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1、991 工 CH7-5 反三角函 1 7-5 Inverse trigonometric functions (反三角函) 我們知道 xsin 、 xcos 、 xtan 、 xcot 、 xsec 與 xcsc 在實上的範圍內是一對一函，因此如果想要定義反三角函，必須先限制三角函的定義域。 ( I ) arcsine Remark: (a) xsin xy12-1-22-2f(x)=sin x(b) 1,12,2:sin x 為 1-1 函 Definition (arcsine function x1sin) 2,21,1:sin111x such that 2,2,sinsin1=yxy
2、yx Remark: (a) xx =)sin(sin1, 2,2x ; xx =)sinsin(1, 1,1x (b) The graph of xy1sin= xy1-122f(x)=sin xy=1y=-10xy1-1220f(x)=sin-1xxy1-1220f(x)=sin xf(x)=sin-1xy=x(c) 2111)sin(xxdxd=， )1,1( x (d) xx11sin)(sin= ,1,1x 991 工 CH7-5 反三角函 2 ( II ) arc cosine Remark: (a) xy12-1-22-2 -f(x)=cos x(b) 1,1,0:cos x 為
3、 1-1 函 Definition (arc cosine function x1cos) ,01,1:cos111x such that ,0,coscos1=yxyyx Remark: (a) xx =)cos(cos1, ,0 x ; xx =)coscos(1, 1,1x (b) The graph of xy1cos= xy1-120f(x)=cos xy=1y=-1xy1-120f(x)=cos-1xxy1-1210f(x)=cos xf(x)=cos-1xy=x(c) 2111cosxxdxd=, )1,1( x (d) xx11cos)(cos= ,1,1x ： 991 工 C
4、H7-5 反三角函 3 Ex1: Evaluate the following expressions: (a) )23(sin1(b) )23(cos1(c) )3(coscos1(d) )21(sinsin(1Solution: Ex2 (a) Suppose )52(sin1= , find cos and tan (b) Find an alternative form )4(cot(cos1xin terms of x (III) arc tangent Remark: (a) xtan xy12-1-23-32f(x)=tan x(b) ),()2,2(:tan x 為 1-1 函
5、 991 工 CH7-5 反三角函 4 Definition (arc tangent function x1tan) )2,2(),(:tan111x such that )2,2(,tantan1=yxyyx Remark: (a) xx =)tan(tan1, )2,2(x ; xx =)tantan(1, ),( x (b) The graph of xy1tan= xy12-1-23-3220f(x)=tan xxy1 2-1-23-3022f(x)=tan-1xxy12-1-23-3220f(x)=tan xf(x)=tan-1xy=x(c) 2111tanxxdxd+=， ),(
6、 x (d) xx11tan)(tan= ,),( x (e) 2tanlim1=xx; 2tanlim1=xx(IV) arc secant Remark: (a) xsec xy12-1-22-2322232f(x)=sec x(b) ),11,()23,)2,0:sec x 為 1-1 函 991 工 CH7-5 反三角函 5 Definition (arc secant function x1sec) )23,)2,0),11,(:sec111x such that =yxyyx ,secsec1)23,)2,0 Remark: (a) xx =)sec(sec1, )23,)2,0x
7、 ; xx =)secsec(1, 1 x (b) The graph of xy1sec= xy12-1-22320f(x)=sec xxy12-1-22320f(x)=sec-1xxy2320f(x)=sec xf(x)=sec -1xy=x(c) 11sec21=xxxdxd， 1 x (一、三象限 ); 1|1sec21=xxxdxd， 1 x (一、二象限 ) (d) xx11sec)(sec+= , 1 x (一、三象限 ); xx11sec)(sec= , 1 x (一、二象限 ) (e) 2seclim1=xx; 23seclim1=xx991 工 CH7-5 反三角函 6 E
8、x4. Evaluate or simplify the following expressions: (a) )31(tan1(b) )2(sec1(c) )sin(tan1xSol: Epute the following derivatives (a) )1(sin21xdxd(b) )cos(sin1xdxdSol: (a) Definition: Six Inverse Trigonometric Functions (1) 2,2,sin)11(sin1=yxyxxy (2) ,0,cos)11(cos1=yxyxxy (3) )2,2(,tan)(tan1=yxyxxy (4)
9、),0(,cot)(cot1=yxyxxy (5) =yxyxxy ,sec)1|(|sec1)23,)2,0 (一、三象限 ) =yxyxxy ,sec)1|(|sec1,2()2,0 (一、二象限 ) (6) =yxyxxy ,csc)1|(|csc123,(2,0( (一、三象限 ) =yxyxxy ,csc)1|(|csc12,0()0,2 (一、四象限 ) 991 工 CH7-5 反三角函 7 Thm7.12: Derivatives of the inverse trigonometric functions 2111)sin(xxdxd=; 2111)cos(xxdxd=, 1|
10、 x 1|1)sec(21=xxxdxd; 1|1)csc(21=xxxdxd, 1| x Ex6. (a) evaluate )32(f ,where )2(tan)(1xxxf= 。 (b) Find an equation of the line tangent to the graph of )2(sec)(1xxg= at the point )3,1(Sol:(a) Table7.13 Integration formulas (1) +=Caxdxxa)(sin1122(2) +=+Caxadxax)(tan11122(3) +=Caxadxaxx)(sec11122991 工 CH7-5 反三角函 8 Ex8. Determine the following integrals. (a) dxx294(b)+dxx 11612(c) 53522541dxxxSolution:
