Mezi cyklometrické funkce patří:
Aby mohla k libovolné funkci existovat inverzní funkce , daná funkce musí být prostá , to znamená, že různým dvěma prvkům musí přiřazovat dvě různé hodnoty. Protože jsou ale goniometrické funkce periodické, tzn. nejsou prosté, musíme nejprve ošetřit jejich definiční obor a také definiční obory goniometrických funkcí. To znamená, že vybereme jen tu podmnožinu definičního oboru dané geometrické funkce, na které je prostá.
Definiční obory cyklometrických a goniometrických funkcí[ 1]
Goniometrické funkce
Cyklometrické funkce
Sinus:
sin
x
{\displaystyle \sin x}
x
∈
⟨
−
π
2
;
π
2
⟩
{\displaystyle x\in \langle \textstyle -{\frac {\pi }{2}};{\frac {\pi }{2}}\rangle }
Arkus sinus:
arcsin
x
{\displaystyle \arcsin x}
x
∈
⟨
−
1
;
1
⟩
{\displaystyle x\in \langle -1;1\rangle }
Cosinus:
cos
x
{\displaystyle \cos x}
x
∈
⟨
0
,
π
⟩
{\displaystyle x\in \langle 0,\pi \rangle }
Arkus cosinus:
arccos
x
{\displaystyle \arccos x}
x
∈
⟨
−
1
;
1
⟩
{\displaystyle x\in \langle -1;1\rangle }
Tangens:
t
g
x
{\displaystyle \mathrm {tg} \,x}
x
∈
(
−
π
2
;
π
2
)
{\displaystyle x\in \textstyle (-{\frac {\pi }{2}};{\frac {\pi }{2}})}
Arkus tangens:
a
r
c
t
g
x
{\displaystyle \mathrm {arctg} \,x}
x
∈
R
{\displaystyle x\in \mathbb {R} }
Cotangens:
c
o
t
g
x
{\displaystyle \mathrm {cotg} \,x}
x
∈
(
0
,
π
)
{\displaystyle x\in (0,\pi )}
Arkus cotangens:
a
r
c
c
o
t
g
x
{\displaystyle \mathrm {arccotg} \,x}
x
∈
R
{\displaystyle x\in \mathbb {R} }
editovat
arcsin
(
sin
x
)
=
x
{\displaystyle \arcsin(\sin x)=x}
|
x
|
≤
π
2
{\displaystyle \ |x|\leq {\frac {\pi }{2}}}
sin
(
arcsin
x
)
=
x
{\displaystyle \sin(\arcsin x)=x}
|
x
|
≤
1
{\displaystyle \ |x|\leq 1}
arccos
(
cos
x
)
=
x
{\displaystyle \arccos(\cos x)=x}
0
≤
x
≤
π
{\displaystyle \ 0\leq x\leq \pi }
cos
(
arccos
x
)
=
x
{\displaystyle \cos(\arccos x)=x}
|
x
|
≤
1
{\displaystyle \ |x|\leq 1}
arctg
(
tg
x
)
=
x
{\displaystyle \operatorname {arctg} (\operatorname {tg} x)=x}
|
x
|
<
π
2
{\displaystyle \ |x|<{\frac {\pi }{2}}}
tg
(
arctg
x
)
=
x
{\displaystyle \operatorname {tg} (\operatorname {arctg} x)=x}
arccotg
(
cotg
x
)
=
x
{\displaystyle \operatorname {arccotg} (\operatorname {cotg} x)=x}
0
<
x
<
π
{\displaystyle \ 0<x<\pi }
cotg
(
arcotg
x
)
=
x
{\displaystyle \operatorname {cotg} (\operatorname {arcotg} x)=x}
editovat
arcsin
x
=
π
2
−
arccos
x
=
arctg
(
x
1
−
x
2
)
=
π
2
−
arccotg
(
x
1
−
x
2
)
arccos
x
=
π
2
−
arcsin
x
=
π
2
−
arctg
(
x
1
−
x
2
)
=
arccotg
(
x
1
−
x
2
)
arctg
x
=
arcsin
(
x
1
+
x
2
)
=
π
2
−
arccos
(
x
1
+
x
2
)
=
π
2
−
arccotg
x
arccotg
x
=
π
2
−
arcsin
(
x
1
+
x
2
)
=
arccos
(
x
1
+
x
2
)
=
π
2
−
arctg
x
{\displaystyle {\begin{aligned}\arcsin x&={\frac {\pi }{2}}-\arccos x=\operatorname {arctg} \left({\frac {x}{\sqrt {1-x^{2}}}}\right)={\frac {\pi }{2}}-\operatorname {arccotg} \left({\frac {x}{\sqrt {1-x^{2}}}}\right)\\[12pt]\arccos x&={\frac {\pi }{2}}-\arcsin x={\frac {\pi }{2}}-\operatorname {arctg} \left({\frac {x}{\sqrt {1-x^{2}}}}\right)=\operatorname {arccotg} \left({\frac {x}{\sqrt {1-x^{2}}}}\right)\\[12pt]\operatorname {arctg} x&=\arcsin \left({\frac {x}{\sqrt {1+x^{2}}}}\right)={\frac {\pi }{2}}-\arccos \left({\frac {x}{\sqrt {1+x^{2}}}}\right)={\frac {\pi }{2}}-\operatorname {arccotg} x\\[12pt]\operatorname {arccotg} x&={\frac {\pi }{2}}-\arcsin \left({\frac {x}{\sqrt {1+x^{2}}}}\right)=\arccos \left({\frac {x}{\sqrt {1+x^{2}}}}\right)={\frac {\pi }{2}}-\operatorname {arctg} x\end{aligned}}}
Dále platí:
arccotg
x
=
{
arctg
1
x
,
pokud platí
x
>
0
,
π
+
arctg
1
x
,
pokud platí
x
<
0.
{\displaystyle \operatorname {arccotg} x={\begin{cases}\operatorname {arctg} \displaystyle {\frac {1}{x}}\,,&{\text{pokud platí }}x>0,\\[12pt]\pi +\operatorname {arctg} \displaystyle {\frac {1}{x}}\,,&{\text{pokud platí }}x<0.\end{cases}}}
editovat
arcsin
(
−
x
)
=
−
arcsin
x
arccos
(
−
x
)
=
π
−
arccos
x
arctg
(
−
x
)
=
−
arctg
x
arccotg
(
−
x
)
=
π
−
arccotg
x
{\displaystyle {\begin{aligned}\arcsin(-x)&=-\arcsin x\\\arccos(-x)&=\pi -\arccos x\\\operatorname {arctg} (-x)&=-\operatorname {arctg} x\\\operatorname {arccotg} (-x)&=\pi -\operatorname {arccotg} x\end{aligned}}}
editovat
arcsin
x
+
arcsin
y
=
{
arcsin
(
x
1
−
y
2
+
y
1
−
x
2
)
,
pokud platí
x
y
≤
0
nebo
x
2
+
y
2
≤
1
,
π
−
arcsin
(
x
1
−
y
2
+
y
1
−
x
2
)
,
pokud platí
x
>
0
,
y
>
0
,
x
2
+
y
2
>
1
,
−
π
−
arcsin
(
x
1
−
y
2
+
y
1
−
x
2
)
,
pokud platí
x
<
0
,
y
<
0
,
x
2
+
y
2
>
1.
{\displaystyle \arcsin x\,+\,\arcsin y={\begin{cases}\arcsin \left(x{\sqrt {1-y^{2}}}+y{\sqrt {1-x^{2}}}\right),&{\text{pokud platí }}xy\leq 0{\text{ nebo }}x^{2}+y^{2}\leq 1,\\[12pt]\pi -\arcsin \left(x{\sqrt {1-y^{2}}}+y{\sqrt {1-x^{2}}}\right),&{\text{pokud platí }}x>0,y>0,x^{2}+y^{2}>1,\\[12pt]-\pi -\arcsin \left(x{\sqrt {1-y^{2}}}+y{\sqrt {1-x^{2}}}\right),&{\text{pokud platí }}x<0,y<0,x^{2}+y^{2}>1.\end{cases}}}
arcsin
x
−
arcsin
y
=
{
arcsin
(
x
1
−
y
2
−
y
1
−
x
2
)
,
pokud platí
x
y
≥
0
nebo
x
2
+
y
2
≤
1
,
π
−
arcsin
(
x
1
−
y
2
−
y
1
−
x
2
)
,
pokud platí
x
>
0
,
y
<
0
,
x
2
+
y
2
>
1
,
−
π
−
arcsin
(
x
1
−
y
2
+
y
1
−
x
2
)
,
pokud platí
x
<
0
,
y
>
0
,
x
2
+
y
2
>
1.
{\displaystyle \arcsin x\,-\,\arcsin y={\begin{cases}\arcsin \left(x{\sqrt {1-y^{2}}}-y{\sqrt {1-x^{2}}}\right),&{\text{pokud platí }}xy\geq 0{\text{ nebo }}x^{2}+y^{2}\leq 1,\\[12pt]\pi -\arcsin \left(x{\sqrt {1-y^{2}}}-y{\sqrt {1-x^{2}}}\right),&{\text{pokud platí }}x>0,y<0,x^{2}+y^{2}>1,\\[12pt]-\pi -\arcsin \left(x{\sqrt {1-y^{2}}}+y{\sqrt {1-x^{2}}}\right),&{\text{pokud platí }}x<0,y>0,x^{2}+y^{2}>1.\end{cases}}}
arccos
x
+
arccos
y
=
{
arccos
(
x
y
−
1
−
x
2
⋅
1
−
y
2
)
,
pokud platí
x
+
y
≥
0
,
2
π
−
arccos
(
x
y
−
1
−
x
2
⋅
1
−
y
2
)
,
pokud platí
x
+
y
<
0.
{\displaystyle \arccos x\,+\,\arccos y={\begin{cases}\arccos \left(xy-{\sqrt {1-x^{2}}}\cdot {\sqrt {1-y^{2}}}\right),&{\text{pokud platí }}x+y\geq 0,\\[12pt]2\pi -\arccos \left(xy-{\sqrt {1-x^{2}}}\cdot {\sqrt {1-y^{2}}}\right),&{\text{pokud platí }}x+y<0.\end{cases}}}
arccos
x
−
arccos
y
=
{
−
arccos
(
x
y
+
1
−
x
2
⋅
1
−
y
2
)
,
pokud platí
x
≥
y
,
arccos
(
x
y
+
1
−
x
2
⋅
1
−
y
2
)
,
pokud platí
x
<
y
.
{\displaystyle \arccos x\,-\,\arccos y={\begin{cases}-\arccos \left(xy+{\sqrt {1-x^{2}}}\cdot {\sqrt {1-y^{2}}}\right),&{\text{pokud platí }}x\geq y,\\[12pt]\arccos \left(xy+{\sqrt {1-x^{2}}}\cdot {\sqrt {1-y^{2}}}\right),&{\text{pokud platí }}x<y.\end{cases}}}
arctg
x
+
arctg
y
=
{
arctg
(
x
+
y
1
−
x
y
)
,
pokud platí
x
y
<
1
,
π
+
arctg
(
x
+
y
1
−
x
y
)
,
pokud platí
x
y
>
1
,
x
>
0
−
π
+
arctg
(
x
+
y
1
−
x
y
)
,
pokud platí
x
y
>
1
,
x
<
0.
{\displaystyle \operatorname {arctg} x\,+\,\operatorname {arctg} y={\begin{cases}\operatorname {arctg} \left(\displaystyle {\frac {x+y}{1-xy}}\right),&{\text{pokud platí }}xy<1,\\[12pt]\pi +\operatorname {arctg} \left(\displaystyle {\frac {x+y}{1-xy}}\right),&{\text{pokud platí }}xy>1,x>0\\[12pt]-\pi +\operatorname {arctg} \left(\displaystyle {\frac {x+y}{1-xy}}\right),&{\text{pokud platí }}xy>1,x<0.\end{cases}}}
arctg
x
−
arctg
y
=
{
arctg
(
x
−
y
1
+
x
y
)
,
pokud platí
x
y
>
−
1
,
π
+
arctg
(
x
−
y
1
+
x
y
)
,
pokud platí
x
y
<
−
1
,
x
>
0
−
π
+
arctg
(
x
−
y
1
+
x
y
)
,
pokud platí
x
y
<
−
1
,
x
<
0.
{\displaystyle \operatorname {arctg} x\,-\,\operatorname {arctg} y={\begin{cases}\operatorname {arctg} \left(\displaystyle {\frac {x-y}{1+xy}}\right),&{\text{pokud platí }}xy>-1,\\[12pt]\pi +\operatorname {arctg} \left(\displaystyle {\frac {x-y}{1+xy}}\right),&{\text{pokud platí }}xy<-1,x>0\\[12pt]-\pi +\operatorname {arctg} \left(\displaystyle {\frac {x-y}{1+xy}}\right),&{\text{pokud platí }}xy<-1,x<0.\end{cases}}}
arccotg
x
+
arccotg
y
=
{
arccotg
(
x
y
−
1
x
+
y
)
,
pokud platí
x
>
−
y
,
π
+
arccotg
(
x
y
−
1
x
+
y
)
,
pokud platí
x
<
−
y
.
{\displaystyle \operatorname {arccotg} x\,+\,\operatorname {arccotg} y={\begin{cases}\operatorname {arccotg} \left(\displaystyle {\frac {xy-1}{x+y}}\right),&{\text{pokud platí }}x>-y,\\[10pt]\pi +\operatorname {arccotg} \left(\displaystyle {\frac {xy-1}{x+y}}\right),&{\text{pokud platí }}x<-y.\end{cases}}}
arcsin
x
+
arccos
x
=
π
2
,
{\displaystyle \arcsin x+\arccos x={\frac {\pi }{2}},}
|
x
|
≤
1
{\displaystyle \ |x|\leq 1}
arctg
x
+
arccotg
x
=
π
2
{\displaystyle \operatorname {arctg} x+\operatorname {arccotg} x={\frac {\pi }{2}}}
editovat
Cyklometrické funkce se dají také vyjádřit použitím logaritmů a komplexních čísel :
arcsin
x
=
−
i
ln
(
i
x
+
1
−
x
2
)
arccos
x
=
π
2
+
i
ln
(
i
x
+
1
−
x
2
)
=
π
2
−
arcsin
x
arctg
x
=
i
2
(
ln
(
1
−
i
x
)
−
ln
(
1
+
i
x
)
)
=
arccotg
1
x
arccotg
x
=
i
2
(
ln
(
x
−
i
)
−
ln
(
x
+
i
)
)
=
arctg
1
x
{\displaystyle {\begin{aligned}\arcsin x&{}=-\mathrm {i} \ln \left(\mathrm {i} x+{\sqrt {1-x^{2}}}\right)&{}\\[10pt]\arccos x&{}={\frac {\pi }{2}}\,+\mathrm {i} \ln \left(\mathrm {i} x+{\sqrt {1-x^{2}}}\right)={\frac {\pi }{2}}-\arcsin x&{}\\[10pt]\operatorname {arctg} x&{}={\frac {\mathrm {i} }{2}}\left(\ln \left(1-\mathrm {i} x\right)-\ln \left(1+\mathrm {i} x\right)\right)=\operatorname {arccotg} {\frac {1}{x}}\\[10pt]\operatorname {arccotg} x&{}={\frac {\mathrm {i} }{2}}\left(\ln \left(x-\mathrm {i} \right)-\ln \left(x+\mathrm {i} \right)\right)=\operatorname {arctg} {\frac {1}{x}}\end{aligned}}}
editovat
Rozvoj cyklometrických funkcí lze psát jako:
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
)
,
je-li
|
z
|
≤
1
arccos
z
=
π
2
−
arcsin
z
=
π
2
−
(
z
+
(
1
2
)
z
3
3
+
(
1
⋅
3
2
⋅
4
)
z
5
5
+
…
)
=
π
2
−
∑
n
=
0
∞
(
2
n
n
)
z
2
n
+
1
4
n
(
2
n
+
1
)
,
je-li
|
z
|
≤
1
arctg
z
=
z
−
z
3
3
+
z
5
5
−
z
7
7
+
…
=
∑
n
=
0
∞
(
−
1
)
n
z
2
n
+
1
2
n
+
1
,
je-li
|
z
|
≤
1
,
z
≠
±
i
arccotg
z
=
π
2
−
arctg
z
=
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
,
je-li
|
z
|
≤
1
,
z
≠
±
i
arcsec
z
=
arccos
(
1
/
z
)
=
π
2
−
(
z
−
1
+
(
1
2
)
z
−
3
3
+
(
1
⋅
3
2
⋅
4
)
z
−
5
5
+
…
)
=
π
2
−
∑
n
=
0
∞
(
2
n
n
)
z
−
(
2
n
+
1
)
4
n
(
2
n
+
1
)
,
je-li
|
z
|
≥
1
arccsc
z
=
arcsin
(
1
/
z
)
=
z
−
1
+
(
1
2
)
z
−
3
3
+
(
1
⋅
3
2
⋅
4
)
z
−
5
5
+
…
=
∑
n
=
0
∞
(
2
n
n
)
z
−
(
2
n
+
1
)
4
n
(
2
n
+
1
)
,
je-li
|
z
|
≥
1
{\displaystyle {\begin{aligned}\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}}\,+\,\dots \ =\sum _{n=0}^{\infty }{\frac {{\binom {2n}{n}}z^{2n+1}}{4^{n}(2n+1)}}\,,\qquad {\text{je-li }}|z|\leq 1\\[10pt]\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}}+\dots \ \right)={\frac {\pi }{2}}\,-\,\sum _{n=0}^{\infty }{\frac {{\binom {2n}{n}}z^{2n+1}}{4^{n}(2n+1)}}\,,\qquad {\text{je-li }}|z|\leq 1\\[10pt]\operatorname {arctg} z&=z-{\frac {z^{3}}{3}}+{\frac {z^{5}}{5}}-{\frac {z^{7}}{7}}+\dots \ =\sum _{n=0}^{\infty }{\frac {(-1)^{n}z^{2n+1}}{2n+1}}\,,\qquad {\text{je-li }}|z|\leq 1,\ z\neq \pm \mathrm {i} \\[10pt]\operatorname {arccotg} z&={\frac {\pi }{2}}-\operatorname {arctg} z\ ={\frac {\pi }{2}}\,-\,\left(z-{\frac {z^{3}}{3}}+{\frac {z^{5}}{5}}-{\frac {z^{7}}{7}}+\dots \ \right)={\frac {\pi }{2}}\,-\,\sum _{n=0}^{\infty }{\frac {(-1)^{n}z^{2n+1}}{2n+1}}\,,\qquad {\text{je-li }}|z|\leq 1,\ z\neq \pm \mathrm {i} \\[10pt]\operatorname {arcsec} z&=\arccos {(1/z)}={\frac {\pi }{2}}\,-\,\left(z^{-1}+\left({\frac {1}{2}}\right){\frac {z^{-3}}{3}}+\left({\frac {1\cdot 3}{2\cdot 4}}\right){\frac {z^{-5}}{5}}+\dots \ \right)={\frac {\pi }{2}}\,-\,\sum _{n=0}^{\infty }{\frac {{\binom {2n}{n}}z^{-(2n+1)}}{4^{n}(2n+1)}}\,,\qquad {\text{je-li }}|z|\geq 1\\[10pt]\operatorname {arccsc} z&=\arcsin {(1/z)}=z^{-1}\,+\,\left({\frac {1}{2}}\right){\frac {z^{-3}}{3}}\,+\,\left({\frac {1\cdot 3}{2\cdot 4}}\right){\frac {z^{-5}}{5}}\,+\,\dots \ =\sum _{n=0}^{\infty }{\frac {{\binom {2n}{n}}z^{-(2n+1)}}{4^{n}(2n+1)}}\,,\qquad {\text{je-li }}|z|\geq 1\end{aligned}}}
V tomto článku byl použit překlad textu z článku Cyklometrická funkcia
↑ WEISSTEIN, Eric W. Inverse Trigonometric Functions. mathworld.wolfram.com [online]. [cit. 2024-10-01]. Dostupné online . (anglicky) 
![{\displaystyle {\begin{aligned}\arcsin x&={\frac {\pi }{2}}-\arccos x=\operatorname {arctg} \left({\frac {x}{\sqrt {1-x^{2}}}}\right)={\frac {\pi }{2}}-\operatorname {arccotg} \left({\frac {x}{\sqrt {1-x^{2}}}}\right)\\[12pt]\arccos x&={\frac {\pi }{2}}-\arcsin x={\frac {\pi }{2}}-\operatorname {arctg} \left({\frac {x}{\sqrt {1-x^{2}}}}\right)=\operatorname {arccotg} \left({\frac {x}{\sqrt {1-x^{2}}}}\right)\\[12pt]\operatorname {arctg} x&=\arcsin \left({\frac {x}{\sqrt {1+x^{2}}}}\right)={\frac {\pi }{2}}-\arccos \left({\frac {x}{\sqrt {1+x^{2}}}}\right)={\frac {\pi }{2}}-\operatorname {arccotg} x\\[12pt]\operatorname {arccotg} x&={\frac {\pi }{2}}-\arcsin \left({\frac {x}{\sqrt {1+x^{2}}}}\right)=\arccos \left({\frac {x}{\sqrt {1+x^{2}}}}\right)={\frac {\pi }{2}}-\operatorname {arctg} x\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3d5ba5d81d93fdc58f5166167a767279188d82e3)
![{\displaystyle \operatorname {arccotg} x={\begin{cases}\operatorname {arctg} \displaystyle {\frac {1}{x}}\,,&{\text{pokud platí }}x>0,\\[12pt]\pi +\operatorname {arctg} \displaystyle {\frac {1}{x}}\,,&{\text{pokud platí }}x<0.\end{cases}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a02b990f57f50d1ba3ce2f7fa90992d99a340c63)
![{\displaystyle \arcsin x\,+\,\arcsin y={\begin{cases}\arcsin \left(x{\sqrt {1-y^{2}}}+y{\sqrt {1-x^{2}}}\right),&{\text{pokud platí }}xy\leq 0{\text{ nebo }}x^{2}+y^{2}\leq 1,\\[12pt]\pi -\arcsin \left(x{\sqrt {1-y^{2}}}+y{\sqrt {1-x^{2}}}\right),&{\text{pokud platí }}x>0,y>0,x^{2}+y^{2}>1,\\[12pt]-\pi -\arcsin \left(x{\sqrt {1-y^{2}}}+y{\sqrt {1-x^{2}}}\right),&{\text{pokud platí }}x<0,y<0,x^{2}+y^{2}>1.\end{cases}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/88439f3b9105a6f11fde74f2c926bf3600268d07)
![{\displaystyle \arcsin x\,-\,\arcsin y={\begin{cases}\arcsin \left(x{\sqrt {1-y^{2}}}-y{\sqrt {1-x^{2}}}\right),&{\text{pokud platí }}xy\geq 0{\text{ nebo }}x^{2}+y^{2}\leq 1,\\[12pt]\pi -\arcsin \left(x{\sqrt {1-y^{2}}}-y{\sqrt {1-x^{2}}}\right),&{\text{pokud platí }}x>0,y<0,x^{2}+y^{2}>1,\\[12pt]-\pi -\arcsin \left(x{\sqrt {1-y^{2}}}+y{\sqrt {1-x^{2}}}\right),&{\text{pokud platí }}x<0,y>0,x^{2}+y^{2}>1.\end{cases}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1aa7e6331938c5fafa6f330dceebc6e47130f49d)
![{\displaystyle \arccos x\,+\,\arccos y={\begin{cases}\arccos \left(xy-{\sqrt {1-x^{2}}}\cdot {\sqrt {1-y^{2}}}\right),&{\text{pokud platí }}x+y\geq 0,\\[12pt]2\pi -\arccos \left(xy-{\sqrt {1-x^{2}}}\cdot {\sqrt {1-y^{2}}}\right),&{\text{pokud platí }}x+y<0.\end{cases}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/148b6e4aa1d53bfbad974058efd66484fd536ce0)
![{\displaystyle \arccos x\,-\,\arccos y={\begin{cases}-\arccos \left(xy+{\sqrt {1-x^{2}}}\cdot {\sqrt {1-y^{2}}}\right),&{\text{pokud platí }}x\geq y,\\[12pt]\arccos \left(xy+{\sqrt {1-x^{2}}}\cdot {\sqrt {1-y^{2}}}\right),&{\text{pokud platí }}x<y.\end{cases}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6712b8114af34dee8b5f19ed6e7623961eb03d24)
![{\displaystyle \operatorname {arctg} x\,+\,\operatorname {arctg} y={\begin{cases}\operatorname {arctg} \left(\displaystyle {\frac {x+y}{1-xy}}\right),&{\text{pokud platí }}xy<1,\\[12pt]\pi +\operatorname {arctg} \left(\displaystyle {\frac {x+y}{1-xy}}\right),&{\text{pokud platí }}xy>1,x>0\\[12pt]-\pi +\operatorname {arctg} \left(\displaystyle {\frac {x+y}{1-xy}}\right),&{\text{pokud platí }}xy>1,x<0.\end{cases}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7f84dd92da2e3514eb7bb7dc98274b64ae60784f)
![{\displaystyle \operatorname {arctg} x\,-\,\operatorname {arctg} y={\begin{cases}\operatorname {arctg} \left(\displaystyle {\frac {x-y}{1+xy}}\right),&{\text{pokud platí }}xy>-1,\\[12pt]\pi +\operatorname {arctg} \left(\displaystyle {\frac {x-y}{1+xy}}\right),&{\text{pokud platí }}xy<-1,x>0\\[12pt]-\pi +\operatorname {arctg} \left(\displaystyle {\frac {x-y}{1+xy}}\right),&{\text{pokud platí }}xy<-1,x<0.\end{cases}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1bcb176f51a1680498dc4dfe2ca53bbdb02a7745)
![{\displaystyle \operatorname {arccotg} x\,+\,\operatorname {arccotg} y={\begin{cases}\operatorname {arccotg} \left(\displaystyle {\frac {xy-1}{x+y}}\right),&{\text{pokud platí }}x>-y,\\[10pt]\pi +\operatorname {arccotg} \left(\displaystyle {\frac {xy-1}{x+y}}\right),&{\text{pokud platí }}x<-y.\end{cases}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9b5cb80e1756739f1e087ee6cbf8718b1f496ccd)
![{\displaystyle {\begin{aligned}\arcsin x&{}=-\mathrm {i} \ln \left(\mathrm {i} x+{\sqrt {1-x^{2}}}\right)&{}\\[10pt]\arccos x&{}={\frac {\pi }{2}}\,+\mathrm {i} \ln \left(\mathrm {i} x+{\sqrt {1-x^{2}}}\right)={\frac {\pi }{2}}-\arcsin x&{}\\[10pt]\operatorname {arctg} x&{}={\frac {\mathrm {i} }{2}}\left(\ln \left(1-\mathrm {i} x\right)-\ln \left(1+\mathrm {i} x\right)\right)=\operatorname {arccotg} {\frac {1}{x}}\\[10pt]\operatorname {arccotg} x&{}={\frac {\mathrm {i} }{2}}\left(\ln \left(x-\mathrm {i} \right)-\ln \left(x+\mathrm {i} \right)\right)=\operatorname {arctg} {\frac {1}{x}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f2a2e9aa38ce919ae5d84dd48c43337f82fbf6a3)
![{\displaystyle {\begin{aligned}\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}}\,+\,\dots \ =\sum _{n=0}^{\infty }{\frac {{\binom {2n}{n}}z^{2n+1}}{4^{n}(2n+1)}}\,,\qquad {\text{je-li }}|z|\leq 1\\[10pt]\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}}+\dots \ \right)={\frac {\pi }{2}}\,-\,\sum _{n=0}^{\infty }{\frac {{\binom {2n}{n}}z^{2n+1}}{4^{n}(2n+1)}}\,,\qquad {\text{je-li }}|z|\leq 1\\[10pt]\operatorname {arctg} z&=z-{\frac {z^{3}}{3}}+{\frac {z^{5}}{5}}-{\frac {z^{7}}{7}}+\dots \ =\sum _{n=0}^{\infty }{\frac {(-1)^{n}z^{2n+1}}{2n+1}}\,,\qquad {\text{je-li }}|z|\leq 1,\ z\neq \pm \mathrm {i} \\[10pt]\operatorname {arccotg} z&={\frac {\pi }{2}}-\operatorname {arctg} z\ ={\frac {\pi }{2}}\,-\,\left(z-{\frac {z^{3}}{3}}+{\frac {z^{5}}{5}}-{\frac {z^{7}}{7}}+\dots \ \right)={\frac {\pi }{2}}\,-\,\sum _{n=0}^{\infty }{\frac {(-1)^{n}z^{2n+1}}{2n+1}}\,,\qquad {\text{je-li }}|z|\leq 1,\ z\neq \pm \mathrm {i} \\[10pt]\operatorname {arcsec} z&=\arccos {(1/z)}={\frac {\pi }{2}}\,-\,\left(z^{-1}+\left({\frac {1}{2}}\right){\frac {z^{-3}}{3}}+\left({\frac {1\cdot 3}{2\cdot 4}}\right){\frac {z^{-5}}{5}}+\dots \ \right)={\frac {\pi }{2}}\,-\,\sum _{n=0}^{\infty }{\frac {{\binom {2n}{n}}z^{-(2n+1)}}{4^{n}(2n+1)}}\,,\qquad {\text{je-li }}|z|\geq 1\\[10pt]\operatorname {arccsc} z&=\arcsin {(1/z)}=z^{-1}\,+\,\left({\frac {1}{2}}\right){\frac {z^{-3}}{3}}\,+\,\left({\frac {1\cdot 3}{2\cdot 4}}\right){\frac {z^{-5}}{5}}\,+\,\dots \ =\sum _{n=0}^{\infty }{\frac {{\binom {2n}{n}}z^{-(2n+1)}}{4^{n}(2n+1)}}\,,\qquad {\text{je-li }}|z|\geq 1\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/218c651495eb86d1ffd661c662d80694af316fa0)
Obrázky, zvuky či videa k tématu cyklometrická funkce na Wikimedia Commons 
