Rhestr Unfathiannau Trigonometrig

Dim ond y nodiant ar gyfer sin a roddir isod, mae'r nodiant ar gyfer y ffwythiannau eraill yn gyffelyb.

Nodiant	Darllener	Disgrifiad	Diffiniad
sin²(x)	"sin sgwâr x"	sin wedi ei sgwario	sin²(x) = (sin(x))²
arcsin(x)	"arcsin x"	ffwythiant gwrthdro sin	arcsin(x) = y  os a dim ond os  sin(y) = x a ${\displaystyle -{\pi \over 2}\leq y\leq {\pi \over 2}}$
(sin(x))−1	"sin x, i'r [pŵer] meinws un"	Cilydd sin	(sin(x))−1 = 1 / sin(x) = csc(x)

Gellir ysgrifennu arcsin(x) yn sin⁻¹(x) yn ogystal; rhaid gofalu rhag drysu hyn â (sin(x))⁻¹.

${\displaystyle {\begin{aligned}\cos(x)&=\sin \left(x+{\frac {\pi }{2}}\right)\\\tan(x)&={\frac {\sin(x)}{\cos(x)}}&\quad \cot(x)&={\frac {\cos(x)}{\sin(x)}}={\frac {1}{\tan(x)}}\\\sec(x)&={\frac {1}{\cos(x)}}&\quad \csc(x)&={\frac {1}{\sin(x)}}\end{aligned}}}$ 

(Gweler ffwythiant trigonometrig am fwy o wybodaeth)

Cyfnodedd

Mae cyfnod o 2π gan y ffwythiannau sin, cosin, secant, a chosecant (cylch llawn): os mae ${\displaystyle k}$  yn unrhyw gyfanrif yna mae

${\displaystyle {\begin{aligned}\sin(x)&=\sin(x+2k\pi )\\\cos(x)&=\cos(x+2k\pi )\\\sec(x)&=\sec(x+2k\pi )\\\csc(x)&=\csc(x+2k\pi )\\\end{aligned}}}$ 

Mae cyfnod o π (hanner cylch) gan y ffwythiannau tangiad a chotangiad:

${\displaystyle {\begin{aligned}\tan(x)&=\tan(x+k\pi )\\\cot(x)&=\cot(x+k\pi )\\\end{aligned}}}$ 

Cymesuredd

${\displaystyle {\begin{aligned}\sin(-x)&=-\sin(x)&\sin \left({\tfrac {\pi }{2}}-x\right)&=\cos(x)&\sin \left(\pi -x\right)&=+\sin(x)\\\cos(-x)&=+\cos(x)&\cos \left({\tfrac {\pi }{2}}-x\right)&=\sin(x)&\cos \left(\pi -x\right)&=-\cos(x)\\\tan(-x)&=-\tan(x)&\tan \left({\tfrac {\pi }{2}}-x\right)&=\cot(x)&\tan \left(\pi -x\right)&=-\tan(x)\\\cot(-x)&=-\cot(x)&\cot \left({\tfrac {\pi }{2}}-x\right)&=\tan(x)&\cot \left(\pi -x\right)&=-\cot(x)\\\sec(-x)&=+\sec(x)&\sec \left({\tfrac {\pi }{2}}-x\right)&=\csc(x)&\sec \left(\pi -x\right)&=-\sec(x)\\\csc(-x)&=-\csc(x)&\csc \left({\tfrac {\pi }{2}}-x\right)&=\sec(x)&\csc \left(\pi -x\right)&=+\csc(x)\end{aligned}}}$ 

Symudiadau

${\displaystyle {\begin{aligned}\sin \left(x+{\tfrac {\pi }{2}}\right)&=+\cos(x)&\sin \left(x+\pi \right)&=-\sin(x)\\\cos \left(x+{\tfrac {\pi }{2}}\right)&=-\sin(x)&\cos \left(x+\pi \right)&=-\cos(x)\\\tan \left(x+{\tfrac {\pi }{2}}\right)&=-\cot(x)&\tan \left(x+\pi \right)&=+\tan(x)\\\cot \left(x+{\tfrac {\pi }{2}}\right)&=-\tan(x)&\cot \left(x+\pi \right)&=+\cot(x)\\\sec \left(x+{\tfrac {\pi }{2}}\right)&=-\csc(x)&\sec \left(x+\pi \right)&=-\sec(x)\\\csc \left(x+{\tfrac {\pi }{2}}\right)&=+\sec(x)&\csc \left(x+\pi \right)&=-\csc(x)\end{aligned}}}$ 

Cyfuniadau llinol

Weithiau mae'n bwysig gwybod bod cyfuniad llinol o donau sin gyda'r un cyfnod (ond gyda gwahanol symudiad cydwedd) yn rhoi ton sin gyda'r un cyfnod. Yn gyffrefinol, mae

${\displaystyle a\sin x+b\cos x={\sqrt {a^{2}+b^{2}}}\cdot \sin(x+\varphi )\,}$ 

lle mae

${\displaystyle \varphi =\left\{{\begin{matrix}{\rm {arctan}}(b/a),&&{\mbox{os mae }}a\geq 0;\;\\\arctan(b/a)\pm \pi ,&&{\mbox{os mae }}a<0.\;\end{matrix}}\right.\;}$ 

Yn gyffredinol, am symudiad cydwedd mympwyol, mae gennym fod

${\displaystyle a\sin x+b\sin(x+\alpha )=c\sin(x+\beta )\,}$ 

lle mae

${\displaystyle c={\sqrt {a^{2}+b^{2}+2ab\cos \alpha }},}$ 

a

${\displaystyle \beta ={\rm {arctan}}\left({\frac {b\sin \alpha }{a+b\cos \alpha }}\right).}$ 

Seilir y canlynol ar theorem Pythagoras:

${\displaystyle {\begin{aligned}\sin ^{2}(x)+\cos ^{2}(x)&=1\\\tan ^{2}(x)+1&=\sec ^{2}(x)\\\cot ^{2}(x)+1&=\csc ^{2}(x)\end{aligned}}}$ 

Gellir deillio'r ail a'r trydydd hafaliad uchod o'r cyntaf trwy rhannu â cos²(x) a sin²(x) yn ôl eu trefn.

Fe'u celwir hefyd yn "fformwlâu adio a thynnu". Gellir eu profi gan ddefnyddio fformwla Euler.

${\displaystyle \sin(x\pm y)=\sin(x)\cos(y)\pm \cos(x)\sin(y)\,}$ 
(Pan y mae "+" ar y chwith, mae "+" ar y de, ac yn gyffelyb gyda "-".)

${\displaystyle \cos(x\pm y)=\cos(x)\cos(y)\mp \sin(x)\sin(y)\,}$ 
(Pan y mae "+" ar y chwith, mae "-" ar y de, ac i'r gwrthwyneb.)

${\displaystyle \tan(x\pm y)={\frac {\tan(x)\pm \tan(y)}{1\mp \tan(x)\tan(y)}}}$ 

Tangiad symiau nifer meidraidd o dermau

Gadewch i x_i = tan(θ_i ), ar gyfer i = 1, ..., n. Gadewch i e_k fod y polynomial cymesur elfennol gyda gradd k yn y newidynnau x_i, i = 1, ..., n, k = 0, ..., n. Yna mae

${\displaystyle \tan(\theta _{1}+\cdots +\theta _{n})={\frac {e_{1}-e_{3}+e_{5}-\cdots }{e_{0}-e_{2}+e_{4}-\cdots }},}$ 

gyda'r nifer o dermau yn dibynnu ar n.

Er enghraifft, mae

${\displaystyle {\begin{aligned}\tan(\theta _{1}+\theta _{2}+\theta _{3})&{}={\frac {e_{1}-e_{3}}{e_{0}-e_{2}}}={\frac {(x_{1}+x_{2}+x_{3})\ -\ (x_{1}x_{2}x_{3})}{1\ -\ (x_{1}x_{2}+x_{1}x_{3}+x_{2}x_{3})}},\\\\\tan(\theta _{1}+\theta _{2}+\theta _{3}+\theta _{4})&{}={\frac {e_{1}-e_{3}}{e_{0}-e_{2}+e_{4}}}\\\\&{}={\frac {(x_{1}+x_{2}+x_{3}+x_{4})\ -\ (x_{1}x_{2}x_{3}+x_{1}x_{2}x_{4}+x_{1}x_{3}x_{4}+x_{2}x_{3}x_{4})}{1\ -\ (x_{1}x_{2}+x_{1}x_{3}+x_{1}x_{4}+x_{2}x_{3}+x_{2}x_{4}+x_{3}x_{4})\ +\ (x_{1}x_{2}x_{3}x_{4})}},\end{aligned}}}$ 

ac yn y blaen. Gellir profi hyn trwy anwythiad mathemategol.

Gellir profi'r canlynol trwy amnewid x = y yn y fformwlâu adio, a defnyddio'r fformwla Pythagoreaidd, neu trwy ddefnyddio fformwla de Moivre gydag n = 2.

${\displaystyle \sin(2x)=2\sin(x)\cos(x)\,}$ 

${\displaystyle \cos(2x)=\cos ^{2}(x)-\sin ^{2}(x)=2\cos ^{2}(x)-1=1-2\sin ^{2}(x)={\frac {1-\tan ^{2}(x)}{1+\tan ^{2}(x)}}\,}$ 

${\displaystyle \tan(2x)={\frac {2\tan(x)}{1-\tan ^{2}(x)}}\,}$ 

${\displaystyle \cot(2x)={\frac {\cot(x)-\tan(x)}{2}}\,}$ 

Gellir defnyddio'r uchod i ganfod triawdau Pythagoraidd. os mae (a, b, c) yw hyd ochrau triongl ongl-sgwâr, yna mae (a² − b², 2ab, c²) hefyd yn ffurfio triongl ongl-sgwâr, lle mae B yw'r ongl a ddyblir. os mae a² − b² yn negatif, cymerwch ei wrthdro a defnyddio ongl cyflenwol 2B yn lle 2B.

${\displaystyle \sin(3x)=3\sin(x)-4\sin ^{3}(x)\,}$ 

${\displaystyle \cos(3x)=4\cos ^{3}(x)-3\cos(x)\,}$ 

${\displaystyle \tan(3x)={\frac {3\tan x-\tan ^{3}x}{1-3\tan ^{2}(x)}}}$ 

Os mai T_n yw'r nfed polynomial Chebyshev, yna mae

${\displaystyle \cos(nx)=T_{n}(\cos(x)).\,}$ 

Os mai S_n yw'r nfed polynomial gwasgar, yna mae

${\displaystyle \sin ^{2}(n\theta )=S_{n}(\sin ^{2}\theta ).\,}$ 

Fformwla de Moivre:

${\displaystyle \cos(nx)+i\sin(nx)=(\cos(x)+i\sin(x))^{n}\,}$ 

${\displaystyle \sin ^{2}(x)={1-\cos(2x) \over 2}}$ 

${\displaystyle \cos ^{2}(x)={1+\cos(2x) \over 2}}$ 

${\displaystyle \sin ^{2}(x)\cos ^{2}(x)={1-\cos(4x) \over 8}}$ 

${\displaystyle \sin ^{3}(x)={\frac {3\sin(x)-\sin(3x)}{4}}}$ 

${\displaystyle \cos ^{3}(x)={\frac {3\cos(x)+\cos(3x)}{4}}}$ 

${\displaystyle \cos \left({\frac {x}{2}}\right)=\pm \,{\sqrt {\frac {1+\cos(x)}{2}}}}$ 

${\displaystyle \sin \left({\frac {x}{2}}\right)=\pm \,{\sqrt {\frac {1-\cos(x)}{2}}}}$ 

${\displaystyle \tan \left({\frac {x}{2}}\right)={\sin(x/2) \over \cos(x/2)}=\pm \,{\sqrt {1-\cos x \over 1+\cos x}}.\qquad \qquad (1)}$ 

${\displaystyle \tan \left({\frac {x}{2}}\right)=\pm \,{\sqrt {(1-\cos x)(1+\cos x) \over (1+\cos x)(1+\cos x)}}=\pm \,{\sqrt {1-\cos ^{2}x \over (1+\cos x)^{2}}}}$ 
${\displaystyle ={\sin x \over 1+\cos x}.}$ 

${\displaystyle \tan \left({\frac {x}{2}}\right)=\pm \,{\sqrt {(1-\cos x)(1-\cos x) \over (1+\cos x)(1-\cos x)}}=\pm \,{\sqrt {(1-\cos x)^{2} \over (1-\cos ^{2}x)}}}$ 
${\displaystyle ={1-\cos x \over \sin x}.}$ 

${\displaystyle \tan \left({\frac {x}{2}}\right)={\frac {\sin(x)}{1+\cos(x)}}={\frac {1-\cos(x)}{\sin(x)}}.}$ 

${\displaystyle \tan \left({x \over 2}\right)=\csc(x)-\cot(x),}$ 

${\displaystyle \cot \left({x \over 2}\right)=\csc(x)+\cot(x).}$ 

${\displaystyle t=\tan \left({\frac {x}{2}}\right),}$ 

${\displaystyle \sin(x)={\frac {2t}{1+t^{2}}}}$

a

${\displaystyle \cos(x)={\frac {1-t^{2}}{1+t^{2}}}}$

a

${\displaystyle e^{ix}={\frac {1+it}{1-it}}.}$

Amnewidiad o t am tan(x/2) yw hyn, gyda'r canlyniad fod sin(x) yn newid yn 2t/(1 + t²) a cos(x) yn (1 − t²)/(1 + t²). Mae hyn yn ddefnyddiol mewn calcwlws ar gyfer integreiddio ffwythiannau cymarebol o sin(x) a cos(x).

${\displaystyle \cos \left(x\right)\cos \left(y\right)={\cos \left(x-y\right)+\cos \left(x+y\right) \over 2}\;}$ 

${\displaystyle \sin \left(x\right)\sin \left(y\right)={\cos \left(x-y\right)-\cos \left(x+y\right) \over 2}\;}$ 

${\displaystyle \sin \left(x\right)\cos \left(y\right)={\sin \left(x-y\right)+\sin \left(x+y\right) \over 2}\;}$ 

(gw. Theorem Ptolemi)

${\displaystyle \cos(x)+\cos(y)=2\cos \left({\frac {x+y}{2}}\right)\cos \left({\frac {x-y}{2}}\right)\;}$ 

${\displaystyle \sin(x)+\sin(y)=2\sin \left({\frac {x+y}{2}}\right)\cos \left({\frac {x-y}{2}}\right)\;}$ 

${\displaystyle \cos(x)-\cos(y)=-2\sin \left({x+y \over 2}\right)\sin \left({x-y \over 2}\right)\;}$ 

${\displaystyle \sin(x)-\sin(y)=2\cos \left({x+y \over 2}\right)\sin \left({x-y \over 2}\right)\;}$ 

fformwla de Moivre

${\displaystyle {\mbox{os mae }}x+y+z=\pi ,}$ 

${\displaystyle {\mbox{yna mae }}\tan(x)+\tan(y)+\tan(z)=\tan(x)\tan(y)\tan(z).\,}$ 

(Os am roi ystyr i'r fformwla tra fod unrhyw un o x, y, a z yn ongl sgwâr, rhaid cymryd mai ∞ yw'r ddau ochr. Nid +∞ neu −∞ yw hyn, ond un pwynt "at anfeidredd" a ychwanegir i'r linell rif real.)

${\displaystyle {\mbox{Os mae }}x+y+z=\pi ={\mbox{hanner cylch,}}\,}$ 

${\displaystyle {\mbox{yna mae }}\sin(2x)+\sin(2y)+\sin(2z)=4\sin(x)\sin(y)\sin(z).\,}$ 

${\displaystyle \arcsin(x)+\arccos(x)=\pi /2\;}$ 

${\displaystyle \arctan(x)+\operatorname {arccot}(x)=\pi /2.\;}$ 

${\displaystyle \arctan(x)+\arctan(1/x)=\left\{{\begin{matrix}\pi /2,&{\mbox{Os mae }}x>0\\-\pi /2,&{\mbox{os mae }}x<0\end{matrix}}\right.}$ 

${\displaystyle \arctan(x)+\arctan(y)=\arctan \left({\frac {x+y}{1-xy}}\right)+\left\{{\begin{matrix}\pi ,&{\mbox{os mae }}x,y>0\\-\pi ,&{\mbox{os mae }}x,y<0\\0,&{\mbox{fel arall }}\end{matrix}}\right.}$ 

${\displaystyle \sin[\arccos(x)]={\sqrt {1-x^{2}}}\,}$ 

${\displaystyle \cos[\arcsin(x)]={\sqrt {1-x^{2}}}\,}$ 

${\displaystyle \sin[\arctan(x)]={\frac {x}{\sqrt {1+x^{2}}}}}$ 

${\displaystyle \cos[\arctan(x)]={\frac {1}{\sqrt {1+x^{2}}}}}$ 

${\displaystyle \tan[\arcsin(x)]={\frac {x}{\sqrt {1-x^{2}}}}}$ 

${\displaystyle \tan[\arccos(x)]={\frac {\sqrt {1-x^{2}}}{x}}}$ 

${\displaystyle e^{ix}=\cos(x)+i\sin(x)\,}$ 

${\displaystyle e^{-ix}=\cos(x)-i\sin(x)\,}$ 

${\displaystyle \cos(x)={\frac {e^{ix}+e^{-ix}}{2}}\;}$ 

${\displaystyle \sin(x)={\frac {e^{ix}-e^{-ix}}{2i}}\;}$ 

lle mae i ² = −1.

Gw. fformwla Euler.

${\displaystyle \sin(\theta )={\frac {e^{i\theta }-e^{-i\theta }}{2i}}\,}$ 

${\displaystyle \cos(\theta )={\frac {e^{i\theta }+e^{-i\theta }}{2}}\,}$ 

${\displaystyle \tan(\theta )={\frac {\sin(\theta )}{\operatorname {cosh} (\theta )}}={\frac {({\frac {e^{i\theta }-e^{-i\theta }}{2i}})}{({\frac {e^{i\theta }+e^{-i\theta }}{2}})}}\,}$ 

${\displaystyle \cot(\theta )={\frac {\cos(\theta )}{\sin(\theta )}}={\frac {({\frac {e^{i\theta }+e^{-i\theta }}{2}})}{({\frac {e^{i\theta }-e^{-i\theta }}{2i}})}}\,}$ 

${\displaystyle \sec(\theta )={\frac {1}{\cos(\theta )}}={\frac {1}{({\frac {e^{i\theta }+e^{-i\theta }}{2}})}}\,}$ 

${\displaystyle \csc(\theta )={\frac {1}{\sin(\theta )}}={\frac {1}{({\frac {e^{i\theta }-e^{-i\theta }}{2i}})}}\,}$ 

${\displaystyle \operatorname {versin} (\theta )=1-\cos(\theta )=1-{\frac {e^{i\theta }+e^{-i\theta }}{2}}\,}$ 

${\displaystyle \operatorname {vercos} (\theta )=1-\sin(\theta )=1-{\frac {e^{i\theta }-e^{-i\theta }}{2i}}\,}$ 

${\displaystyle \operatorname {exsec} (\theta )=\operatorname {sec} (\theta )-1\ ={\frac {1}{\cos(\theta )}}-1={\frac {1}{({\frac {e^{i\theta }+e^{-i\theta }}{2}})}}-1\,}$ 

${\displaystyle \operatorname {excsc} (\theta )=\operatorname {csc} (\theta )-1\ ={\frac {1}{\sin(\theta )}}-1={\frac {1}{({\frac {e^{i\theta }-e^{-i\theta }}{2i}})}}-1\,}$ 

${\displaystyle \operatorname {sinh} (\theta )=-i\sin(i\theta )={\frac {e^{\theta }-e^{-\theta }}{2}}\,}$ 

${\displaystyle \operatorname {cosh} (\theta )=\cos(i\theta )={\frac {e^{\theta }+e^{-\theta }}{2}}\,}$ 

${\displaystyle \operatorname {tanh} (\theta )=-i\tan(i\theta )={\frac {\operatorname {sinh} (\theta )}{\operatorname {cosh} (\theta )}}={\frac {e^{\theta }-e^{-\theta }}{e^{\theta }+e^{-\theta }}}={\frac {e^{2\theta }-1}{e^{2\theta }+1}}\,}$ 

${\displaystyle \operatorname {coth} (\theta )=i\operatorname {cot} (i\theta )={\frac {\operatorname {cosh} (\theta )}{\operatorname {sinh} (\theta )}}={\frac {e^{\theta }+e^{-\theta }}{e^{\theta }-e^{-\theta }}}={\frac {e^{2\theta }+1}{e^{2\theta }-1}}\,}$ 

${\displaystyle \operatorname {sech} (\theta )={\frac {1}{\operatorname {cosh} (\theta )}}=\operatorname {sec} (i\theta )={\frac {2}{e^{\theta }+e^{-\theta }}}\,}$ 

${\displaystyle \operatorname {csch} (\theta )={\frac {1}{\operatorname {sinh} (\theta )}}=i\cos(i\theta )={\frac {2}{e^{\theta }-e^{-\theta }}}\,}$ 

${\displaystyle \operatorname {versinh} (\theta )=1-\cos(i\theta )=1-{\frac {e^{\theta }+e^{-\theta }}{2}}\,}$ 

${\displaystyle \operatorname {vercosh} (\theta )=1+i\sin(i\theta )=1-{\frac {e^{\theta }-e^{-\theta }}{2}}\,}$ 

${\displaystyle \operatorname {exsech} (\theta )=\operatorname {sech} (\theta )-1={\frac {1}{\operatorname {cosh} (\theta )}}-1=\operatorname {sec} (i\theta )={\frac {2}{e^{\theta }+e^{-\theta }}}-1\,}$ 

${\displaystyle \operatorname {excsch} (\theta )=\operatorname {csch} (\theta )-1={\frac {1}{\operatorname {sinh} (\theta )}}-1=i\cos(i\theta )={\frac {2}{e^{\theta }-e^{-\theta }}}-1\,}$ 

${\displaystyle \arcsin(\theta )=-i\ln(i\theta +{\sqrt {1-\theta ^{2}}})\,}$ 

${\displaystyle \arccos(\theta )=-i\ln(\theta +i{\sqrt {1-\theta ^{2}}})\,}$ 

${\displaystyle \arctan(\theta )={\frac {\ln({\frac {i+\theta }{i-\theta }})i}{2}}\,}$ 

${\displaystyle \operatorname {arccot}(\theta )=\arctan(-\theta )={\frac {i\ln({\frac {i-\theta }{i+\theta }})}{2}}\,}$ 

${\displaystyle \operatorname {arcsec}(\theta )=\arccos(\theta ^{-1})=-i\ln(\theta ^{-1}+{\sqrt {1-\theta ^{-2}}}i)\,}$ 

${\displaystyle \operatorname {arccsc}(\theta )=\arcsin(\theta ^{-1})=-i\ln(i\theta ^{-1}+{\sqrt {1-\theta ^{-2}}})\,}$ 

${\displaystyle \operatorname {arcversin} (\theta )=\arccos(1-\theta )=-i\ln(1-\theta +i{\sqrt {1-(1-\theta )^{2}}})\,}$ 

${\displaystyle \operatorname {arcvercos} (\theta )=\operatorname {arcsin} (1-\theta )=-i\ln(i-i\theta +{\sqrt {1-(1-\theta )^{2}}})\,}$ 

${\displaystyle \operatorname {arcexsec} (\theta )=\operatorname {arcsec}(1+\theta )=-i\ln((\theta +1)^{-1}+i{\sqrt {1-(1+\theta )^{2}}})\,}$ 

${\displaystyle \operatorname {arcexcsc} (\theta )=\operatorname {arccsc}(1+\theta )=-i\ln(i(\theta +1)^{-1}+{\sqrt {1-(1+\theta )^{2}}})\,}$ 

${\displaystyle \operatorname {arcsinh} (\theta )=\ln(\theta +{\sqrt {\theta ^{2}+1}})\,}$ 

${\displaystyle \operatorname {arccosh} (\theta )=\ln(\theta +{\sqrt {\theta ^{2}-1}})\,}$ 

${\displaystyle \operatorname {arctanh} (\theta )={\frac {\ln({\frac {i+\theta }{i-\theta }})}{2}}\,}$ 

${\displaystyle \operatorname {arccoth} (\theta )=\operatorname {arctanh} (-\theta )={\frac {\ln({\frac {i-\theta }{i+\theta }})}{2}}\,}$ 

${\displaystyle \operatorname {arcsech} (\theta )=\operatorname {arccosh} (\theta ^{-1})=\ln(\theta ^{-1}+{\sqrt {\theta ^{-2}-1}})\,}$ 

${\displaystyle \operatorname {arccsch} (\theta )=\operatorname {arcsinh} (\theta ^{-1})=\ln(\theta ^{-1}+{\sqrt {\theta ^{-2}+1}})\,}$ 

${\displaystyle \operatorname {arcversinh} (\theta )=\operatorname {arccosh} (\theta )-1=\ln(\theta +{\sqrt {\theta ^{2}-1}})-1\,}$ 

${\displaystyle \operatorname {arcvercosh} (\theta )=\operatorname {arcsinh} (\theta )-1=\ln(\theta +{\sqrt {\theta ^{2}+1}})-1\,}$ 

${\displaystyle \operatorname {arcexsech} (\theta )=\operatorname {arcsech} (\theta +1)=\operatorname {arccosh} ((\theta +1)^{-1})=\ln((\theta +1)^{-1}+{\sqrt {(\theta +1)^{-2}-1}})\,}$ 

${\displaystyle \operatorname {arcexcsch} (\theta )=\operatorname {arccsch} (\theta +1)=\operatorname {arcsinh} ((\theta +1)^{-1})=\ln((\theta +1)^{-1}+{\sqrt {(\theta +1)^{-2}+1}})\,}$