Inversās trigonometriskās funkcijas

Inversās trigonometriskās funkcijas jeb ciklometriskās funkcijas ir trigonometrisko funkciju inversās funkcijas. Tām ir sašaurināti definīcijas apgabali, pie tam tā, lai šajā apgabalā katra funkcijas vērtība tiktu iegūta tikai vienu reizi. Pastāv inversā sinusa, kosinusa, tangensa, kotangensa, sekansa un kosekansa funkcijas. Inversās trigonometriskās funkcijas tiek izmantotas, lai aprēķinātu leņķus. Tās plaši tiek izmantotas navigācijā, fizikā, inženierijā u.c.

Visbiežāk inversās trigonometriskās funkcijas pieraksta, parastajai trigonometriskajai funkcijai pieliekot priekšā arc- (latīņu: arcus — ‘loks’) — arcsin x, arccos x utt. Vēl tās var tikt pierakstītas kā sin⁻¹ (x), cos⁻¹ (x), tan⁻¹ (x) utt., taču šajā gadījumā tas var tikt sajaukts ar parasto trigonometrisko funkciju, kas kāpināta −1 pakāpē. Pastāv vēl dažādi to pieraksti.

Inverso trigonometrisko funkciju uzskaitījums

Nosaukums	Pieraksts	Definīcija	x definīcijas apgabals	Galvenās vērtības diapazons (radiānos)	Galvenās vērtības diapazons (grādos)
Arksinuss	y = arcsin x	x = sin y	−1 ≤ x ≤ 1	−π/2 ≤ y ≤ π/2	−90° ≤ y ≤ 90°
Arkkosinuss	y = arccos x	x = cos y	−1 ≤ x ≤ 1	0 ≤ y ≤ π	0° ≤ y ≤ 180°
Arktangenss	y = arctg x	x = tg y	visi reālie skaitļi	−π/2 < y < π/2	−90° < y < 90°
Arkkotangenss	y = arcctg x	x = ctg y	visi reālie skaitļi	0 < y < π	0° < y < 180°
Arksekanss	y = arcsec x	x = sec y	x ≤ −1 vai 1 ≤ x	0 ≤ y < π/2 vai π/2 < y ≤ π	0° ≤ y < 90° vai 90° < y ≤ 180°
Arkkosekanss	y = arccsc x	x = csc y	x ≤ −1 vai 1 ≤ x	−π/2 ≤ y < 0 vai 0 < y ≤ π/2	-90° ≤ y < 0° vai 0° < y ≤ 90°

Atvasinājumi

Atvasinājumi kompleksām z vērtībām.

${\begin{aligned}{\frac {d}{dz}}\arcsin z&{}={\frac {1}{\sqrt {1-z^{2}}}};\quad z\neq -1,+1\\{\frac {d}{dz}}\arccos z&{}={\frac {-1}{\sqrt {1-z^{2}}}};\quad z\neq -1,+1\\{\frac {d}{dz}}\operatorname {arctg} z&{}={\frac {1}{1+z^{2}}};\quad z\neq -i,+i\\{\frac {d}{dz}}\operatorname {arcctg} z&{}={\frac {-1}{1+z^{2}}};\quad z\neq -i,+i\\{\frac {d}{dz}}\operatorname {arcsec} z&{}={\frac {1}{z^{2}\,{\sqrt {1-z^{-2}}}}};\quad z\neq -1,0,+1\\{\frac {d}{dz}}\operatorname {arccsc} z&{}={\frac {-1}{z^{2}\,{\sqrt {1-z^{-2}}}}};\quad z\neq -1,0,+1\end{aligned}}$${\begin{aligned}{\frac {d}{dz}}\arcsin z&{}={\frac {1}{\sqrt {1-z^{2}}}};\quad z\neq -1,+1\\{\frac {d}{dz}}\arccos z&{}={\frac {-1}{\sqrt {1-z^{2}}}};\quad z\neq -1,+1\\{\frac {d}{dz}}\operatorname {arctg} z&{}={\frac {1}{1+z^{2}}};\quad z\neq -i,+i\\{\frac {d}{dz}}\operatorname {arcctg} z&{}={\frac {-1}{1+z^{2}}};\quad z\neq -i,+i\\{\frac {d}{dz}}\operatorname {arcsec} z&{}={\frac {1}{z^{2}\,{\sqrt {1-z^{-2}}}}};\quad z\neq -1,0,+1\\{\frac {d}{dz}}\operatorname {arccsc} z&{}={\frac {-1}{z^{2}\,{\sqrt {1-z^{-2}}}}};\quad z\neq -1,0,+1\end{aligned}}$

Tikai x reālām vērtībām:

${\begin{aligned}{\frac {d}{dx}}\operatorname {arcsec} x&{}={\frac {1}{|x|\,{\sqrt {x^{2}-1}}}};\qquad |x|>1\\{\frac {d}{dx}}\operatorname {arccsc} x&{}={\frac {-1}{|x|\,{\sqrt {x^{2}-1}}}};\qquad |x|>1\end{aligned}}$${\begin{aligned}{\frac {d}{dx}}\operatorname {arcsec} x&{}={\frac {1}{|x|\,{\sqrt {x^{2}-1}}}};\qquad |x|>1\\{\frac {d}{dx}}\operatorname {arccsc} x&{}={\frac {-1}{|x|\,{\sqrt {x^{2}-1}}}};\qquad |x|>1\end{aligned}}$

Noteiktie, neīstie integrāļi

${\begin{aligned}\arcsin x&{}=\int _{0}^{x}{\frac {1}{\sqrt {1-z^{2}}}}\,dz,\qquad |x|\leq 1\\\arccos x&{}=\int _{x}^{1}{\frac {1}{\sqrt {1-z^{2}}}}\,dz,\qquad |x|\leq 1\\\operatorname {arctg} x&{}=\int _{0}^{x}{\frac {1}{z^{2}+1}}\,dz,\\\operatorname {arcctg} x&{}=\int _{x}^{\infty }{\frac {1}{z^{2}+1}}\,dz,\\\operatorname {arcsec} x&{}=\int _{1}^{x}{\frac {1}{z{\sqrt {z^{2}-1}}}}\,dz,\qquad x\geq 1\\\operatorname {arcsec} x&{}=\pi +\int _{x}^{-1}{\frac {1}{z{\sqrt {z^{2}-1}}}}\,dz,\qquad x\leq -1\\\operatorname {arccsc} x&{}=\int _{x}^{\infty }{\frac {1}{z{\sqrt {z^{2}-1}}}}\,dz,\qquad x\geq 1\\\operatorname {arccsc} x&{}=\int _{-\infty }^{x}{\frac {1}{z{\sqrt {z^{2}-1}}}}\,dz,\qquad x\leq -1\end{aligned}}$${\begin{aligned}\arcsin x&{}=\int _{0}^{x}{\frac {1}{\sqrt {1-z^{2}}}}\,dz,\qquad |x|\leq 1\\\arccos x&{}=\int _{x}^{1}{\frac {1}{\sqrt {1-z^{2}}}}\,dz,\qquad |x|\leq 1\\\operatorname {arctg} x&{}=\int _{0}^{x}{\frac {1}{z^{2}+1}}\,dz,\\\operatorname {arcctg} x&{}=\int _{x}^{\infty }{\frac {1}{z^{2}+1}}\,dz,\\\operatorname {arcsec} x&{}=\int _{1}^{x}{\frac {1}{z{\sqrt {z^{2}-1}}}}\,dz,\qquad x\geq 1\\\operatorname {arcsec} x&{}=\pi +\int _{x}^{-1}{\frac {1}{z{\sqrt {z^{2}-1}}}}\,dz,\qquad x\leq -1\\\operatorname {arccsc} x&{}=\int _{x}^{\infty }{\frac {1}{z{\sqrt {z^{2}-1}}}}\,dz,\qquad x\geq 1\\\operatorname {arccsc} x&{}=\int _{-\infty }^{x}{\frac {1}{z{\sqrt {z^{2}-1}}}}\,dz,\qquad x\leq -1\end{aligned}}$

Funkciju izvirzījumi

Funkciju izvirzījumi pakāpju rindās:

$\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 }{\frac {{\binom {2n}{n}}z^{2n+1}}{4^{n}(2n+1)}};\qquad |z|\leq 1$$\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 }{\frac {{\binom {2n}{n}}z^{2n+1}}{4^{n}(2n+1)}};\qquad |z|\leq 1$

$\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}}+\cdots \ \right)={\frac {\pi }{2}}-\sum _{n=0}^{\infty }{\frac {{\binom {2n}{n}}z^{2n+1}}{4^{n}(2n+1)}};\qquad |z|\leq 1$$\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}}+\cdots \ \right)={\frac {\pi }{2}}-\sum _{n=0}^{\infty }{\frac {{\binom {2n}{n}}z^{2n+1}}{4^{n}(2n+1)}};\qquad |z|\leq 1$

$\operatorname {arctg} 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}};\qquad |z|\leq 1\qquad z\neq i,-i$$\operatorname {arctg} 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}};\qquad |z|\leq 1\qquad z\neq i,-i$

$\operatorname {arcctg} z={\frac {\pi }{2}}-\operatorname {arctg} z\ ={\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}};\qquad |z|\leq 1\qquad z\neq i,-i$$\operatorname {arcctg} z={\frac {\pi }{2}}-\operatorname {arctg} z\ ={\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}};\qquad |z|\leq 1\qquad z\neq i,-i$

$\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}}+\cdots \ \right)={\frac {\pi }{2}}-\sum _{n=0}^{\infty }{\frac {{\binom {2n}{n}}z^{-(2n+1)}}{4^{n}(2n+1)}};\qquad |z|\geq 1$$\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}}+\cdots \ \right)={\frac {\pi }{2}}-\sum _{n=0}^{\infty }{\frac {{\binom {2n}{n}}z^{-(2n+1)}}{4^{n}(2n+1)}};\qquad |z|\geq 1$

$\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}}+\cdots \ =\sum _{n=0}^{\infty }{\frac {{\binom {2n}{n}}z^{-(2n+1)}}{4^{n}(2n+1)}};\qquad |z|\geq 1$$\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}}+\cdots \ =\sum _{n=0}^{\infty }{\frac {{\binom {2n}{n}}z^{-(2n+1)}}{4^{n}(2n+1)}};\qquad |z|\geq 1$

Nenoteiktie integrāļi

Reālām un kompleksām x vērtībām:

${\begin{aligned}\int \arcsin x\,dx&{}=x\,\arcsin x+{\sqrt {1-x^{2}}}+C\\\int \arccos x\,dx&{}=x\,\arccos x-{\sqrt {1-x^{2}}}+C\\\int \operatorname {arctg} x\,dx&{}=x\,\operatorname {arctg} x-{\frac {1}{2}}\ln \left(1+x^{2}\right)+C\\\int \operatorname {arcctg} x\,dx&{}=x\,\operatorname {arcctg} x+{\frac {1}{2}}\ln \left(1+x^{2}\right)+C\\\int \operatorname {arcsec} x\,dx&{}=x\,\operatorname {arcsec} x-\ln \left[x\left(1+{\sqrt {{x^{2}-1} \over x^{2}}}\right)\right]+C\\\int \operatorname {arccsc} x\,dx&{}=x\,\operatorname {arccsc} x+\ln \left[x\left(1+{\sqrt {{x^{2}-1} \over x^{2}}}\right)\right]+C\end{aligned}}$${\begin{aligned}\int \arcsin x\,dx&{}=x\,\arcsin x+{\sqrt {1-x^{2}}}+C\\\int \arccos x\,dx&{}=x\,\arccos x-{\sqrt {1-x^{2}}}+C\\\int \operatorname {arctg} x\,dx&{}=x\,\operatorname {arctg} x-{\frac {1}{2}}\ln \left(1+x^{2}\right)+C\\\int \operatorname {arcctg} x\,dx&{}=x\,\operatorname {arcctg} x+{\frac {1}{2}}\ln \left(1+x^{2}\right)+C\\\int \operatorname {arcsec} x\,dx&{}=x\,\operatorname {arcsec} x-\ln \left[x\left(1+{\sqrt {{x^{2}-1} \over x^{2}}}\right)\right]+C\\\int \operatorname {arccsc} x\,dx&{}=x\,\operatorname {arccsc} x+\ln \left[x\left(1+{\sqrt {{x^{2}-1} \over x^{2}}}\right)\right]+C\end{aligned}}$

Reālām un kompleksām x ≥ 1 vērtībām:

${\begin{aligned}\int \operatorname {arcsec} x\,dx&{}=x\,\operatorname {arcsec} x-\ln \left(x+{\sqrt {x^{2}-1}}\right)+C\\\int \operatorname {arccsc} x\,dx&{}=x\,\operatorname {arccsc} x+\ln \left(x+{\sqrt {x^{2}-1}}\right)+C\end{aligned}}$${\begin{aligned}\int \operatorname {arcsec} x\,dx&{}=x\,\operatorname {arcsec} x-\ln \left(x+{\sqrt {x^{2}-1}}\right)+C\\\int \operatorname {arccsc} x\,dx&{}=x\,\operatorname {arccsc} x+\ln \left(x+{\sqrt {x^{2}-1}}\right)+C\end{aligned}}$

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