In matematica , le funzioni trigonometriche inverse sono un insieme di funzioni strettamente collegate alle funzioni trigonometriche . Le funzioni inverse principali sono elencate nella seguente tabella.
Nome
Notazione usuale
Definizione
Dominio
Codominio
arcoseno
y
=
arcsin
(
x
)
{\displaystyle y=\arcsin(x)}
x
=
sin
(
y
)
{\displaystyle x=\sin(y)}
[
−
1
;
+
1
]
{\displaystyle \left[-1;+1\right]}
−
π
2
≤
y
≤
π
2
{\displaystyle -{\frac {\pi }{2}}\leq y\leq {\frac {\pi }{2}}}
arcocoseno
y
=
arccos
(
x
)
{\displaystyle y=\arccos(x)}
x
=
cos
(
y
)
{\displaystyle x=\cos(y)}
[
−
1
;
+
1
]
{\displaystyle \left[-1;+1\right]}
0
≤
y
≤
π
{\displaystyle 0\leq y\leq \pi }
arcotangente
y
=
arctan
(
x
)
{\displaystyle y=\arctan(x)}
x
=
tan
(
y
)
{\displaystyle x=\tan(y)}
R
{\displaystyle \mathbb {R} }
−
π
2
<
y
<
π
2
{\displaystyle -{\frac {\pi }{2}}<y<{\frac {\pi }{2}}}
arcocosecante
y
=
arccsc
(
x
)
{\displaystyle y=\operatorname {arccsc}(x)}
x
=
cosec
(
y
)
,
y
=
arcsin
(
1
x
)
{\displaystyle x=\operatorname {cosec} (y),y=\arcsin \left({\frac {1}{x}}\right)}
(
−
∞
;
−
1
]
∪
[
1
;
+
∞
)
{\displaystyle \left(-\infty ;-1\right]\cup \left[1;+\infty \right)}
−
π
2
≤
y
<
0
∨
0
<
y
≤
π
2
{\displaystyle -{\frac {\pi }{2}}\leq y<0\vee 0<y\leq {\frac {\pi }{2}}}
arcosecante
y
=
arcsec
(
x
)
{\displaystyle y=\operatorname {arcsec}(x)}
x
=
sec
(
y
)
,
y
=
arccos
(
1
x
)
{\displaystyle x=\sec(y),y=\arccos \left({\frac {1}{x}}\right)}
(
−
∞
;
−
1
]
∪
[
1
;
+
∞
)
{\displaystyle \left(-\infty ;-1\right]\cup \left[1;+\infty \right)}
0
≤
y
<
π
2
∨
π
2
<
y
≤
π
{\displaystyle 0\leq y<{\frac {\pi }{2}}\vee {\frac {\pi }{2}}<y\leq \pi }
arcocotangente
y
=
arccot
(
x
)
{\displaystyle y=\operatorname {arccot}(x)}
x
=
cot
(
y
)
,
y
=
arctan
(
1
x
)
{\displaystyle x=\cot(y),y=\arctan \left({\frac {1}{x}}\right)}
R
{\displaystyle \mathbb {R} }
0
<
y
<
π
{\displaystyle 0<y<\pi }
Talvolta vengono utilizzate le notazioni
sin
−
1
{\displaystyle \sin ^{-1}}
cos
−
1
{\displaystyle \cos ^{-1}}
arcsin
(
x
)
{\displaystyle \arcsin(x)}
1
/
sin
(
x
)
{\displaystyle 1/\sin(x)}
Nei linguaggi di programmazione al computer le funzioni arcsin, arccos, arctan sono generalmente chiamate asin, acos, atan. Molti linguaggi di programmazione forniscono anche la funzione con due argomenti atan2 , che calcola l'arcotangente di y /x dati y ed x , ma in un intervallo di [-π,π].
Analogamente al seno ed al coseno, le funzioni trigonometriche inverse si possono in alternativa definire in termini di serie infinite.
arcsin
z
=
z
+
(
1
2
)
z
3
3
+
(
1
⋅
3
2
⋅
4
)
z
5
5
+
(
1
⋅
3
⋅
5
2
⋅
4
⋅
6
)
z
7
7
+
⋯
=
∑
n
=
0
∞
(
2
n
n
)
z
2
n
+
1
4
n
(
2
n
+
1
)
,
|
z
|
<
1
{\displaystyle \arcsin z=z+\left({\frac {1}{2}}\right){\frac {z^{3}}{3}}+\left({\frac {1\cdot 3}{2\cdot 4}}\right){\frac {z^{5}}{5}}+\left({\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6}}\right){\frac {z^{7}}{7}}+\cdots =\sum _{n=0}^{\infty }{2n \choose n}{\frac {z^{2n+1}}{4^{n}(2n+1)}}\ ,\quad \left|z\right|<1}
arccos
z
=
π
2
−
arcsin
z
=
π
2
−
[
z
+
(
1
2
)
z
3
3
+
(
1
⋅
3
2
⋅
4
)
z
5
5
+
(
1
⋅
3
⋅
5
2
⋅
4
⋅
6
)
z
7
7
+
⋯
]
=
π
2
−
∑
n
=
0
∞
(
2
n
n
)
z
2
n
+
1
4
n
(
2
n
+
1
)
,
|
z
|
<
1
{\displaystyle \arccos z={\frac {\pi }{2}}-\arcsin z={\frac {\pi }{2}}-\left[z+\left({\frac {1}{2}}\right){\frac {z^{3}}{3}}+\left({\frac {1\cdot 3}{2\cdot 4}}\right){\frac {z^{5}}{5}}+\left({\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6}}\right){\frac {z^{7}}{7}}+\cdots \right]={\frac {\pi }{2}}-\sum _{n=0}^{\infty }{2n \choose n}{\frac {z^{2n+1}}{4^{n}(2n+1)}}\ ,\quad \left|z\right|<1}
arctan
z
=
z
−
z
3
3
+
z
5
5
−
z
7
7
+
⋯
=
∑
n
=
0
∞
(
−
1
)
n
z
2
n
+
1
2
n
+
1
,
|
z
|
<
1
{\displaystyle \arctan z=z-{\frac {z^{3}}{3}}+{\frac {z^{5}}{5}}-{\frac {z^{7}}{7}}+\cdots =\sum _{n=0}^{\infty }{\frac {(-1)^{n}z^{2n+1}}{2n+1}}\ ,\quad \left|z\right|<1}
arccsc
z
=
arcsin
(
1
z
)
=
1
z
+
(
1
2
)
1
3
z
3
+
(
1
⋅
3
2
⋅
4
)
1
5
z
5
+
(
1
⋅
3
⋅
5
2
⋅
4
⋅
6
)
1
7
z
7
+
⋯
=
∑
n
=
0
∞
(
2
n
n
)
1
4
n
(
2
n
+
1
)
z
2
n
+
1
,
|
z
|
>
1
{\displaystyle \operatorname {arccsc} z=\arcsin \left({\frac {1}{z}}\right)={\frac {1}{z}}+\left({\frac {1}{2}}\right){\frac {1}{3z^{3}}}+\left({\frac {1\cdot 3}{2\cdot 4}}\right){\frac {1}{5z^{5}}}+\left({\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6}}\right){\frac {1}{7z^{7}}}+\cdots =\sum _{n=0}^{\infty }{2n \choose n}{\frac {1}{4^{n}(2n+1)z^{2n+1}}}\ ,\quad \left|z\right|>1}
arcsec
z
=
arccos
(
1
z
)
=
π
2
−
[
1
z
+
(
1
2
)
1
3
z
3
+
(
1
⋅
3
2
⋅
4
)
1
5
z
5
+
(
1
⋅
3
⋅
5
2
⋅
4
⋅
6
)
1
7
z
7
+
⋯
]
=
π
2
−
∑
n
=
0
∞
(
2
n
n
)
1
4
n
(
2
n
+
1
)
z
2
n
+
1
,
|
z
|
>
1
{\displaystyle \operatorname {arcsec} z=\arccos \left({\frac {1}{z}}\right)={\frac {\pi }{2}}-\left[{\frac {1}{z}}+\left({\frac {1}{2}}\right){\frac {1}{3z^{3}}}+\left({\frac {1\cdot 3}{2\cdot 4}}\right){\frac {1}{5z^{5}}}+\left({\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6}}\right){\frac {1}{7z^{7}}}+\cdots \right]={\frac {\pi }{2}}-\sum _{n=0}^{\infty }{2n \choose n}{\frac {1}{4^{n}(2n+1)z^{2n+1}}}\ ,\quad \left|z\right|>1}
arccot
z
=
arctan
(
1
x
)
=
π
2
−
[
z
−
z
3
3
+
z
5
5
−
z
7
7
+
⋯
]
=
π
2
−
∑
n
=
0
∞
(
−
1
)
n
z
2
n
+
1
2
n
+
1
,
|
z
|
<
1
{\displaystyle \operatorname {arccot} z=\arctan \left({\frac {1}{x}}\right)={\frac {\pi }{2}}-\left[z-{\frac {z^{3}}{3}}+{\frac {z^{5}}{5}}-{\frac {z^{7}}{7}}+\cdots \right]={\frac {\pi }{2}}-\sum _{n=0}^{\infty }{\frac {(-1)^{n}z^{2n+1}}{2n+1}}\ ,\quad \left|z\right|<1}
Queste funzioni si possono anche definire dimostrando che sono integrali di altre funzioni.
arcsin
(
x
)
=
∫
0
x
1
1
−
z
2
d
z
,
|
x
|
<
1
{\displaystyle \arcsin \left(x\right)=\int _{0}^{x}{\frac {1}{\sqrt {1-z^{2}}}}\,\mathrm {d} z,\quad |x|<1}
arccos
(
x
)
=
∫
x
1
1
1
−
z
2
d
z
,
|
x
|
<
1
{\displaystyle \arccos \left(x\right)=\int _{x}^{1}{\frac {1}{\sqrt {1-z^{2}}}}\,\mathrm {d} z,\quad |x|<1}
arctan
(
x
)
=
∫
0
x
1
z
2
+
1
d
z
,
∀
x
∈
R
{\displaystyle \arctan \left(x\right)=\int _{0}^{x}{\frac {1}{z^{2}+1}}\,\mathrm {d} z,\quad \forall x\in \mathbb {R} }
arccot
(
x
)
=
∫
x
∞
1
z
2
+
1
d
z
,
z
>
0
{\displaystyle \operatorname {arccot} \left(x\right)=\int _{x}^{\infty }{\frac {1}{z^{2}+1}}\,\mathrm {d} z,\quad z>0}
arcsec
(
x
)
=
∫
x
1
1
|
z
|
z
2
−
1
d
z
,
x
>
1
{\displaystyle \operatorname {arcsec} \left(x\right)=\int _{x}^{1}{\frac {1}{|z|{\sqrt {z^{2}-1}}}}\,\mathrm {d} z,\quad x>1}
arccsc
(
x
)
=
∫
x
∞
−
1
|
z
|
z
2
−
1
d
z
,
x
>
1
{\displaystyle \operatorname {arccsc} \left(x\right)=\int _{x}^{\infty }{\frac {-1}{|z|{\sqrt {z^{2}-1}}}}\,\mathrm {d} z,\quad x>1}
È possibile esprimere queste funzioni usando i logaritmi naturali . Ciò permette di estendere in modo naturale il loro dominio all'intero piano complesso .
arcsin
x
=
−
i
ln
(
i
x
+
1
−
x
2
)
=
arccsc
1
x
{\displaystyle \arcsin x\,=\,-i\,\ln \left(i\,x+{\sqrt {1-x^{2}}}\right)\,=\,\operatorname {arccsc} {\frac {1}{x}}}
arccos
x
=
−
i
ln
(
x
+
x
2
−
1
)
=
π
2
+
i
ln
(
i
x
+
1
−
x
2
)
=
π
2
−
arcsin
x
=
arcsec
1
x
{\displaystyle \arccos x\,=\,-i\,\ln \left(x+{\sqrt {x^{2}-1}}\right)\,=\,{\frac {\pi }{2}}\,+i\ln \left(i\,x+{\sqrt {1-x^{2}}}\right)\,=\,{\frac {\pi }{2}}-\arcsin x\,=\,\operatorname {arcsec} {\frac {1}{x}}}
arctan
x
=
i
2
(
ln
(
1
−
i
x
)
−
ln
(
1
+
i
x
)
)
=
arccot
1
x
{\displaystyle \arctan x\,=\,{\frac {i}{2}}\left(\ln \left(1-i\,x\right)-\ln \left(1+i\,x\right)\right)\,=\,\operatorname {arccot} {\frac {1}{x}}}
arccsc
x
=
−
i
ln
(
1
−
1
x
2
+
i
x
)
=
arcsin
1
x
{\displaystyle \operatorname {arccsc} x\,=\,-i\,\ln \left({\sqrt {1-{\frac {1}{x^{2}}}}}+{\frac {i}{x}}\right)\,=\,\arcsin {\frac {1}{x}}}
arcsec
x
=
−
i
ln
(
1
x
2
−
1
+
1
x
)
=
i
ln
(
1
−
1
x
2
+
i
x
)
+
π
2
=
π
2
−
arccsc
x
=
arccos
1
x
{\displaystyle \operatorname {arcsec} x\,=\,-i\,\ln \left({\sqrt {{\frac {1}{x^{2}}}-1}}+{\frac {1}{x}}\right)\,=\,i\,\ln \left({\sqrt {1-{\frac {1}{x^{2}}}}}+{\frac {i}{x}}\right)+{\frac {\pi }{2}}\,=\,{\frac {\pi }{2}}-\operatorname {arccsc} x\,=\,\arccos {\frac {1}{x}}}
arccot
x
=
i
2
(
ln
(
1
−
i
x
)
−
ln
(
1
+
i
x
)
)
=
arctan
1
x
{\displaystyle \operatorname {arccot} x\,=\,{\frac {i}{2}}\left(\ln \left(1-{\frac {i}{x}}\right)-\ln \left(1+{\frac {i}{x}}\right)\right)\,=\,\arctan {\frac {1}{x}}}
Queste relazioni si possono dimostrare elementarmente tramite l'espansione delle funzioni trigonometriche alla forma esponenziale.
arcsin
x
=
θ
{\displaystyle \arcsin x\,=\,\theta }
e
i
θ
−
e
−
i
θ
2
i
=
x
{\displaystyle {\frac {e^{i\,\theta }-e^{-i\,\theta }}{2i}}\,=\,x}
Sia
k
=
e
i
θ
{\displaystyle k=e^{i\,\theta }}
k
−
1
k
2
i
=
x
{\displaystyle {\frac {k-{\frac {1}{k}}}{2i}}\,=\,x}
k
2
−
2
i
k
x
−
1
=
0
{\displaystyle k^{2}-2\,i\,k\,x-1\,=\,0}
k
{\displaystyle k}
k
=
i
x
±
1
−
x
2
=
e
i
θ
{\displaystyle k\,=\,i\,x\pm {\sqrt {1-x^{2}}}\,=\,e^{i\,\theta }}
θ
=
arcsin
x
=
−
i
ln
(
i
x
+
1
−
x
2
)
{\displaystyle \theta \,=\,\arcsin \,x\,=\,-i\ln \left(i\,x+{\sqrt {1-x^{2}}}\right)}
Derivate delle funzioni trigonometriche inverse
modifica
Le derivate delle funzioni trigonometriche inverse valgono:
d
d
x
arcsin
x
=
1
1
−
x
2
{\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}\arcsin x\,=\,{\frac {1}{\sqrt {1-x^{2}}}}}
d
d
x
arccos
x
=
−
1
1
−
x
2
{\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}\arccos x\,=\,-{\frac {1}{\sqrt {1-x^{2}}}}}
d
d
x
arctan
x
=
1
1
+
x
2
{\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}\arctan x\,=\,{\frac {1}{1+x^{2}}}}
d
d
x
arccsc
x
=
−
1
x
x
2
−
1
{\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}\operatorname {arccsc} x\,=\,-{\frac {1}{x\,{\sqrt {x^{2}-1}}}}}
d
d
x
arcsec
x
=
1
x
x
2
−
1
{\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}\operatorname {arcsec} x\,=\,{\frac {1}{x\,{\sqrt {x^{2}-1}}}}}
d
d
x
arccot
x
=
−
1
1
+
x
2
{\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}\operatorname {arccot} x\,=\,-{\frac {1}{1+x^{2}}}}
Questi risultati si ottengono facilmente derivando la forma logaritmica mostrata sopra.
Integrali indefiniti delle funzioni trigonometriche inverse
modifica
∫
arcsin
x
d
x
=
x
arcsin
x
+
1
−
x
2
+
C
{\displaystyle \int \arcsin x\,\mathrm {d} x\,=\,x\,\arcsin x+{\sqrt {1-x^{2}}}+C}
∫
arccos
x
d
x
=
x
arccos
x
−
1
−
x
2
+
C
{\displaystyle \int \arccos x\,\mathrm {d} x\,=\,x\,\arccos x-{\sqrt {1-x^{2}}}+C}
∫
arctan
x
d
x
=
x
arctan
x
−
1
2
ln
(
1
+
x
2
)
+
C
{\displaystyle \int \arctan x\,\mathrm {d} x\,=\,x\,\arctan x-{\frac {1}{2}}\ln \left(1+x^{2}\right)+C}
∫
arccsc
x
d
x
=
x
arccsc
x
+
ln
(
x
+
x
2
−
1
)
+
C
{\displaystyle \int \operatorname {arccsc} x\,\mathrm {d} x\,=\,x\,\operatorname {arccsc} x+\ln \left(x+{\sqrt {x^{2}-1}}\right)+C}
∫
arcsec
x
d
x
=
x
arcsec
x
−
ln
(
x
+
x
2
−
1
)
+
C
{\displaystyle \int \operatorname {arcsec} x\,\mathrm {d} x\,=\,x\,\operatorname {arcsec} x-\ln \left(x+{\sqrt {x^{2}-1}}\right)+C}
∫
arccot
x
d
x
=
x
arccot
x
+
1
2
ln
(
1
+
x
2
)
+
C
{\displaystyle \int \operatorname {arccot} x\,\mathrm {d} x\,=\,x\,\operatorname {arccot} x+{\frac {1}{2}}\ln \left(1+x^{2}\right)+C}
Tutti questi integrali si ricavano integrazione per parti e le derivate elencate al paragrafo precedente.
![{\displaystyle \left[-1;+1\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0e7099e3c51db3b988242091d5f1c36169f3168f)
![{\displaystyle \left(-\infty ;-1\right]\cup \left[1;+\infty \right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fd97d33078b35928d00f9a8b1fdcb8186f7793fc)
![{\displaystyle \arccos z={\frac {\pi }{2}}-\arcsin z={\frac {\pi }{2}}-\left[z+\left({\frac {1}{2}}\right){\frac {z^{3}}{3}}+\left({\frac {1\cdot 3}{2\cdot 4}}\right){\frac {z^{5}}{5}}+\left({\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6}}\right){\frac {z^{7}}{7}}+\cdots \right]={\frac {\pi }{2}}-\sum _{n=0}^{\infty }{2n \choose n}{\frac {z^{2n+1}}{4^{n}(2n+1)}}\ ,\quad \left|z\right|<1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/57984222c9f8da61954ec836e74923b39a6427b0)
![{\displaystyle \operatorname {arcsec} z=\arccos \left({\frac {1}{z}}\right)={\frac {\pi }{2}}-\left[{\frac {1}{z}}+\left({\frac {1}{2}}\right){\frac {1}{3z^{3}}}+\left({\frac {1\cdot 3}{2\cdot 4}}\right){\frac {1}{5z^{5}}}+\left({\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6}}\right){\frac {1}{7z^{7}}}+\cdots \right]={\frac {\pi }{2}}-\sum _{n=0}^{\infty }{2n \choose n}{\frac {1}{4^{n}(2n+1)z^{2n+1}}}\ ,\quad \left|z\right|>1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/099efa57d57d46cfee9cb09bbdb19b66f620ca10)
![{\displaystyle \operatorname {arccot} z=\arctan \left({\frac {1}{x}}\right)={\frac {\pi }{2}}-\left[z-{\frac {z^{3}}{3}}+{\frac {z^{5}}{5}}-{\frac {z^{7}}{7}}+\cdots \right]={\frac {\pi }{2}}-\sum _{n=0}^{\infty }{\frac {(-1)^{n}z^{2n+1}}{2n+1}}\ ,\quad \left|z\right|<1}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5faf0fc12d3f1b02ffb2752cb672a8bb28c6daa7)
Wikimedia Commons contiene immagini o altri file su funzione trigonometrica inversa
