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21.2 Differentiating sin and cos 691 Proof that lim0 cos1 0 cos1 cos1 cos1 0 0 cos1 cos21 1 0 cos1 1 cos2 1 0 cos1 sin2 1 0 cos1 sin sin 0 cos1 But lim sin 1 and 0 so sin 0 0 cos1 1 1 lim cos1 100. 0 We have now shown that ! sin cos , as conjectured! The Chain Rule tells us that ! sin[] cos[] . Now that we have proven sin cos , the derivative of cos is easy to tackle by using the fact that cos and sin are related to one another by a horizontal shift. Looking back at the graphs of cos and its derivative it is easy to speculate that the derivative of cos is sin . Proof that the Derivative of cos is sin Observe that sin 2 cos ; replacing by 2 shifts the sine graph left 2 units. Similarly, cos 2 sin ; replacing by 2 shifts the cosine graph left 2 units. sin x = f(x) cos x = g(x) —2 —3 — 2 — 1 2 —1 2 3 2 2 x —2 —3 — 2 1 — —1 3 2 x 2 2 Figure 21.8 692 CHAPTER 21 Differentiation of Trigonometric Functions ! cos ! sin 2 cos 1 (by the Chain Rule) sin Combining this result with the Chain Rule gives us ! sin[] cos[] or informally ! sin[%&] cos[%&] [%&] ! cos[] sin[] ! cos[%&] sin[%&] [%&], where mess is a function of . Using the Chain Rule and either the Product or Quotient Rule allows us to Þnd the derivatives of tan , csc , sec , and cot . EXERCISE 21.2 EXERCISE 21.3 Show that Show that ! tan sec2 . ! sec sec tan . EXAMPLE 21.1 Differentiate the following. (a) 3 sin2 (b) 7 cos23 5 (c) tan SOLUTIONS (a) This is the product of 3 and sin[%&]. 3 sin23 cos22 3 sin262 cos2 (b) 7 cos23 5 7[cos3 5]2, so basically this is 7[%&]2 and its derivative is 14[%&] [%&], where the %& is cos3 5. Then the Chain Rule must be applied to cos3 5). 14 cos3 5sin3 53 42 cos3 5sin3 5 (c) tan2tan212. This is basically [%&]12, so its derivative is 1 [%&]12 [%&].Weknow%& tanstuff,so[%&] sec2stuffstuff. 12tan212 sec222 sec22 tan2 21.2 Differentiating sin and cos 693 An Excursion: Polynomial Approximations of Trigonometric Functions We spent a fair amount of energy proving that lim0 sin 1. Now that this fact is ours, we can get some mileage out of it. From lim0 sin 1 it follows that sin for very small values of .9 This approximation is excellent for very small values of , but as gets increasingly far from zero the approximation becomes poor and eventually is useless. For example, sin 0.1 0.0998334 ’’’0.1 and sin 0.02 0.0199986 ’’’0.02, but sin 3 0.1411200 ’’’, which is not close to 3. y y = x sin x x Figure 21.9 Historically the approximation sin for very small values of was used in ancient times, well before the development of calculus. Using it, sin 1 can be approximated with a greatdegreeofaccuracy.Fromthere, usingadditionformulasandknowledgeaboutspeciÞc triangles, trigonometric tables can be built up. The approximation sin , when used to estimate sin 1, gives sin 1 sin 180 180 0.01745329 ’’’, where the actual value of sin 1 0.0174524 ’’’. Theapproximationsin , thetangentlineapproximationofsin at 0, givesan estimate for sin that is too large if is positive and too small if is negative. (See Figure 21.9). This approximation can be improved upon. The approximation sin 36, the best cubic approximation to sin around 0, was used well before the seventeenth century. This gives a higher degree of accuracy for near zero. Using it to approximate sin 1 gives sin 1 sin180180 61803 0.0174524064. This is identical to the numerical approximation supplied by a calculator. The third degree polynomial approximation of sin around 0 can be improved upon by using a Þfth degree approximation, which can in turn be improved upon by using a seventh degree polynomial. (We use only polynomials of odd degree because sine is an odd function.) As the degree of the polynomial used to approximate sine increases, the accuracy of the approximation near 0 increases and the interval around 0 for which the approximation is reasonable enlarges as well. Amazingly enough, if we continue along in this way we can come up with an inÞnite ÒpolynomialÓ that is exactly equal to sin everywhere. These polynomial expansions are known as Taylor series, after the British mathematician Brook Taylor. (We take up Taylor series in Chapter 30). 9 This statement is true only if is given in radians. 694 CHAPTER 21 Differentiation of Trigonometric Functions By the seventeenth century mathematicians were using inÞnite polynomial expansions of functions, trigonometric and others. The polynomial expansion of sin is given by 3 5 7 sin 3 2 5 4 3 2 7 6 5 4 3 2 . If we use factorial notation, letting ! 12’’’3 2 1, this can be written as 3 5 7 sin 3! 5! 7! . The Þrst several terms can be used to approximate sin for small. For instance, sin0.10.1 0.13 0.15 , so sin0.10.1 0.13 0.09983 ’’’. Compare this with sin0.10.0998334166 ’’’. Using one more term of the polyno-mial gives sin0.10.1 0.13 0.15 0.0998334167. A good match! P R O B L E M S F O R S E C T I O N 2 1 . 2 1. Using the derivatives of sine and cosine and either the Product Rule or the Quotient Rule, show that ! tan sec2 . 2. Show that ! sec sec tan . 3. Find the Þrst and second derivatives of the following. (a) 5 cos (c) 0.5 tan (b) 3 sin2 (d) "2 sin cos 4. Differentiate the following. (a) cos2 (d) sin34 (b) cos2 (c) tan2 (e) 7[cos53] 5. Consider the function &0.3 sin . (a) For what values of does have its local maxima and local minima? (b) Is a periodic function? (c) Sketch the graph of &0.3 sin . 21.3 Applications 695 (d) What is the maximum value of for &0.3 sin for 0? At what -value is this maximum attained? Your answers must be exact, not numerical approximations from a calculator. Give justiÞcation that this value is indeed the maximum. Evaluate the following derivatives. is a differentiable function. 6. (a) ! sin (b) ! cos (c) ! sin 7. (a) ! cos (b) ! tan (c) ! tan Evaluate. 8. ! sin3 ln 3 9. ! cos2sin 10. ! sin3cos 2 11. ! sin2 12. ! 2cos7 13. ! &3 cos27 14. Find !! in terms of and . sin cos 15. Find . (a) sin (b) 3 tan32 (c) tan 3sec3 16. Why have we been telling you that radians are more appropriate than degrees when using calculus? Suppose is measured in degrees. Then cos cos radians cos where the argument is now in radians. Find the derivative. Is the derivative sin ? 21.3 APPLICATIONS Optimization EXAMPLE 21.2 What angle of launch will propel an object (such as a cannonball or a baseball) farthest horizontally? Thisquestionisvitallyimportanttoengineersandsportsmenalike.Ifweconsideronly the force of gravity (ignoring air resistance, the Coriolis effect, etc.), then it can be shown that the path the object will take is a parabola. In Section 20.7 we showed that if an object is ... - tailieumienphi.vn
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