Eercise 8 sin Show that the function y = 3 + 4cos inator is not zero. Eercise 9 Suppose an isosceles triangle has two equal sies of length a an equal base angles θ. Show that the perimeter of the triangle is a( + cosθ. Deuce that, of all isosceles triangles with fie perimeter, the triangle of largest area is equilateral.

13 A guie for teachers Years an {3} Integrating trigonometric functions Since integration is the reverse of ifferentiation, we have immeiately that cos = sin +C an sin = cos +C. Thus, for eample, we can fin the area uner the sine curve between = 0 an = π, as shown on the following graph. Area = 0 sin = [ cos ] π 0 =. y y = sin 0 More generally, since sin(a + b = a cos(a + b an cos(a + b = a sin(a + b, we obtain, for a 0, cos(a +b = a sin(a +b+c an sin(a +b = cos(a +b+c. a From (tan = sec, we obtain sec = tan +C.

14 {4} The calculus of trigonometric functions Eample Fin ( + cos ( sin3 + sec. Solution 0 ( + cos = [ + sin ] π 0 = π 6 0 ( sin3 + sec = [ 3 cos3 + ] π tan 6 = Eercise 0 a b c Fin 3 π 6 (sin + cos3. Differentiate sin, an hence fin 0 cos. Use the ientity + tan = sec to fin tan. Special integrals The two integrals cos θ θ an sin θ θ require some special attention. They are hanle in similar ways. To procee, we make use of two trigonometric ientities (a ouble-angle formula an the Pythagorean ientity: cos θ sin θ = cosθ an cos θ + sin θ =. Aing these two ientities, we have cos θ = +cosθ, an so we can replace cos θ in the integral with ( + cosθ. Thus cos θ θ = ( + cosθθ = (θ + sinθ +C.

15 A guie for teachers Years an {5} Eercise Fin sin θ θ. (Follow the metho use for cos θ, but subtract the two ientities rather than aing them. Simple harmonic motion Simple harmonic motion (SHM is a special case of motion in a straight line which occurs in several eamples in nature. A particle P is sai to be unergoing simple harmonic motion when it moves backwars an forwars about a fie point (the centre of motion so that its acceleration is irecte back towars the centre of motion an proportional to its isplacement from the centre. O P Hence the isplacement of the particle P will satisfy the equation t = n, where n is a positive constant. This is an eample of a secon orer ifferential equation. It can be shown that the general solution to this equation is (t = A sinnt + B cosnt, where A an B are constants. In the case where the particle starts at the origin, so = 0 when t = 0, we have B = 0 an so the function (t = A sinnt is a solution to the ifferential equation. We can easily check this: (t = A sinnt = t = n A sinnt = n. In the general case, since any trigonometric epression of the form A sinθ + B cosθ can written in the form C sin(θ + α, we can write the general solution as (t = C sin(nt + α, where C an α are constants. The constant α is calle the phase shift of the motion (an as we saw above can be taken as 0 if the particle begins at the origin. From our knowlege of the trigonometric functions, we see that the amplitue of the motion is C an the perio is π n.

16 {6} The calculus of trigonometric functions We net erive a formula for the velocity of the particle, with the help of a very useful epression for acceleration: t = (v. We write the ifferential equation as (v = n an integrate with respect to to obtain v = K n, where K is a constant. If the amplitue of the motion is C, then when = C the velocity is 0, an so K = n C. Hence, we have v = n (C. Eample A particle is moving in simple harmonic motion. Fin a formula for the isplacement (t of the particle (with in metres an t in secons given that: the perio of the motion is 6 secons the particle passes through the centre of oscillation when t = the particle has a velocity of π m/s when t = 4. What is the amplitue of the motion? Solution The general equation for simple harmonic motion is = C sin(nt + α. Since the perio is 6, we have π n = 6, giving n = π 8. Also, since = 0 when t =, we have C sin(n + α = 0. As C 0, we may take n + α = 0, an so α = n = π 4. Finally, the velocity is given by v = t Cπ ( π = Cn cos(nt + α = 8 cos 8 t π. 4 Now, v = π when t = 4, giving Cπ 8 = π, an so C = 6. Thus the equation for isplacement is = 6 ( π sin 8 t π 4 an the amplitue is 6.

17 A guie for teachers Years an {7} Eercise Fin the spee of a particle moving in SHM when it is passing through the centre, if its perio is π secons an its amplitue is metres. Eample In a certain bay, there is a low tie of 6 metres at am an a high tie of 0 metres at 8am. Assuming that the tie motion is simple harmonic, fin an epression for the height at time t after am, an fin the first time after am when the tie is 9 metres. Solution 0m 8m 6m Let t be the time in hours since am, an let be the height of the tie in metres at time t. Since the centre of the motion is at 8 m an the amplitue is m, we can epress the height as = 8 + sin(nt + α. The perio is 4 hours, so π n = 4, giving n = π. Also, since = 6 when t = 0, we may 7 take α = π. Hence the height of the tie is ( π = 8 + sin 7 t π. (We can check that when t = 7, = 0, as epecte. ( π Now, when = 9, we have sin 7 t π =. The smallest solution is t = 4, that is, 3 4 hours 40 minutes after am. So the require time is 5:40am.

18 {8} The calculus of trigonometric functions Hooke s law An inetensible string is one which can bear a mass without altering its length. In practice, all strings are etensible, however the etension is usually very negligible. In the case of a string which is etensible, Hooke s law provies a simple relationship between the tension in the string an the etension it eperiences. It states that the tension in an elastic string (or spring is irectly proportional to the etension of the string beyon its natural length. Thus, if T is the tension in the spring, then T = k, where is the etension an k is a positive constant, sometimes calle the stiffness constant for the spring. Suppose that we have a particle of mass m attache to the bottom of a vertical spring with stiffness constant k. Since the system is stationary, the tension in the spring is given by mg, where g is the acceleration ue to gravity. By Hooke s law, the tension in the spring is also equal to ke, where e is the etension of the natural length of the spring after the mass is ae. We let the position of the particle at this stage be O, an then further epress the mass by a isplacement from O an release it. The tension T in the spring is now T = k(e + = mg + k. The resultant ownwar force is then F = mg T = k. m + This force prouces an acceleration: by Newton s secon law of motion, F = m t. So we have t = k m. This is the ifferential equation for simple harmonic motion with n = k. Hence, the m perio of the motion is given by π n = π m k. We can conclue that the larger the mass, the longer the perio, an the stronger the spring (that is, the larger the stiffness constant, the shorter the perio.

19 A guie for teachers Years an {9} Links forwar Inverse trigonometric functions The sine an cosine functions are not one-to-one, an therefore they o not possess inverses. To overcome this problem, we have to restrict their omains, an fin inverses for these functions with restricte omains. These issues are iscusse in the moule Functions II. We will briefly iscuss the sine function, an leave cosine as a irecte eercise. Note that sin0 = sinπ, an so our chosen omain cannot inclue both 0 an π. We can see that sin is a one-to-one function on the interval [ π, π ], but not on any larger interval containing the origin. We restrict the omain of y = sin to [ π, π ]. y y = sin (, 0 (, This restricte function, with omain [ π, π ] an range [,], is one-to-one. Hence, it has an inverse function enote by f ( = sin, which is rea as inverse sine of. (This inverse function is also often enote by arcsin. It is important not to confuse sin with (sin = ; these are two completely sin ifferent functions.

20 {0} The calculus of trigonometric functions The graph of y = sin is rawn as follows. y y = sin- (, 0 (, The omain of sin is [,] an its range is [ π, π ]. We can see from the graph that sin is an o function, that is, sin ( = sin. We can also see that it is an increasing function. An obvious (an interesting question to ask is What is its erivative? The erivative of inverse sine We recall from the moule Introuction to ifferential calculus that y y =. Let y = sin. Then = sin y, an so Hence y = cos y = sin y (since cos y 0 =. (sin =. We observe from the graph of y = sin that the graient of the curve is positive. We can also see that the graient approaches infinity as we approach = or =.

21 A guie for teachers Years an {} Using the chain rule, it is easy to show that ( sin a = a an so a = sin a +C, which gives us a new an important integral. Eercise 3 a Restrict the omain of y = cos to the interval [0,π]. Draw a sketch to show that this restricte function is one-to-one, an write own the omain an range of its inverse y = cos. b Eplain graphically why cos ( = π cos (, an fin cos (. c Show that the erivative of cos is. Let f ( = sin +cos, for [,]. Fin the erivative of f, an conclue that f is a constant function an fin its value. The hyperbolic functions In the 7th century, the mathematician Johann Bernoulli (an others stuie the curve prouce by a hanging chain. It ha been (incorrectly thought by some that the curve was a parabola, but Bernoulli showe that its equation is very ifferent. The curve is often referre to as a catenary (from the Latin wor for chain an, with appropriate choice of origin an scale, has the equation C ( = e + e. y 5 4 y = The catenary curve.

22 {} The calculus of trigonometric functions If we efine S( = e e, then it is easy to see that S ( = C ( an C ( = S(. Moreover, C (0 = an S(0 = 0, an so these functions bear some similarity to the trigonometric functions cos an sin although of course they are not perioic. (They o, however, have a comple perio. Hence, by analogy, C ( is written as cosh (pronounce cosh of an S( is written as sinh (usually pronounce shine of. These function have other similarities to the trigonometric functions. For eample, analogous to cos + sin =, we have cosh sinh =. There are many other interesting analogues. These functions are often calle the hyperbolic functions (hence the h, because this last ientity enables us to parameterise half of the hyperbola a y b = by = a cosh t, y = b sinh t. The set of functions consisting of polynomials, rational functions, eponential an logarithmic functions, trigonometric functions an hyperbolic functions is often referre to as the set of elementary functions, since they are the most commonly occurring an well stuie of all functions. Other functions which are not combinations of the above are sometimes referre to as special functions. f ( = 0 e t t. An eample of a special function is History an applications Some of the history of trigonometry was covere in earlier moules. The famous mathematician Euler, an his contemporaries Jacob an Johann Bernoulli, applie the ieas of calculus to the trigonometric functions proucing remarkable results. The mathematician Joseph Fourier, after whom Fourier series are name, applie the ieas of trigonometric series to solve physical problems in particular, he looke at the istribution of heat in a metal bar. One of the main moern applications of the trigonometric functions is to the analysis of signals an waves. Many ifferent types of waves arise in the stuy of alternating currents an signals.

23 A guie for teachers Years an {3} Sine Square Triangle Sawtooth All of our moern telecommunications an electronic evices were only mae possible by an unerstaning of the physics of electricity an the moern evelopment of electrical engineering. To give just a little insight into this, we are going to show how to buil a sawtooth wave using trigonometric functions. The basic ientities cos A cosb = ( cos(a B + cos(a + B sin A sinb = ( cos(a B cos(a + B sin A cosb = ( sin(a B + sin(a + B can be establishe by epaning the right-han sies. Using these ientities, we can easily prove that, if m,n are positive integers, then π 0 if m n, sinm sinn = π if m = n. A similar formula hols for cosm cosn. We can also show that π cosm sinn = 0, for any integers m,n.

24 {4} The calculus of trigonometric functions We will attempt to construct the function y =, for [ π,π], using an infinite sum of sine functions. We choose sine because the function y = is o, an so is the sine function. Thus we write: = b sin + b sin + b 3 sin3 + = b n sinn. n= Multiplying both sies by sin an integrating from π to π, we have π sin = b sin sin + b sin sin + = b π, so b = π π π sin. Using the same iea, we can see that b n = π π sinn. This integral can be foun using a technique known as integration by parts, giving π b n = n cosnπ. The first few coefficients are b =, b =, b 3 = 3,... an so we have = (sin sin + 3 sin3 4 sin4 + 5 sin5 sin6 +, 6 at least for π < < π. It nees to be mentione here that there are serious issues to be eamine in regar to convergence, since we have an infinite series, but these issues are beyon the scope of this moule. Graphically, we can chop the series off after a finite number of terms an plot. The following graph shows the sum of the first 0 terms.

25 A guie for teachers Years an {5} y 3 y = f( f( = cos(n sin(n n n = Since the trigonometric functions are perioic, the series is converging to a perioic function, which takes copies of the line y = an repeats them every π. This is an approimation of the sawtooth function shown at the start of this section. Note what happens near an at the jump iscontinuity! Answers to eercises Eercise tan ( sin lim = lim = =. 0 0 cos Eercise sin3 + sin7 ( 3 lim = lim sin3 ( 7 + lim sin7 = =. Eercise 3 a b Multiplying top an bottom by + cos gives cos sin ( lim 0 = lim 0 ( + cos = sin ( lim lim 0 0 Hence, for small, we have + cos =. cos, an so cos.

26 {6} The calculus of trigonometric functions Eercise 4 a We use two compoun-angle formulas: cos(a B = cos A cosb + sin A sinb cos(a + B = cos A cosb sin A sinb. Let C = A + B an D = A B. Then A = (C + D an B = (C D. So it follows that ( C + D cosc cosd = sin A sinb = sin sin ( C D. b cos( + h cos (cos = lim h 0 h h sin( + = lim sin h h 0 h ( = lim sin( + h ( sin h lim = sin. h 0 h 0 h Eercise 5 ( sin (tan = = cos + sin cos cos = cos = sec. Eercise 6 a (cosec = (sin = (sin cos = cos sin = cosec cot. b c (sec = (cos = (cos sin = sin cos = tan sec. (cot = ( cos sin = sin cos sin = cosec. Eercise 7 log e ( + sin cos = ( loge ( + sin log e (cos = cos + sin + sin cos = cos + sin + sin ( + sin cos = + sin ( + sin cos = = sec. cos

27 A guie for teachers Years an {7} Eercise 8 The erivative of the function is y = ( sin 3 + 4cos = 3cos + 4 (3 + 4cos. Since cos, it follows that y > 0 whenever 3+4cos 0. Hence, y is an increasing function wherever it is efine. Eercise 9 The perimeter of the triangle is P = a + a cosθ = a( + cosθ, so a = P ( + cosθ. The area of the triangle is A = Substituting for a gives A = ( a cosθ ( a sinθ = a cosθ sinθ = a sinθ. P sinθ 8( + cosθ. We want to fin θ in the range 0 to π then such that A θ = 0. By the quotient rule, if A θ = 0, 6P cosθ ( + cosθ + 6P sinθ ( + cosθ sinθ = 0. It follows after some calculation that (+cosθ (cosθ = 0. So cosθ = or cosθ =. Hence, for 0 < θ < π, the only solution is θ = π 3. The triangle is equilateral. Eercise 0 a b c 3 π 6 (sin + cos3 = [ cos + 3 sin3 ] π 3 We have ( sin = sin + cos. Hence 0 π 6 = 6. [ ] π cos = sin sin = π 0 0. (sec tan = = tan +C.

28 {8} The calculus of trigonometric functions Eercise We have sin θ = ( cosθ, an so sin θ θ = (θ sinθ +C. Eercise Since the perio is π, we have π n = π, so n =. The amplitue is C =. Hence, from v = n (C, when = 0, v = ±. Thus the spee is m/s. Eercise 3 a y (0, y = cos 0 (π, b This restricte function has omain [0,π] an range [,]. So its inverse has omain [,] an range [0,π]. y (,π 3 y = cos- ( (, 0 0 (,0 y = cos- ( (, π 3 Since cos π 3 =, we have cos ( = π 3 an so cos ( = π cos ( = π π 3 = π 3.

29 A guie for teachers Years an {9} c Let y = cos. Then = cos y, an so y = sin y = cos y =. (Note that 0 y π an so 0 sin y. Hence, y =. Let f ( = sin + cos, for [,]. Then f ( = = 0. So f ( = C, for some constant C. Now f (0 = π an so sin + cos = π.

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Here the units used are radians and sin x = sin(x radians). Recall that sin x and cos x are defined and continuous everywhere and

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