Arkus sínus a arkus kosínus Arkus tangens a arkus kotangens Arkus sekans a arkus kosekans Cyklometrické funkce jsou inverzní zobrazení ke goniometrickým funkcím .
Definice Mezi cyklometrické funkce patří:
Arkus sinus – arcsin {\displaystyle \arcsin } Arkus kosinus – arccos {\displaystyle \arccos } Arkus tangens – a r c t g {\displaystyle \mathrm {arctg} } Arkus kotangens – a r c c o t g {\displaystyle \mathrm {arccotg} } Arkus sekans – arcsec {\displaystyle \operatorname {arcsec} } Arkus kosekans – arccsc {\displaystyle \operatorname {arccsc} } 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 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} }
Vztahy mezi cyklometrickými a goniometrickými funkcemi sin a arcsin 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} cos a arccos 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} tg a arctg 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} cotg a arccotg 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} Vztahy mezi cyklometrickými funkcemi 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}}} Vztahy mezi cyklometrickými funkcemi se vzájemně opačnými argumenty 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}}} Součty a rozdíly cyklometrických funkcí arcsin x + arcsin y 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 − 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 + 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 − 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 + 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 − 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 + 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 arcsin x + arccos x = π 2 , {\displaystyle \arcsin x+\arccos x={\frac {\pi }{2}},} | x | ≤ 1 {\displaystyle \ |x|\leq 1} arctg x + arccotg x arctg x + arccotg x = π 2 {\displaystyle \operatorname {arctg} x+\operatorname {arccotg} x={\frac {\pi }{2}}} Vyjádření cyklometrických funkcí v logaritmickém tvaru 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}}} Vztahy mezi trigonometrickými funkcemi a cyklometrickými funkcemi Vztahy goniometrických a cyklometrických funkcí je možné jednoduše odvodit z pravoúhlého trojúhelníka ze znalosti Pythagorovy věty .
θ {\displaystyle \theta } sin θ {\displaystyle \sin \theta } cos θ {\displaystyle \cos \theta } tg θ {\displaystyle \operatorname {tg} \theta } Diagram arcsin x {\displaystyle \arcsin x} sin ( arcsin x ) = x {\displaystyle \sin(\arcsin x)=x} cos ( arcsin x ) = 1 − x 2 {\displaystyle \cos(\arcsin x)={\sqrt {1-x^{2}}}} tg ( arcsin x ) = x 1 − x 2 {\displaystyle \operatorname {tg} (\arcsin x)={\frac {x}{\sqrt {1-x^{2}}}}} arccos x {\displaystyle \arccos x} sin ( arccos x ) = 1 − x 2 {\displaystyle \sin(\arccos x)={\sqrt {1-x^{2}}}} cos ( arccos x ) = x {\displaystyle \cos(\arccos x)=x} tg ( arccos x ) = 1 − x 2 x {\displaystyle \operatorname {tg} (\arccos x)={\frac {\sqrt {1-x^{2}}}{x}}} arctg x {\displaystyle \operatorname {arctg} x} sin ( arctg x ) = x 1 + x 2 {\displaystyle \sin(\operatorname {arctg} x)={\frac {x}{\sqrt {1+x^{2}}}}} cos ( arctg x ) = 1 1 + x 2 {\displaystyle \cos(\operatorname {arctg} x)={\frac {1}{\sqrt {1+x^{2}}}}} tg ( arctg x ) = x {\displaystyle \operatorname {tg} (\operatorname {arctg} x)=x} arccotg x {\displaystyle \operatorname {arccotg} x} sin ( arccotg x ) = 1 1 + x 2 {\displaystyle \sin(\operatorname {arccotg} x)={\frac {1}{\sqrt {1+x^{2}}}}} cos ( arccotg x ) = x 1 + x 2 {\displaystyle \cos(\operatorname {arccotg} x)={\frac {x}{\sqrt {1+x^{2}}}}} tg ( arccotg x ) = 1 x {\displaystyle \operatorname {tg} (\operatorname {arccotg} x)={\frac {1}{x}}} arcsec x {\displaystyle \operatorname {arcsec} x} sin ( arcsec x ) = x 2 − 1 x {\displaystyle \sin(\operatorname {arcsec} x)={\frac {\sqrt {x^{2}-1}}{x}}} cos ( arcsec x ) = 1 x {\displaystyle \cos(\operatorname {arcsec} x)={\frac {1}{x}}} tg ( arcsec x ) = x 2 − 1 {\displaystyle \operatorname {tg} (\operatorname {arcsec} x)={\sqrt {x^{2}-1}}} arccsc x {\displaystyle \operatorname {arccsc} x} sin ( arccsc x ) = 1 x {\displaystyle \sin(\operatorname {arccsc} x)={\frac {1}{x}}} cos ( arccsc x ) = x 2 − 1 x {\displaystyle \cos(\operatorname {arccsc} x)={\frac {\sqrt {x^{2}-1}}{x}}} tg ( arccsc x ) = 1 x 2 − 1 {\displaystyle \operatorname {tg} (\operatorname {arccsc} x)={\frac {1}{\sqrt {x^{2}-1}}}}
Vyjádření nekonečným rozvojem 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}}} Literatura Rektorys, K. a spol.: Přehled užité matematiky I. , Prometheus, Praha, 2003, 7. vydání. ISBN 80-7196-179-5 Bartch, Hans-Jochen: Matematické vzorce , SNTL, Praha 1987, 2. revidované vydání Externí odkazy Obrázky, zvuky či videa k tématu cyklometrická funkce na Wikimedia Commons Reference V tomto článku byl použit překlad textu z článku Cyklometrická funkcia na slovenské Wikipedii.
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![{\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 
