ࡱ>  'bjbjޥ 8ǴǴ-  jjFFFZZZ8$dZ*?(!777>>>>>>>$@bCX>F7m6l777>jj!>7=7=7=7jVF!>7=7>7=7=7=!=QR8 7=>>0*?7=C;~C7="CFY=<777=77777>>7=777*?7777C777777777 0: 1.3. Inverse Trigonometric Functions In this section we look at the derivatives of the inverse trigonometric functions and the corresponding integrals. The inverse trigonometric functions are the inverses of the regular trigonometric functions. Restrictions are placed on the trigonometric functions to make the one-to-one. Here is a summary followed by more details on each one. Inverse Sine. y = sin-1x = arcsin x is the inverse function to x = sin y with the restriction -  \eq \f((,2) ( y (  \eq \f((,2). In other words (7) y = sin-1 x is that angle y such that -  \eq \f((,2) ( y (  \eq \f((,2) and sin y = x Here is the graph of x = sin y with the part where  \eq \f((,2)(y(  \eq \f((,2) shown in red. Also is a table of values of y=sin-1x and its graph.  Since x = sin y is an odd function, it follows that y = sin-1x is also, i.e. sin-1(- x) = - sin-1x Inverse Cosine. y = cos-1x = arccos x is the inverse function to x = cos y with the restriction 0 ( y ( (. In other words (9) y = cos-1 x is that angle y such that 0 ( y ( ( and cos y = x Here is the graph of x = cos y with the part where 0 ( y ( ( shown in red. Also a table of values of y=cos1x and its graph  There is a connection between cos-1 and sin-1. Proposition 1. cos-1(- x) = ( - cos-1x cos-1x =  \eq \f((,2) - sin-1x Proof. To show the first let y = cos-1x and z = cos-1(- x). Then cos y = x and cosz = - x. We need to show z = ( - y. However, cos((y)=cosy= - x. So z=(y which was to be shown. To show the secind first consider the case where 0 ( x ( 1. If we let y = cos-1x and z = sin-1x then cos y = x and sin z = x. We need to show y =  \eq \f((,2) - z. However, cos( \eq \f((,2) - z) = sin z = x. So y= \eq \f((,2) - z which was to be shown. Now consider the case 1 ( x ( 0. Then 0 ( - x ( 1. So cosx=(cos-1(- x) = ( - ( \eq \f((,2) - sin-1(- x)) = ( - ( \eq \f((,2) + sin-1(x)) =  \eq \f((,2)sin1x. // Inverse Tangent. y = tan-1x = arctan x is the inverse function to x = tan y with the restriction -  \eq \f((,2) < y <  \eq \f((,2). In other words y = tan-1 x is that angle y such that -  \eq \f((,2) < y <  \eq \f((,2) and tan y = x Here is the derivative of y = sin-1x and the corresponding integral. Inverse Cotangent. y = cot-1x = arccot x is the inverse function to x = tan y with the restriction 0 < y < (. In other words y = cot-1 x is that angle y such that 0 < y < ( and cot y = x Inverse Secant. y = sec-1x = arcsec x is the inverse function to x = sec y where y lies in one of the two intervals 0 ( y <  \eq \f((,2) and ( ( y <  \eq \f(3(,2) In other words y = sec-1 x is that angle y such that 0 ( y <  \eq \f((,2) or ( ( y <  \eq \f(3(,2) and sec y = x Inverse Cosecant. y = csc-1x = arccsc x is the inverse function to x = csc y where y lies in one of the two intervals 0 ( y <  \eq \f((,2) and ( ( y <  \eq \f(3(,2) In other words y = csc-1 x is that angle y such that 0 ( y <  \eq \f((,2) or ( ( y <  \eq \f(3(,2) and csc y = x Here is a summary of the derivatives of the inverse trigonometric functions. Let's see why some of Proposition 1. Show that  \eq \f(d,dx) \b(sin-1x) =  \eq \f(1,\r(,1 - x2))  \eq \f(d,dx) \b(cos-1x) =  \eq \f(- 1,\r(,1 - x2)) Proof. Let y = sin-1x so that x = sin y. Use implicit differentiation. Take the derivative of both sides of this last formula with respect to x to get 1 = (cos y)  \eq \f(dy,dx). Thus  \eq \f(dy,dx) =  eq \f(1,cos y) =  eq \f(1,\r(,1 - sin2y)) =  \eq \f(1,\r(,1 - x2)) which proves the sin-1 formula. The cos-1 follows from the sin-1 formula and cos-1x =  \eq \f((,2) - sin-1x. // Using the chain rule we get the following analogues of (10) (15) when x is replaced by an expression involving x. Example 1. Find  \eq \f(dy,dx) if y = sin-1(3x 4) Solution. Using (16) one has  \eq \f(dy,dx) =  \eq \f(d,dx) sin-1(3x 4) =  \eq \f(1,\r(,1 - (3x - 4)2))  \eq \b(\f(d,dx) (3x - 4)) =  \eq \f(3,\r(,1 - (3x - 4)2)). Example 2. Find  \eq \f(dy,dx) if y = sin-1 \eq \b(\r(,sin x)) Solution. Using (16) one has  \eq \f(dy,dx) =  \eq \f(d,dx) sin-1 \eq \b(\r(,sin x)) =  \eq \f(1,\r(,1 - (\r(,sin x))2))  \eq \b(\f(d,dx) (sin x)1/2) =  \eq \b(\f(1,\r(,1 - sin x))) (1/2) (sin x)-1/2 \b(\f(d,dx)(sin x)) =  \eq \f(cos x,2 \r(,1 - sin x) \r(,sin x)). Example 3. Find  \eq \f(dy,dx) if y = cos-1 \eq \b(\f(1 + 2cos x,2 + cos x)) Solution. Using (17) one has  \eq \f(dy,dx) =  \eq \f(d,dx) cos-1 \eq \b(\f(1 + 2cos x,2 + cos x)) =  \eq \f(- 1,\r(,1 - \b(\f(1 + 2cos x,2 + cos x))2))  \eq \b(\f(d,dx) \b(\f(1 + 2cos x,2 + cos x))) =  \eq \f(- (2 + cos x),\r(,(2 + cos x)2 - (1 + 2cos x)2))  \eq \b(\f((2 + cos x)\f(d,dx)(1 + 2cos x) - (1 + 2cos x)\f(d,dx)(2 + cos x),(2 + cos x)2)) =  \eq \f(- (2 + cos x),\r(,4 + 4cos x + cos2 x - 1 - 4cos x - 4cos2 x))  \eq \b(\f((2 + cos x)\b(- 2sin x) - (1 + 2cos x)\b(- sin x),(2 + cos x)2)) =  \eq \f(- 1,\r(,3 - 3cos2 x))  \eq \b(\f(- 4 sin x - 2sin x cos x + sin x + 2sin x cos x,2 + cos x)) =  \eq \f(3 sin x,\r(,3 - 3cos2 x) (2 + cos x)) =  \eq \f(\r(,3) sin x,\r(,1 - cos2 x) (2 + cos x)) =  \eq \f(\r(,3),2 + cos x) For the last formula we are assuming x lies in an interval where sin x is non-negative. If x lies in an interval in which sin x is negative then the answer would be  \eq \f(- \r(,3),2 + cos x). Each of the derivative formulas (10) (15) has a corresponding integration formula. However the integration formulas for the co functions don't give new integration formulas. Example 4. Find  \eq \i(,, \f(1,\r(,6 - 2x2))) dx Solution.  \eq \i(,, \f(1,\r(,6 - 2x2))) dx =  \eq \i(,, \f(1,\r(,6) \r(,1 - \f(x2,3)))) dx =  \eq \f(1,\r(,6))  \eq \i(,, \f(1,\r(,1 - \b(\f(x,\r(,3)))2))) dx. Let u =  \eq \f(x,\r(,3)). Then du =  \eq \f(dx,\r(,3)) and dx=  \eq \r(,3) du and the integral becomes  \eq \f(1,\r(,6))  \eq \i(,, \f(\r(,3),\r(,1 - u2))) du =  \eq \f(1,\r(,2)) sin-1u =  \eq \f(1,\r(,2))sin1\b(\f(x,\r(,3))).     1.1 - PAGE4 1.3 - PAGE1 (1) y = sin-1 x = that number y such that sin y = x and -  \eq \f((,2) ( y (  \eq \f((,2) (2) y = cos-1 x = that number y such that cos y = x and 0 ( y ( ( (3) y = tan-1 x = that number y such that tan y = x and -  \eq \f((,2) < y <  \eq \f((,2) (4) y = cot-1 x = that number y such that cot y = x and 0 < y < ( (5) y = sec-1 x = that number y such that sec y = x and 0 ( y <  \eq \f((,2) or - ( ( y <  \eq \f(3(,2) (6) y = csc-1 x = that number y such that csc y = x and 0 < y (  \eq \f((,2) or - ( < y (  \eq \f(3(,2) y x 1 xy = sin-1x001 \eq \f((,2) ( 1.57- 1-  \eq \f((,2) ( - 1.57 \eq \f(\r(,2),2) \eq \f((,4) ( 0.78-  \eq \f(\r(,2),2)-  \eq \f((,4) ( - 0.78 y x 1 xy = cos-1x - 1( -  \eq \f(\r(,2),2) \eq \f(3(,2) 0 \eq \f((,2) ( 1.57  \eq \f(\r(,2),2) \eq \f((,4) ( 0.78 10 (10)  \eq \f(d,dx) sin-1 x =  \eq \f(1,\r(,1 - x2)) (11)  \eq \f(d,dx) cos-1 x =  \eq \f(- 1,\r(,1 - x2)) (12) eq \f(d ,dx) tan-1 x =  \eq \f(1,1 + x2) (13) eq \f(d ,dx) cot-1 x =  \eq \f(- 1,1 - x2) (14) eq \f(d ,dx) sec-1 x =  \eq \f(1,x \r(,x2 - 1)) (15) eq \f(d ,dx) csc-1 x =  \eq \f(- 1,x \r(,x2 - 1)) (16)  \eq \f(d,dx) sin-1 u(x) =  \eq \f(1,\r(,1 - u(x)2))  \eq \f(du,dx) (17)  \eq \f(d,dx) cos-1 u(x) =  \eq \f(- 1,\r(,1 - u(x)2))  \eq \f(du,dx) (18) eq \f(d ,dx) tan-1 u(x) =  \eq \f(1,1 + u(x)2)  \eq \f(du,dx) (19) eq \f(d ,dx) cot-1 u(x) =  \eq \f(- 1,1 - u(x)2)  \eq \f(du,dx) (20) eq \f(d ,dx) sec-1 u(x) =  \eq \f(1, u(x) \r(,u(x)2 - 1))  \eq \f(du,dx) (21) eq \f(d ,dx) csc-1 u(x) =  \eq \f(- 1, u(x) \r(,u(x)2 - 1))  \eq \f(du,dx) (22)  eq \i(,, ) \eq \f(1,\r(,1 - x2)) dx = sin-1 x (23)  eq \i(,, ) \eq \f(1,1 + x2) dx = tan-1 x (24)  eq \i(,, ) \eq \f(1,x \r(,x2 - 1)) dx = sec-1 x %&'       »zzvhb jh=D jphN=h=D6CJaJhN=h=DCJaJjhN=h=DCJUaJ h=D6 h=DH*] h=D] h=D6] ht}] ht}5] jh&_6U]mHnHuh0S jh36U]mHnHuhdh=DhN=.&  y     k  F ( P pgd )gd )xgd )gd&_xgd5gdotgd3gd&_xgd3    % ' ( ) 8 9 C F G O P S T U V W X Y Z [ \ d e h i m r s v w x y } ̽̽視 jphN=h6CJaJhN=hCJaJhjhUhoth&_ hN=6 jhN= jphN=hN=6CJaJhN=hN=CJaJjhN=hN=CJUaJhN= h=DH*h=D h=D6hbh )0          $ % , - R S Y [ \ j k l o ʻշ|w|w|w|rw|kghd hbhd hbH* hb6hb h_kh&_jh&_UmHnHujh&_UmHnHtHu h_kH* h_k6h_kh&_h=D jphN=h6CJaJhN=hCJaJjhN=hCJUaJhjhU hot6hot jhot'o q t u v }        " # 0 1 2 3 4 5 6 7 @ A D E F G L ɾh3jh&_UmHnHtHu h&_H*h ) jph&_6 jh&_ h&_6h&_ h&_H*] h&_6] h&_] h&_5] jh )5U]mHnHuhd hot6hot hotH*7L \ ] d e | } ~        $ & ' , . / 6 7 ? @ C D ָֿֿֿֿֿ֦֮֗} jph_kh )6CJaJh_kh )CJaJjh_kh )CJUaJ jph )6 h )6 h )5 h )h ) h )H*jh3UmHnHuhdh ) h&_H* h3H*h3 jph&_6 jh&_ h&_6h&_1D L N O P V n o u w x } ~ IJKLMN]^dfglmsuvh_kh )CJaJjh_kh )CJUaJ jh ) jph )6 h )5 h )6 h )H*h )I:;<=>?KLOPQR^_bciknostxyҾҹҾҹҾҹ h )H* jph )6 jh ) h )6h )jh_kh )CJUaJh_kh )CJaJ jph_kh )6CJaJI!":;CDGHKLOPXY\]qrxz{|»әәә jphN=h )6CJaJhN=h )CJaJjhN=h )CJUaJ h )H*] h )6] h )] h )5] hEDwh ) h )6 h )H*h )jh_kh )CJUaJ jph_kh )6CJaJh_kh )CJaJ2#%&,./9:VW^_yz}~½¸± h )H* jph )6 h )H*] h )6] h )] h )5] h(eh(e h(eH* h(e6h(e h(eh ) jphN=h )6CJaJhN=h )CJaJjhN=h )CJUaJh ) h )66x NC\)  8gdh(Qgdh(Q$gdh(Qgd$gdt}gdYBgdVgd )gd )"#GHIJMNVWZ[`abcdehiqrsvw³£³ h )H*h )CJaJ jph )6 jphN=h )6CJaJhN=h )CJaJjhN=h )CJUaJ jh ) h )6h ) h )H*] h )6] h )]>   45<=DEijklopxy|}˿˥˝˥˘˥ h )H* jph )6 jh ) h )H*] h )6] h )] h )5] h )6h )jhN=h )CJUaJ jphN=h )6CJaJh )CJaJhN=h )CJaJ<  wxҾҾҾҨҡҔ~~ hotH* hot6hotjhotUh "bhzd0 hzd05 h(eh(ejh )UmHnHuh )CJaJ h )6 jh ) jph )6h ) jphN=h )6CJaJhN=h )CJaJjhN=h )CJUaJ0    !"+,34  !"#%&4;=O̶̶̶̶̶̶̶̶̶̾jh "bUhYB h "bH* h "b6h "bhzd0 hzd05 jhV5U]mHnHu hVH* hV6jhVUhVHOQfhxz{!()1678<=CEGHMNWY_clmuz{|ĻijĠ h6jhUhh(Qh h\ 5 h\ H* h\ 6h\ jh\ U h "b5hYBh "b jph_khV6CJaJh_khVCJaJjh_khVCJUaJ hV6hV hVH*6  $%+-.36>?@ABCLTXabjopqtu}~hh(QhN= h5 hH* h6jhUh h\ H*jh\ Uh\ h\ 6K!'(*+./7;<KLVWYZ\defmnvԾԾԾԾùùù괯jhh(QUhh(Q h l5 hh(Q5 h*6 hBB&6h*jhBB&U hBB&H*hBB& hH* h6jhUh hN=6hN=jhN=U?v{|}#$GHQRTUWXYZefgiz{hyhh(QEH H* hh(Q5 hh(QH*jhh(QUhh(Q hh(Q6V   "#$%'(/0CDSTZ[\]ijqrstvwxy   !&'0134;<JKXYZ[fgijqrjhh(QU hh(QH*hh(Q hh(Q6\5?t      ! # $ 3 4 C D E [!!" 8xgd0S  8gdtgd @  8gdh(Q%&HIop>?GHIPQjklprst}  ǿǺǺǿǺǺǿǺǺǿDZǿDZǺhNy h @H* h @6jh @Uh @ h l5 h @5 h`hh(Qh\jh lUmHnHujhh(QUhh(Q hh(Q6F  #$'(019:BDGHPRZ[`befqrsu        Խh[ hNyh @h hS@6 hS@H*hS@ hNyH*jhNyU hNy6jh @U h @6hNyhNyEHH*hNyh @C      ! 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