Defnyddir y nodiant canlynol ar gyfer pob un o'r chwech ffwythiant trigonometrig (sin, cosin (cos), tangiad (tan), cotangiad (cot), secant (sec), a chosecant (csc). 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
−
π
2
≤
y
≤
π
2
{\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 .
cos
(
x
)
=
sin
(
x
+
π
2
)
tan
(
x
)
=
sin
(
x
)
cos
(
x
)
cot
(
x
)
=
cos
(
x
)
sin
(
x
)
=
1
tan
(
x
)
sec
(
x
)
=
1
cos
(
x
)
csc
(
x
)
=
1
sin
(
x
)
{\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)
golygu
Mae cyfnod o 2π gan y ffwythiannau sin, cosin, secant, a chosecant (cylch llawn): os mae
k
{\displaystyle k}
gyfanrif yna mae
sin
(
x
)
=
sin
(
x
+
2
k
π
)
cos
(
x
)
=
cos
(
x
+
2
k
π
)
sec
(
x
)
=
sec
(
x
+
2
k
π
)
csc
(
x
)
=
csc
(
x
+
2
k
π
)
{\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:
tan
(
x
)
=
tan
(
x
+
k
π
)
cot
(
x
)
=
cot
(
x
+
k
π
)
{\displaystyle {\begin{aligned}\tan(x)&=\tan(x+k\pi )\\\cot(x)&=\cot(x+k\pi )\\\end{aligned}}}
sin
(
−
x
)
=
−
sin
(
x
)
sin
(
π
2
−
x
)
=
cos
(
x
)
sin
(
π
−
x
)
=
+
sin
(
x
)
cos
(
−
x
)
=
+
cos
(
x
)
cos
(
π
2
−
x
)
=
sin
(
x
)
cos
(
π
−
x
)
=
−
cos
(
x
)
tan
(
−
x
)
=
−
tan
(
x
)
tan
(
π
2
−
x
)
=
cot
(
x
)
tan
(
π
−
x
)
=
−
tan
(
x
)
cot
(
−
x
)
=
−
cot
(
x
)
cot
(
π
2
−
x
)
=
tan
(
x
)
cot
(
π
−
x
)
=
−
cot
(
x
)
sec
(
−
x
)
=
+
sec
(
x
)
sec
(
π
2
−
x
)
=
csc
(
x
)
sec
(
π
−
x
)
=
−
sec
(
x
)
csc
(
−
x
)
=
−
csc
(
x
)
csc
(
π
2
−
x
)
=
sec
(
x
)
csc
(
π
−
x
)
=
+
csc
(
x
)
{\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}}}
sin
(
x
+
π
2
)
=
+
cos
(
x
)
sin
(
x
+
π
)
=
−
sin
(
x
)
cos
(
x
+
π
2
)
=
−
sin
(
x
)
cos
(
x
+
π
)
=
−
cos
(
x
)
tan
(
x
+
π
2
)
=
−
cot
(
x
)
tan
(
x
+
π
)
=
+
tan
(
x
)
cot
(
x
+
π
2
)
=
−
tan
(
x
)
cot
(
x
+
π
)
=
+
cot
(
x
)
sec
(
x
+
π
2
)
=
−
csc
(
x
)
sec
(
x
+
π
)
=
−
sec
(
x
)
csc
(
x
+
π
2
)
=
+
sec
(
x
)
csc
(
x
+
π
)
=
−
csc
(
x
)
{\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}}}
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
a
sin
x
+
b
cos
x
=
a
2
+
b
2
⋅
sin
(
x
+
φ
)
{\displaystyle a\sin x+b\cos x={\sqrt {a^{2}+b^{2}}}\cdot \sin(x+\varphi )\,}
lle mae
φ
=
{
a
r
c
t
a
n
(
b
/
a
)
,
os mae
a
≥
0
;
arctan
(
b
/
a
)
±
π
,
os mae
a
<
0.
{\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
a
sin
x
+
b
sin
(
x
+
α
)
=
c
sin
(
x
+
β
)
{\displaystyle a\sin x+b\sin(x+\alpha )=c\sin(x+\beta )\,}
lle mae
c
=
a
2
+
b
2
+
2
a
b
cos
α
,
{\displaystyle c={\sqrt {a^{2}+b^{2}+2ab\cos \alpha }},}
a
β
=
a
r
c
t
a
n
(
b
sin
α
a
+
b
cos
α
)
.
{\displaystyle \beta ={\rm {arctan}}\left({\frac {b\sin \alpha }{a+b\cos \alpha }}\right).}
Unfathiannau Pythagoreaidd
golygu
Seilir y canlynol ar theorem Pythagoras :
sin
2
(
x
)
+
cos
2
(
x
)
=
1
tan
2
(
x
)
+
1
=
sec
2
(
x
)
cot
2
(
x
)
+
1
=
csc
2
(
x
)
{\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.
Unfathiannau swm neu wahaniaeth onglau
golygu
Fe'u celwir hefyd yn "fformwlâu adio a thynnu". Gellir eu profi gan ddefnyddio fformwla Euler .
sin
(
x
±
y
)
=
sin
(
x
)
cos
(
y
)
±
cos
(
x
)
sin
(
y
)
{\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 "-".)
cos
(
x
±
y
)
=
cos
(
x
)
cos
(
y
)
∓
sin
(
x
)
sin
(
y
)
{\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.)
tan
(
x
±
y
)
=
tan
(
x
)
±
tan
(
y
)
1
∓
tan
(
x
)
tan
(
y
)
{\displaystyle \tan(x\pm y)={\frac {\tan(x)\pm \tan(y)}{1\mp \tan(x)\tan(y)}}}
Tangiad symiau nifer meidraidd o dermau
golygu
Gadewch i x i i i = 1, ..., n . Gadewch i e k polynomial cymesur elfennol gyda gradd k yn y newidynnau x i i = 1, ..., n , k = 0, ..., n . Yna mae
tan
(
θ
1
+
⋯
+
θ
n
)
=
e
1
−
e
3
+
e
5
−
⋯
e
0
−
e
2
+
e
4
−
⋯
,
{\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
tan
(
θ
1
+
θ
2
+
θ
3
)
=
e
1
−
e
3
e
0
−
e
2
=
(
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
(
θ
1
+
θ
2
+
θ
3
+
θ
4
)
=
e
1
−
e
3
e
0
−
e
2
+
e
4
=
(
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
)
,
{\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.
sin
(
2
x
)
=
2
sin
(
x
)
cos
(
x
)
{\displaystyle \sin(2x)=2\sin(x)\cos(x)\,}
cos
(
2
x
)
=
cos
2
(
x
)
−
sin
2
(
x
)
=
2
cos
2
(
x
)
−
1
=
1
−
2
sin
2
(
x
)
=
1
−
tan
2
(
x
)
1
+
tan
2
(
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)}}\,}
tan
(
2
x
)
=
2
tan
(
x
)
1
−
tan
2
(
x
)
{\displaystyle \tan(2x)={\frac {2\tan(x)}{1-\tan ^{2}(x)}}\,}
cot
(
2
x
)
=
cot
(
x
)
−
tan
(
x
)
2
{\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 (a 2 − b 2 , 2ab , c 2 ) hefyd yn ffurfio triongl ongl-sgwâr, lle mae B yw'r ongl a ddyblir. os mae a 2 − b 2 yn negatif, cymerwch ei wrthdro a defnyddio ongl cyflenwol 2B yn lle 2B .
sin
(
3
x
)
=
3
sin
(
x
)
−
4
sin
3
(
x
)
{\displaystyle \sin(3x)=3\sin(x)-4\sin ^{3}(x)\,}
cos
(
3
x
)
=
4
cos
3
(
x
)
−
3
cos
(
x
)
{\displaystyle \cos(3x)=4\cos ^{3}(x)-3\cos(x)\,}
tan
(
3
x
)
=
3
tan
x
−
tan
3
x
1
−
3
tan
2
(
x
)
{\displaystyle \tan(3x)={\frac {3\tan x-\tan ^{3}x}{1-3\tan ^{2}(x)}}}
Os mai Tn yw'r n fed polynomial Chebyshev , yna mae
cos
(
n
x
)
=
T
n
(
cos
(
x
)
)
.
{\displaystyle \cos(nx)=T_{n}(\cos(x)).\,}
Os mai S n n fed polynomial gwasgar , yna mae
sin
2
(
n
θ
)
=
S
n
(
sin
2
θ
)
.
{\displaystyle \sin ^{2}(n\theta )=S_{n}(\sin ^{2}\theta ).\,}
Fformwla de Moivre :
cos
(
n
x
)
+
i
sin
(
n
x
)
=
(
cos
(
x
)
+
i
sin
(
x
)
)
n
{\displaystyle \cos(nx)+i\sin(nx)=(\cos(x)+i\sin(x))^{n}\,}
sin
2
(
x
)
=
1
−
cos
(
2
x
)
2
{\displaystyle \sin ^{2}(x)={1-\cos(2x) \over 2}}
cos
2
(
x
)
=
1
+
cos
(
2
x
)
2
{\displaystyle \cos ^{2}(x)={1+\cos(2x) \over 2}}
sin
2
(
x
)
cos
2
(
x
)
=
1
−
cos
(
4
x
)
8
{\displaystyle \sin ^{2}(x)\cos ^{2}(x)={1-\cos(4x) \over 8}}
sin
3
(
x
)
=
3
sin
(
x
)
−
sin
(
3
x
)
4
{\displaystyle \sin ^{3}(x)={\frac {3\sin(x)-\sin(3x)}{4}}}
cos
3
(
x
)
=
3
cos
(
x
)
+
cos
(
3
x
)
4
{\displaystyle \cos ^{3}(x)={\frac {3\cos(x)+\cos(3x)}{4}}}
cos
(
x
2
)
=
±
1
+
cos
(
x
)
2
{\displaystyle \cos \left({\frac {x}{2}}\right)=\pm \,{\sqrt {\frac {1+\cos(x)}{2}}}}
sin
(
x
2
)
=
±
1
−
cos
(
x
)
2
{\displaystyle \sin \left({\frac {x}{2}}\right)=\pm \,{\sqrt {\frac {1-\cos(x)}{2}}}}
tan
(
x
2
)
=
sin
(
x
/
2
)
cos
(
x
/
2
)
=
±
1
−
cos
x
1
+
cos
x
.
(
1
)
{\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)}
tan
(
x
2
)
=
±
(
1
−
cos
x
)
(
1
+
cos
x
)
(
1
+
cos
x
)
(
1
+
cos
x
)
=
±
1
−
cos
2
x
(
1
+
cos
x
)
2
{\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}}}}
=
sin
x
1
+
cos
x
.
{\displaystyle ={\sin x \over 1+\cos x}.}
tan
(
x
2
)
=
±
(
1
−
cos
x
)
(
1
−
cos
x
)
(
1
+
cos
x
)
(
1
−
cos
x
)
=
±
(
1
−
cos
x
)
2
(
1
−
cos
2
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)}}}
=
1
−
cos
x
sin
x
.
{\displaystyle ={1-\cos x \over \sin x}.}
tan
(
x
2
)
=
sin
(
x
)
1
+
cos
(
x
)
=
1
−
cos
(
x
)
sin
(
x
)
.
{\displaystyle \tan \left({\frac {x}{2}}\right)={\frac {\sin(x)}{1+\cos(x)}}={\frac {1-\cos(x)}{\sin(x)}}.}
tan
(
x
2
)
=
csc
(
x
)
−
cot
(
x
)
,
{\displaystyle \tan \left({x \over 2}\right)=\csc(x)-\cot(x),}
cot
(
x
2
)
=
csc
(
x
)
+
cot
(
x
)
.
{\displaystyle \cot \left({x \over 2}\right)=\csc(x)+\cot(x).}
t
=
tan
(
x
2
)
,
{\displaystyle t=\tan \left({\frac {x}{2}}\right),}
sin
(
x
)
=
2
t
1
+
t
2
{\displaystyle \sin(x)={\frac {2t}{1+t^{2}}}}
a
cos
(
x
)
=
1
−
t
2
1
+
t
2
{\displaystyle \cos(x)={\frac {1-t^{2}}{1+t^{2}}}}
a
e
i
x
=
1
+
i
t
1
−
i
t
.
{\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 + t 2 ) a cos(x ) yn (1 − t 2 )/(1 + t 2 ). Mae hyn yn ddefnyddiol mewn calcwlws ar gyfer integreiddio ffwythiannau cymarebol o sin(x ) a cos(x ).
Unfathiannau lluoswm-i-swm
golygu
Unfathiannau swm-i-lluoswm
golygu
cos
(
x
)
+
cos
(
y
)
=
2
cos
(
x
+
y
2
)
cos
(
x
−
y
2
)
{\displaystyle \cos(x)+\cos(y)=2\cos \left({\frac {x+y}{2}}\right)\cos \left({\frac {x-y}{2}}\right)\;}
sin
(
x
)
+
sin
(
y
)
=
2
sin
(
x
+
y
2
)
cos
(
x
−
y
2
)
{\displaystyle \sin(x)+\sin(y)=2\sin \left({\frac {x+y}{2}}\right)\cos \left({\frac {x-y}{2}}\right)\;}
cos
(
x
)
−
cos
(
y
)
=
−
2
sin
(
x
+
y
2
)
sin
(
x
−
y
2
)
{\displaystyle \cos(x)-\cos(y)=-2\sin \left({x+y \over 2}\right)\sin \left({x-y \over 2}\right)\;}
sin
(
x
)
−
sin
(
y
)
=
2
cos
(
x
+
y
2
)
sin
(
x
−
y
2
)
{\displaystyle \sin(x)-\sin(y)=2\cos \left({x+y \over 2}\right)\sin \left({x-y \over 2}\right)\;}
fformwla de Moivre
os mae
x
+
y
+
z
=
π
,
{\displaystyle {\mbox{os mae }}x+y+z=\pi ,}
yna mae
tan
(
x
)
+
tan
(
y
)
+
tan
(
z
)
=
tan
(
x
)
tan
(
y
)
tan
(
z
)
.
{\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 .)
Os mae
x
+
y
+
z
=
π
=
hanner cylch,
{\displaystyle {\mbox{Os mae }}x+y+z=\pi ={\mbox{hanner cylch,}}\,}
yna mae
sin
(
2
x
)
+
sin
(
2
y
)
+
sin
(
2
z
)
=
4
sin
(
x
)
sin
(
y
)
sin
(
z
)
.
{\displaystyle {\mbox{yna mae }}\sin(2x)+\sin(2y)+\sin(2z)=4\sin(x)\sin(y)\sin(z).\,}
Ffwythiannau trigonometrig gwrthdro
golygu
arcsin
(
x
)
+
arccos
(
x
)
=
π
/
2
{\displaystyle \arcsin(x)+\arccos(x)=\pi /2\;}
arctan
(
x
)
+
arccot
(
x
)
=
π
/
2.
{\displaystyle \arctan(x)+\operatorname {arccot}(x)=\pi /2.\;}
arctan
(
x
)
+
arctan
(
1
/
x
)
=
{
π
/
2
,
Os mae
x
>
0
−
π
/
2
,
os mae
x
<
0
{\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.}
arctan
(
x
)
+
arctan
(
y
)
=
arctan
(
x
+
y
1
−
x
y
)
+
{
π
,
os mae
x
,
y
>
0
−
π
,
os mae
x
,
y
<
0
0
,
fel arall
{\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.}
sin
[
arccos
(
x
)
]
=
1
−
x
2
{\displaystyle \sin[\arccos(x)]={\sqrt {1-x^{2}}}\,}
cos
[
arcsin
(
x
)
]
=
1
−
x
2
{\displaystyle \cos[\arcsin(x)]={\sqrt {1-x^{2}}}\,}
sin
[
arctan
(
x
)
]
=
x
1
+
x
2
{\displaystyle \sin[\arctan(x)]={\frac {x}{\sqrt {1+x^{2}}}}}
cos
[
arctan
(
x
)
]
=
1
1
+
x
2
{\displaystyle \cos[\arctan(x)]={\frac {1}{\sqrt {1+x^{2}}}}}
tan
[
arcsin
(
x
)
]
=
x
1
−
x
2
{\displaystyle \tan[\arcsin(x)]={\frac {x}{\sqrt {1-x^{2}}}}}
tan
[
arccos
(
x
)
]
=
1
−
x
2
x
{\displaystyle \tan[\arccos(x)]={\frac {\sqrt {1-x^{2}}}{x}}}
golygu
sin
(
θ
)
=
e
i
θ
−
e
−
i
θ
2
i
{\displaystyle \sin(\theta )={\frac {e^{i\theta }-e^{-i\theta }}{2i}}\,}
cos
(
θ
)
=
e
i
θ
+
e
−
i
θ
2
{\displaystyle \cos(\theta )={\frac {e^{i\theta }+e^{-i\theta }}{2}}\,}
tan
(
θ
)
=
sin
(
θ
)
cosh
(
θ
)
=
(
e
i
θ
−
e
−
i
θ
2
i
)
(
e
i
θ
+
e
−
i
θ
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}})}}\,}
cot
(
θ
)
=
cos
(
θ
)
sin
(
θ
)
=
(
e
i
θ
+
e
−
i
θ
2
)
(
e
i
θ
−
e
−
i
θ
2
i
)
{\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}})}}\,}
sec
(
θ
)
=
1
cos
(
θ
)
=
1
(
e
i
θ
+
e
−
i
θ
2
)
{\displaystyle \sec(\theta )={\frac {1}{\cos(\theta )}}={\frac {1}{({\frac {e^{i\theta }+e^{-i\theta }}{2}})}}\,}
csc
(
θ
)
=
1
sin
(
θ
)
=
1
(
e
i
θ
−
e
−
i
θ
2
i
)
{\displaystyle \csc(\theta )={\frac {1}{\sin(\theta )}}={\frac {1}{({\frac {e^{i\theta }-e^{-i\theta }}{2i}})}}\,}
versin
(
θ
)
=
1
−
cos
(
θ
)
=
1
−
e
i
θ
+
e
−
i
θ
2
{\displaystyle \operatorname {versin} (\theta )=1-\cos(\theta )=1-{\frac {e^{i\theta }+e^{-i\theta }}{2}}\,}
vercos
(
θ
)
=
1
−
sin
(
θ
)
=
1
−
e
i
θ
−
e
−
i
θ
2
i
{\displaystyle \operatorname {vercos} (\theta )=1-\sin(\theta )=1-{\frac {e^{i\theta }-e^{-i\theta }}{2i}}\,}
exsec
(
θ
)
=
sec
(
θ
)
−
1
=
1
cos
(
θ
)
−
1
=
1
(
e
i
θ
+
e
−
i
θ
2
)
−
1
{\displaystyle \operatorname {exsec} (\theta )=\operatorname {sec} (\theta )-1\ ={\frac {1}{\cos(\theta )}}-1={\frac {1}{({\frac {e^{i\theta }+e^{-i\theta }}{2}})}}-1\,}
excsc
(
θ
)
=
csc
(
θ
)
−
1
=
1
sin
(
θ
)
−
1
=
1
(
e
i
θ
−
e
−
i
θ
2
i
)
−
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\,}
sinh
(
θ
)
=
−
i
sin
(
i
θ
)
=
e
θ
−
e
−
θ
2
{\displaystyle \operatorname {sinh} (\theta )=-i\sin(i\theta )={\frac {e^{\theta }-e^{-\theta }}{2}}\,}
cosh
(
θ
)
=
cos
(
i
θ
)
=
e
θ
+
e
−
θ
2
{\displaystyle \operatorname {cosh} (\theta )=\cos(i\theta )={\frac {e^{\theta }+e^{-\theta }}{2}}\,}
tanh
(
θ
)
=
−
i
tan
(
i
θ
)
=
sinh
(
θ
)
cosh
(
θ
)
=
e
θ
−
e
−
θ
e
θ
+
e
−
θ
=
e
2
θ
−
1
e
2
θ
+
1
{\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}}\,}
coth
(
θ
)
=
i
cot
(
i
θ
)
=
cosh
(
θ
)
sinh
(
θ
)
=
e
θ
+
e
−
θ
e
θ
−
e
−
θ
=
e
2
θ
+
1
e
2
θ
−
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}}\,}
sech
(
θ
)
=
1
cosh
(
θ
)
=
sec
(
i
θ
)
=
2
e
θ
+
e
−
θ
{\displaystyle \operatorname {sech} (\theta )={\frac {1}{\operatorname {cosh} (\theta )}}=\operatorname {sec} (i\theta )={\frac {2}{e^{\theta }+e^{-\theta }}}\,}
csch
(
θ
)
=
1
sinh
(
θ
)
=
i
cos
(
i
θ
)
=
2
e
θ
−
e
−
θ
{\displaystyle \operatorname {csch} (\theta )={\frac {1}{\operatorname {sinh} (\theta )}}=i\cos(i\theta )={\frac {2}{e^{\theta }-e^{-\theta }}}\,}
versinh
(
θ
)
=
1
−
cos
(
i
θ
)
=
1
−
e
θ
+
e
−
θ
2
{\displaystyle \operatorname {versinh} (\theta )=1-\cos(i\theta )=1-{\frac {e^{\theta }+e^{-\theta }}{2}}\,}
vercosh
(
θ
)
=
1
+
i
sin
(
i
θ
)
=
1
−
e
θ
−
e
−
θ
2
{\displaystyle \operatorname {vercosh} (\theta )=1+i\sin(i\theta )=1-{\frac {e^{\theta }-e^{-\theta }}{2}}\,}
exsech
(
θ
)
=
sech
(
θ
)
−
1
=
1
cosh
(
θ
)
−
1
=
sec
(
i
θ
)
=
2
e
θ
+
e
−
θ
−
1
{\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\,}
excsch
(
θ
)
=
csch
(
θ
)
−
1
=
1
sinh
(
θ
)
−
1
=
i
cos
(
i
θ
)
=
2
e
θ
−
e
−
θ
−
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\,}
arcsin
(
θ
)
=
−
i
ln
(
i
θ
+
1
−
θ
2
)
{\displaystyle \arcsin(\theta )=-i\ln(i\theta +{\sqrt {1-\theta ^{2}}})\,}
arccos
(
θ
)
=
−
i
ln
(
θ
+
i
1
−
θ
2
)
{\displaystyle \arccos(\theta )=-i\ln(\theta +i{\sqrt {1-\theta ^{2}}})\,}
arctan
(
θ
)
=
ln
(
i
+
θ
i
−
θ
)
i
2
{\displaystyle \arctan(\theta )={\frac {\ln({\frac {i+\theta }{i-\theta }})i}{2}}\,}
arccot
(
θ
)
=
arctan
(
−
θ
)
=
i
ln
(
i
−
θ
i
+
θ
)
2
{\displaystyle \operatorname {arccot}(\theta )=\arctan(-\theta )={\frac {i\ln({\frac {i-\theta }{i+\theta }})}{2}}\,}
arcsec
(
θ
)
=
arccos
(
θ
−
1
)
=
−
i
ln
(
θ
−
1
+
1
−
θ
−
2
i
)
{\displaystyle \operatorname {arcsec}(\theta )=\arccos(\theta ^{-1})=-i\ln(\theta ^{-1}+{\sqrt {1-\theta ^{-2}}}i)\,}
arccsc
(
θ
)
=
arcsin
(
θ
−
1
)
=
−
i
ln
(
i
θ
−
1
+
1
−
θ
−
2
)
{\displaystyle \operatorname {arccsc}(\theta )=\arcsin(\theta ^{-1})=-i\ln(i\theta ^{-1}+{\sqrt {1-\theta ^{-2}}})\,}
arcversin
(
θ
)
=
arccos
(
1
−
θ
)
=
−
i
ln
(
1
−
θ
+
i
1
−
(
1
−
θ
)
2
)
{\displaystyle \operatorname {arcversin} (\theta )=\arccos(1-\theta )=-i\ln(1-\theta +i{\sqrt {1-(1-\theta )^{2}}})\,}
arcvercos
(
θ
)
=
arcsin
(
1
−
θ
)
=
−
i
ln
(
i
−
i
θ
+
1
−
(
1
−
θ
)
2
)
{\displaystyle \operatorname {arcvercos} (\theta )=\operatorname {arcsin} (1-\theta )=-i\ln(i-i\theta +{\sqrt {1-(1-\theta )^{2}}})\,}
arcexsec
(
θ
)
=
arcsec
(
1
+
θ
)
=
−
i
ln
(
(
θ
+
1
)
−
1
+
i
1
−
(
1
+
θ
)
2
)
{\displaystyle \operatorname {arcexsec} (\theta )=\operatorname {arcsec}(1+\theta )=-i\ln((\theta +1)^{-1}+i{\sqrt {1-(1+\theta )^{2}}})\,}
arcexcsc
(
θ
)
=
arccsc
(
1
+
θ
)
=
−
i
ln
(
i
(
θ
+
1
)
−
1
+
1
−
(
1
+
θ
)
2
)
{\displaystyle \operatorname {arcexcsc} (\theta )=\operatorname {arccsc}(1+\theta )=-i\ln(i(\theta +1)^{-1}+{\sqrt {1-(1+\theta )^{2}}})\,}
arcsinh
(
θ
)
=
ln
(
θ
+
θ
2
+
1
)
{\displaystyle \operatorname {arcsinh} (\theta )=\ln(\theta +{\sqrt {\theta ^{2}+1}})\,}
arccosh
(
θ
)
=
ln
(
θ
+
θ
2
−
1
)
{\displaystyle \operatorname {arccosh} (\theta )=\ln(\theta +{\sqrt {\theta ^{2}-1}})\,}
arctanh
(
θ
)
=
ln
(
i
+
θ
i
−
θ
)
2
{\displaystyle \operatorname {arctanh} (\theta )={\frac {\ln({\frac {i+\theta }{i-\theta }})}{2}}\,}
arccoth
(
θ
)
=
arctanh
(
−
θ
)
=
ln
(
i
−
θ
i
+
θ
)
2
{\displaystyle \operatorname {arccoth} (\theta )=\operatorname {arctanh} (-\theta )={\frac {\ln({\frac {i-\theta }{i+\theta }})}{2}}\,}
arcsech
(
θ
)
=
arccosh
(
θ
−
1
)
=
ln
(
θ
−
1
+
θ
−
2
−
1
)
{\displaystyle \operatorname {arcsech} (\theta )=\operatorname {arccosh} (\theta ^{-1})=\ln(\theta ^{-1}+{\sqrt {\theta ^{-2}-1}})\,}
arccsch
(
θ
)
=
arcsinh
(
θ
−
1
)
=
ln
(
θ
−
1
+
θ
−
2
+
1
)
{\displaystyle \operatorname {arccsch} (\theta )=\operatorname {arcsinh} (\theta ^{-1})=\ln(\theta ^{-1}+{\sqrt {\theta ^{-2}+1}})\,}
arcversinh
(
θ
)
=
arccosh
(
θ
)
−
1
=
ln
(
θ
+
θ
2
−
1
)
−
1
{\displaystyle \operatorname {arcversinh} (\theta )=\operatorname {arccosh} (\theta )-1=\ln(\theta +{\sqrt {\theta ^{2}-1}})-1\,}
arcvercosh
(
θ
)
=
arcsinh
(
θ
)
−
1
=
ln
(
θ
+
θ
2
+
1
)
−
1
{\displaystyle \operatorname {arcvercosh} (\theta )=\operatorname {arcsinh} (\theta )-1=\ln(\theta +{\sqrt {\theta ^{2}+1}})-1\,}
arcexsech
(
θ
)
=
arcsech
(
θ
+
1
)
=
arccosh
(
(
θ
+
1
)
−
1
)
=
ln
(
(
θ
+
1
)
−
1
+
(
θ
+
1
)
−
2
−
1
)
{\displaystyle \operatorname {arcexsech} (\theta )=\operatorname {arcsech} (\theta +1)=\operatorname {arccosh} ((\theta +1)^{-1})=\ln((\theta +1)^{-1}+{\sqrt {(\theta +1)^{-2}-1}})\,}
arcexcsch
(
θ
)
=
arccsch
(
θ
+
1
)
=
arcsinh
(
(
θ
+
1
)
−
1
)
=
ln
(
(
θ
+
1
)
−
1
+
(
θ
+
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}})\,}
![{\displaystyle \sin[\arccos(x)]={\sqrt {1-x^{2}}}\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/89020e4173e7c25657019ade3629e832bb572ff7)
![{\displaystyle \cos[\arcsin(x)]={\sqrt {1-x^{2}}}\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6123652c97881e3910234616ea5ee831deae4a73)
![{\displaystyle \sin[\arctan(x)]={\frac {x}{\sqrt {1+x^{2}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7d3ed8cedb4c73b01b65314a15152647c9000086)
![{\displaystyle \cos[\arctan(x)]={\frac {1}{\sqrt {1+x^{2}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d2e67d269ac585855446bba315fc260582bc38be)
![{\displaystyle \tan[\arcsin(x)]={\frac {x}{\sqrt {1-x^{2}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/35f9ce679263ce5306237e3342d22abeb3d9d346)
![{\displaystyle \tan[\arccos(x)]={\frac {\sqrt {1-x^{2}}}{x}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/22e131c5707de0e5ab51c45a4c518f8b137ffb99)
