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# CALCULUS Math 2B

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## Popular in Mathematics (M)

This 7 page Class Notes was uploaded by Adam Crona on Saturday September 12, 2015. The Class Notes belongs to Math 2B at University of California - Irvine taught by Staff in Fall. Since its upload, it has received 45 views. For similar materials see /class/201871/math-2b-university-of-california-irvine in Mathematics (M) at University of California - Irvine.

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Date Created: 09/12/15
Jim Lambers IVIath 213 Fall Quarter 2004 05 Lecture 17 Notes These notes correspond to Section 75 in the text Inverse Trigonometric Functions We now wish to define inverses of the trigonometric functions starting with the sine function sinJ Unfortunately sinJ is not one to one on the domain 00 lt J lt 00 it oscillates between 1 and 1 However we can define an inverse function if we restrict ourselves to the domain rrE rrE on which sinJ one to one De nition 1 Inverse Sine Function The inverse sine function denoted by sirfl de ned by the equivalence sin lJy 22gt sinyJ ggygg 1gJgl 1 The sine and inverse sine functions satisfy the cancellation equations sin 1sinJ J E 3 J g E 2 2 2 and sinsin l J 1 3 J g 1 We can use the cancellation equations and the Chain Rule to compute the derivative of i sin 1 Since siny J it follows that do cost 7 1 4 JW which yields 19 1 cosy L41 V Since cos y 2 0 for 7r2 g y g rrE we can use the identity sin2 y cos2 y 1 to conclude cost 11 sin2 y 6 and then use the relation J sing to obtain 1 1 1 7sin 1ltJlt1 I 11 x 1 J2 39 The inverse cosine function written cos 1 J is defined similarly Unlike the sine function the cosine function is onetoone on the domain 0 7r De nition 2 Inverse Cosine Function The inverse cosine function denoted by cos lJu de ned by the equivalence cos Jy ltgt cosyJ Ogygrr 1gJgl 8 The cosine and inverse cosine functions satisfy the cancellation equations cos 1cosJ 0 3 J 3 7r 9 and coscos 1J J 1 lt J g 1 10 The graphs of sin 1 J and cos 1 J are shown in Figure 1 Using the same approach with sin J y5iff1X ycosilx quot 35 7 7 15 7 7 3 7 7 1 7 7 25 7 7 05 7 7 2 7 gt 0 gt15 7 7 7057 1 7 391 05 7 7 715 7 7 n 1 1 1 1 1 1 1 15 1 05 0 05 1 15 15 1 05 0 05 1 15 X X Figure 1 Graphs of the inverse sine function sin J and the inverse cosine function cos 1 J we can easin compute the derivative of cos 1 J 1cos 1 dJ 1 The tangent function tanJ is onetoone on the domain 7r2 rrE Note that we must use the open interval in this case since tanJ has vertical asynlptotes at J jar2 The inverse tangent function is therefore defined follows De nition 3 Inverse Tangent Function The inverse tangent function denoted by tan l de ned by the equivalence 1 7r 7r tan Jy ltgt tanyJ ltylt OOltJ39ltOO 12 The tangent and inverse tangent functions satisfy the cancellation equations tan 1tanJ J g lt J lt 13 and tantan 1J 39 00 lt J lt 00 14 The vertical asymptotes of tanJ translate to horizontal asymptotes in tan l specifically we have T 7 T lirn tan 1J 397 lim tan 1J L 1a xgtoo 2 39 law 2 The graph of tan l J is shown in Figure 2 To obtain the derivative of tan l we proceed with sin 1 Differentiating both sides of the relation tang J with respect to J yields t dy 2 1 16 sec yd and it follows from the identity sec2 9 1 tan2 9 that dz 1 J 2 1 dJ 1 tan 9 Since tang J we can conclude that d 7 1 7tan 17 72 00 lt J lt 00 18 dJ 1 J The differentiation rules introduced in this lecture can be reversed to obtain two new integration rules Q 1 sirfl J C 19 and 1 7 2 r 1 1 J2117 tan JC 20 By applying the Chain Rule to differentiate tanJa we can generalize the previous integration rule follows 1 1 71 J v 2 a2 dJ Stan C 21 Figure 2 Graph of inverse tangent function tan 1 The horizontal asymptotes at y 213 are indicated by the dashed lines Example 1 Evaluate the integral 2 1 it 1 22 1 41 J Solution Using the substitution u 21 we have du 2 11 and therefore 2 4 1 1 1 7 d 7 it do 23 i014JZJ 2 01u2l since u 0 when J 0 and u 22 2 4 when 7 2 We then have 141 1141l 71171 2 01Wdu tan alo tan 4 tan 0 tan 406629 24 Example 2 Evaluate 21r11gt 25 Solution Using the substitution u 12 we have du 21 11 and therefore 21 11 du t 7 v sin u C sin 112 C 26 C The quot erlnatirm j in with trigonometric identities can be used to simplify o certain expressions involving both trigonometric and inverse trigonometric functions the next two examples illustrate 1 Example 3 Simplify the expression sintarf Solution There are two approaches we can use The first approach an algebraic one uses the fact that tan1 sin1 cos1 and sec1 1 cos 1 well the identity sec2 1 1 tanz1 We have sin tan 1 1 sintaif1 1 71costan l 1 costan 1 1 1 tantan 1 71 sectan 1 xsec2tanquot1 1 1 tanZtanquot1 1 j 27 v1 12 The second approach is geometric Consider a right triangle where one of the angles is equal to 9 tan 1 Then from the cancellation equation tantan 11 1 we have tanB Since tanB is the ratio of the length of the side opposite the angle 9 to the length of the adjacent side we can assume that the length of the opposite side is 1 and the length of the adjacent side is 1 Since sintan 1 1 sin 9 and sinB is defined to be the ratio of the length of the opposite side to the length of the hypotenuse we simply need to compute this ratio The opposite side has length 1 and since the adjacent side has length 1 it follows that the hypotenuse has length v1 12 Therefore 71 sintan 28 Example 4 Simplify the expression tansin l Solution Proceeding in the last example we let 9 sin 1 so by the cancellation equation sinsin l 1 1 we have sin 9 Since sine is the ratio of opposite to hypotenuse we can choose the opposite side to have length 1 and the hypotenuse to have length 1 Since tangent is equal to ratio of opposite to adjacent and the adjacent side must have length v1 12 we have tansin 11 tan 9 29 Another approach is to use the fact that tan 9 sin 9 cos 9 and use the cancellation sinsin l 1 1 to obtain 1 7 sin sin 1 1 tansin 11 71 0 cossin 1 cossin 1 We can then use the identity cos2 9 1 sinz 9 to obtain 139 tansin 11 1 cossin 1 31 1 1 sinsirf Cl Example 5 Consider the diagram in Figure 1 Where should the point P be placed so to maximize the angle 9 Solution Let 1 be the distance along the base from the left edge to the point P Then because the tangent is defined to be the ratio of the length of the opposite side to the length of the adjacent side it follows that 5 t 7 32 an a J 9 Similarly we have 2 ti 3 7 an 3 J Because 9 is maximized when the angles 139 and 3 are minimized we can determine the correct placement of P by finding the value of 1 such that 7 5 7 2 ycr6tan l7tan 1 7 4 1 J 1 is a minimum To find the minimum we must determine where y 0 Differentiating with respect to 1 we obtain 1 1 1 W J i 7 12 1i2 5 1 6 Figure 3 Figure for exanlpie in which P nlust be placed to maximize 9 1 1 2 133 x2 13fx23 J2 Setting 1 0 yields the equation 2 3 124 x225 Crossmultiplying we obtain the sinlper equation 29 25 5M3 12 4 which upon expanding yields 2x25059 6xx2205x2 30x65

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