Аркуссинус е математическа функция , която се определя като обратна на функцията синус в интервала
[
−
π
/
2
,
π
/
2
]
{\displaystyle [-\pi /2\;,\;\pi /2]}
Графика, изобразяваща функцията
f
(
x
)
=
arcsin
(
x
)
{\displaystyle f(x)=\arcsin(x)}
arcsin
(
sin
(
x
)
)
=
x
{\displaystyle \arcsin(\sin(x))=x}
arcsin
(
x
)
=
arctan
(
x
1
−
x
2
)
{\displaystyle \arcsin(x)=\arctan \left({\frac {x}{\sqrt {1-x^{2}}}}\right)}
arcsin
(
x
)
=
1
2
arccos
(
1
−
2
x
2
)
,
0
≤
x
≤
1
{\displaystyle \arcsin(x)={\dfrac {1}{2}}\arccos \left(1-2x^{2}\right){\text{,}}\ \ 0\leq x\leq 1}
d
d
x
arcsin
(
a
x
+
b
)
=
a
1
−
(
a
x
+
b
)
2
{\displaystyle {\frac {d}{dx}}\arcsin(ax+b)={\frac {a}{\sqrt {1-(ax+b)^{2}}}}}
За
a
=
1
{\displaystyle a=1}
b
=
0
{\displaystyle b=0}
d
d
x
arcsin
(
x
)
=
1
1
−
x
2
{\displaystyle {\frac {d}{dx}}\arcsin(x)={\frac {1}{\sqrt {1-x^{2}}}}}
arcsin
(
−
1
)
=
−
π
2
{\displaystyle \arcsin(-1)=-{\frac {\pi }{2}}}
arcsin
(
−
1
2
2
)
=
−
1
4
π
{\displaystyle \arcsin \left(-{\frac {1}{2}}{\sqrt {2}}\right)=-{\frac {1}{4}}\pi }
arcsin
(
0
)
=
0
{\displaystyle \arcsin(0)=0\!}
arcsin
(
1
2
2
)
=
1
4
π
{\displaystyle \arcsin \left({\frac {1}{2}}{\sqrt {2}}\right)={\frac {1}{4}}\pi }
arcsin
(
1
)
=
π
2
{\displaystyle \arcsin(1)={\frac {\pi }{2}}}
∫
0
x
d
t
1
−
t
2
=
arcsin
(
x
)
{\displaystyle \int _{0}^{x}{\frac {\mathrm {d} t}{\sqrt {1-t^{2}}}}=\arcsin(x)}
∫
d
x
m
2
−
x
2
=
arcsin
(
x
m
)
+
C
{\displaystyle \int {\frac {\mathrm {d} x}{\sqrt {m^{2}-x^{2}}}}=\arcsin \left({\frac {x}{m}}\right)+C}
∫
arcsin
(
m
x
)
d
x
=
m
x
arcsin
(
m
x
)
+
1
−
m
2
x
2
m
+
C
{\displaystyle \int \arcsin(mx)\mathrm {d} x={\frac {mx\arcsin(mx)+{\sqrt {1-m^{2}x^{2}}}}{m}}+C}
Нека разгледаме следния интеграл:
I
=
∫
0
x
d
t
1
−
t
2
{\displaystyle {\mathcal {I}}=\int _{0}^{x}{\frac {\mathrm {d} t}{\sqrt {1-t^{2}}}}}
От биномната теорема получаваме:
I
=
∫
0
x
∑
k
=
0
∞
(
2
k
)
!
t
2
k
d
t
2
2
k
(
k
!
)
2
{\displaystyle {\mathcal {I}}=\int _{0}^{x}\sum _{k=0}^{\infty }{\frac {(2k)!t^{2k}\mathrm {d} t}{2^{2k}(k!)^{2}}}}
=
∑
k
=
0
∞
(
2
k
)
!
2
2
k
(
k
!
)
2
∫
0
x
t
2
k
d
t
{\displaystyle \ \ =\sum _{k=0}^{\infty }{\frac {(2k)!}{2^{2k}(k!)^{2}}}\int _{0}^{x}t^{2k}\mathrm {d} t}
Освен това знаем, че:
I
=
∫
0
x
d
t
1
−
t
2
=
arcsin
(
x
)
{\displaystyle {\mathcal {I}}=\int _{0}^{x}{\frac {\mathrm {d} t}{\sqrt {1-t^{2}}}}=\arcsin(x)}
Следователно:
arcsin
(
x
)
=
∑
k
=
0
∞
(
2
k
)
!
2
2
k
(
k
!
)
2
∫
0
x
t
2
k
d
t
{\displaystyle \arcsin(x)=\sum _{k=0}^{\infty }{\frac {(2k)!}{2^{2k}(k!)^{2}}}\int _{0}^{x}t^{2k}\mathrm {d} t}
=
∑
k
=
0
∞
(
2
k
)
!
2
2
k
(
k
!
)
2
t
2
k
+
1
2
k
+
1
|
0
x
{\displaystyle \qquad \quad \ \ \ \ =\sum _{k=0}^{\infty }{\frac {(2k)!}{2^{2k}(k!)^{2}}}{\frac {t^{2k+1}}{2k+1}}{\Bigg |}_{0}^{x}}
Откъдето вече лесно се вижда, че:
arcsin
(
x
)
=
∑
k
=
0
∞
(
2
k
)
!
x
2
k
+
1
2
2
k
(
k
!
)
2
(
2
k
+
1
)
,
−
1
≤
x
≤
1
{\displaystyle {\begin{aligned}\arcsin(x)=\sum _{k=0}^{\infty }{\frac {(2k)!x^{2k+1}}{2^{2k}(k!)^{2}(2k+1)}}{\text{,}}\ -1\leq x\leq 1\end{aligned}}}
=
x
+
1
6
x
3
+
3
40
x
5
+
5
112
x
7
+
⋯
{\displaystyle \qquad \quad \ \ \ \ =x+{\frac {1}{6}}x^{3}+{\frac {3}{40}}x^{5}+{\frac {5}{112}}x^{7}+\cdots }
![{\displaystyle [-\pi /2\;,\;\pi /2]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/32afeaaebd01441648d269f6a936baf7206f8e11)
