Lecture 6 : Inverse Trigonometric Functions Inverse Sine
Inverse Sine Function (arcsin x = sin 1x) The trigonometric function sinxis not one-to-one functions, hence in order to create an inverse, we must restrict its domain The restricted sine function is given by f(x) = 8
Restricted Sine Function
Inverse Sine Function (arcsin x = sin 1x) We see from the graph of the restricted sine function (or from its derivative) that the function is one-to-one and hence has an inverse, shown in red in the diagram below Hp 2,1L H-p 4,-1 2 L H1,p 2L H-1 2,-p 4 L-p 2-p 4 p 4 p 2-1 5-1 0-0 5 0 5 1 0 1 5 This inverse function, f 1(x), is denoted by f 1
Inverse functions - University of Arkansas
arcsin(sinx) = xif 0
Section 55 Inverse Trigonometric Functions and Their Graphs
Section 5 5 Inverse Trigonometric Functions and Their Graphs DEFINITION: The inverse sine function, denoted by sin 1 x (or arcsinx), is de ned to be the inverse of the restricted sine function
Inverse trigonometric functions (Sect 76) Review
The derivative of arcsin is given by arcsin0(x) = 1 √ 1 − x2 Proof: For x ∈ [−1,1] holds arcsin0(x) = 1 sin0 arcsin(x) = 1 cos arcsin(x) For x ∈ [−1,1] we get arcsin(x) = y ∈ hπ 2, π 2 i, and the cosine is positive in that interval, then cos(y) = + q 1 − sin2(y), hence arcsin0(x) = 1 q 1 − sin2 arcsin(x) ⇒ arcsin 0(x) = 1
Inverse trigonometric functions (Sect 76)
The derivative of arcsin is given by arcsin0(x) = 1 √ 1−x2 Proof: For x ∈ [−1,1] holds arcsin0(x) = 1 sin0 arcsin(x) = 1 cos arcsin(x) For x ∈ [−1,1] we get arcsin(x) = y ∈ hπ 2, π 2 i, and the cosine is positive in that interval, then cos(y) = + q 1−sin2(y), hence arcsin0(x) = 1 q 1−sin2 arcsin(x) ⇒ arcsin 0(x) = 1 √ 1
The complex inverse trigonometric and hyperbolic functions
2 The inverse trigonometric functions: arcsin and arccos The arcsine function is the solution to the equation: z = sinw = eiw − e−iw 2i ∗In our conventions, the real inverse tangent function, Arctan x, is a continuous single-valued function that varies smoothly from − 1 2π to +2π as x varies from −∞ to +∞ In contrast, Arccotx
Formule trigonometrice a b a b c b a c - Math
53 arcsinx+arcsiny= 2 6 6 6 4 arcsin(x p 1 y2 + y 1 x2); daca xy 0 sau x2 + y2 1;
Lecture 23: Improper integrals - Harvard University
Solution: The antiderivative is arcsin(x) In this case, it is not the point x = 0 which produces the difficulty It is the point x = 1 Take a > 0 and evaluate Z 1−a 0 1 √ 1− x2 dx = arcsin(x)1−a 0 = arcsin(1− a)− arcsin(0) Now arcsin(1 − a) has no problem at limit a → 0 Since arcsin(1) = π/2 exists We get therefore the
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Formule trigonometrice
1. sin=a
c ; cos=b c ; tg=a b ; ctg=b a (a; b- catetele,c- ipotenuza triunghiului dreptunghic,- unghiul, opus cateteia).2. tg=sin
cos; ctg=cos sin:3. tgctg= 1:
4. sin
2 = cos; sin() =sin:5. cos
2 =sin; cos() =cos: 6. tg 2 =ctg; ctg 2 =tg:7. sec
2 =cosec; cosec 2 = sec:8. sin
2+ cos2= 1:
9. 1 + tg
2= sec2:
10. 1 + ctg
2= cosec2:
11. sin() = sincossincos:
12. cos() = coscossinsin:
13. tg() =tgtg
1tgtg:
14. ctg() =ctgctg1
ctgctg:15. sin2= 2sincos:
16. cos2= cos2sin2:
17. tg2=2tg
1tg2:18. ctg2=ctg21
2ctg:19. sin3= 3sin4sin3:
20. cos3= 4cos33cos:
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1 21.sin 2 =s 1cos 2 22.
cos 2 =s
1 + cos
2 23.tg 2 =s 1cos
1 + cos:
24. tg
2 =sin1 + cos=1cos
sin: 25.ctg 2 =s
1 + cos
1cos:26. ctg
2 =sin1cos=1 + cos
sin:27. 1 + cos= 2cos2
228. 1cos= 2sin2
229. sinsin= 2sin
2 cos 230. cos+ cos= 2cos+
2 cos 231. coscos=2sin+
2 sin 232. tgtg=sin()
coscos:33. ctgctg=sin()
sinsin:34. sinsin=1
2 [cos()cos(+)]:35. sincos=1
2 [sin(+) + sin()]:36. coscos=1
2 [cos(+) + cos()]: 0Copyrightc
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237. Ecuatii trigonometrice elementare:
sinx=a;jaj 1;x= (1)narcsina+n; cosx=a;jaj 1;x=arccosa+ 2n; tgx=a; x= arctga+n; ctgx=a; x= arcctga+n9 >>>>>;n2Z:38. arcsinx+ arccosx=
2 ;jxj 1:39. arctgx+ arcctgx=
240. arcsecx+ arccosecx=
2 ;jxj 1:41. sin(arcsinx) =x; x2[1;+1]:
42. arcsin(sinx) =x; x2
2 243. cos(arccosx) =x; x2[1;+1]:
44. arccos(cosx) =x; x2[0;]:
45. tg(arctgx) =x; x2R:
46. arctg(tgx) =x; x2
2 247. ctg(arcctgx) =x; x2R:
48. arcctg(ctgx) =x; x2(0;):
49. arcsinx= arccosp
1x2= arctgx
p1x2= arcctgp
1x2 x ;0< x <1:50. arccosx= arcsinp
1x2= arctgp
1x2 x = arcctgx p1x2;0< x <1:
51. arctgx= arcsinx
p1 +x2= arccos1
p1 +x2= arcctg1
x ;0< x <+1:52. arcctgx= arcsin1
p1 +x2= arccosx
p1 +x2= arctg1
x ;0< x <+1:53. arcsinx+arcsiny=2
6664arcsin(xp
1y2+yp
1x2);dacaxy0 saux2+y21;
arcsin(xp1y2+yp
1x2);dacax >0; y >0 six2+y2>1;
arcsin(xp1y2+yp
1x2);dacax <0;y <0 six2+y2>1:
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354. arcsinxarcsiny=2
6664arcsin(xp
1y2yp1x2);dacaxy0 saux2+y21;
arcsin(xp 1y2yp1x2);dacax >0;y <0 six2+y2>1;
arcsin(xp 1y2yp1x2);dacax <0;y >0 six2+y2>1:
55. arccosx+ arccosy=2
64arccos(xyq
(1x2)(1y2));dacax+y0;2arccos(xyq
(1x2)(1y2));dacax+y <0:56. arccosxarccosy=2
64arccos(xy+q
(1x2)(1y2));dacaxy; arccos(xy+q (1x2)(1y2));dacax < y:57. arctgx+ arctgy=2
666666664arctg
x+y1xy;dacaxy <1;
+ arctgx+y1xy;dacax >0 sixy >1;
+ arctgx+y1xy;dacax <0 sixy >1:
58. arctgxarctgy=2
666666664arctg
xy1 +xy;dacaxy >1;
+ arctgxy1 +xy;dacax >0 sixy <1;
+ arctgxy1 +xy;dacax <0 sixy <1:
59. 2arcsinx=2
6666666664arcsin(2xp
1x2);dacajxj p
2 2 arcsin(2xp1x2);dacap
2 2 < x1; arcsin(2xp1x2);daca1x
2 2
60. 2arccosx=2
4arccos(2x21) cand 0x1;
2arccos(2x21) cand1x <0:
61. 2arctgx=2
666666664arctg
2x1x2;dacajxj<1;
+ arctg2x1x2;dacax >1;
+ arctg2x1x2;dacax <1:
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4 62.1 2 arcsinx=2 6
66664arcsins
1p 1x2 2 ;daca 0x1; arcsins 1p 1x2 2 ;daca1x <0: 63.1 2 arccosx= arccoss 1 +x 2 ;daca1x1: 64.
1 2 arctgx=2 6