بما أن الدوال المثلثية الست غير متباينة ، تم اقتصارها حتى تكون لها دوال عكسية. لذلك، تكون مديات الدوال العكسية مجموعات فرعية لمديات الدوال الأصلية.
فمثلا، على سبيل المثال، باستخدام الدالة بمعنى الدوال متعددة القيم ، تمامًا كما يمكن تعريف دالة الجذر التربيعي y = √x y 2 = x y = arcsin(x )y ) = x .
إسم
ترميز
تعريف
مجال الدالة
مدى الدالة راديان )
مدى الدالةدرجات )
قوس جيب الزاوية y = arcsin(x ) x = sin (y )
−
1
≤
x
≤
1
{\displaystyle -1\leq x\leq 1}
−
π
2
≤
y
≤
π
2
{\displaystyle -{\frac {\pi }{2}}\leq y\leq {\frac {\pi }{2}}}
−
90
∘
≤
y
≤
90
∘
{\displaystyle -90^{\circ }\leq y\leq 90^{\circ }}
قوس جيب تمام الزاوية y = arccos(x ) x = cos (y )
−
1
≤
x
≤
1
{\displaystyle -1\leq x\leq 1}
0
≤
y
≤
π
{\displaystyle 0\leq y\leq \pi }
0
≤
y
≤
180
∘
{\displaystyle 0\leq y\leq 180^{\circ }}
قوس ظل الزاوية y = arctan(x ) x = tan (y )كل الأعداد الحقيقية (
R
{\displaystyle \mathbb {R} }
−
π
2
<
y
<
π
2
{\displaystyle -{\frac {\pi }{2}}<y<{\frac {\pi }{2}}}
−
90
∘
<
y
<
90
∘
{\displaystyle -90^{\circ }<y<90^{\circ }}
قوس ظل تمام الزاوية y = arccot(x ) x = cot (y )كل الأعداد الحقيقية (
R
{\displaystyle \mathbb {R} }
0
<
y
<
π
{\displaystyle 0<y<\pi }
0
<
y
<
180
∘
{\displaystyle 0<y<180^{\circ }}
قوس قاطع الزاوية y = arcsec(x ) x = sec (y )
x
≤
−
1
{\displaystyle x\leq -1}
x
≥
1
{\displaystyle x\geq 1}
0
≤
y
<
π
2
{\displaystyle 0\leq y<{\frac {\pi }{2}}}
π
2
<
y
≤
π
{\displaystyle {\frac {\pi }{2}}<y\leq \pi }
0
≤
y
<
90
∘
{\displaystyle 0\leq y<90^{\circ }}
90
∘
<
y
≤
180
∘
{\displaystyle 90^{\circ }<y\leq 180^{\circ }}
قوس قاطع تمام الزاوية y = arccsc(x ) x = csc (y )
x
≤
−
1
{\displaystyle x\leq -1}
x
≥
1
{\displaystyle x\geq 1}
−
π
2
≤
y
<
0
{\displaystyle -{\frac {\pi }{2}}\leq y<0}
0
<
y
≤
π
2
{\displaystyle 0<y\leq {\frac {\pi }{2}}}
−
90
∘
≤
y
<
0
{\displaystyle -90^{\circ }\leq y<0}
0
<
y
≤
90
∘
{\displaystyle 0<y\leq 90^{\circ }}
عدل
θ
{\displaystyle \theta }
sin
(
θ
)
{\displaystyle \sin(\theta )}
cos
(
θ
)
{\displaystyle \cos(\theta )}
tan
(
θ
)
{\displaystyle \tan(\theta )}
رسم توضيحي
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}}}}
tan
(
arcsin
(
x
)
)
=
x
1
−
x
2
{\displaystyle \tan(\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}
tan
(
arccos
(
x
)
)
=
1
−
x
2
x
{\displaystyle \tan(\arccos(x))={\frac {\sqrt {1-x^{2}}}{x}}}
arctan
(
x
)
{\displaystyle \arctan(x)}
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
(
arctan
(
x
)
)
=
x
{\displaystyle \tan(\arctan(x))=x}
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}}}
tan
(
arccsc
(
x
)
)
=
1
x
2
−
1
{\displaystyle \tan(\operatorname {arccsc}(x))={\frac {1}{\sqrt {x^{2}-1}}}}
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}}}
tan
(
arcsec
(
x
)
)
=
x
2
−
1
{\displaystyle \tan(\operatorname {arcsec}(x))={\sqrt {x^{2}-1}}}
arccot
(
x
)
{\displaystyle \operatorname {arccot}(x)}
sin
(
arccot
(
x
)
)
=
1
1
+
x
2
{\displaystyle \sin(\operatorname {arccot}(x))={\frac {1}{\sqrt {1+x^{2}}}}}
cos
(
arccot
(
x
)
)
=
x
1
+
x
2
{\displaystyle \cos(\operatorname {arccot}(x))={\frac {x}{\sqrt {1+x^{2}}}}}
tan
(
arccot
(
x
)
)
=
1
x
{\displaystyle \tan(\operatorname {arccot}(x))={\frac {1}{x}}}
عدل
زوايا متتامة:
arccos
(
x
)
=
π
2
−
arcsin
(
x
)
arccot
(
x
)
=
π
2
−
arctan
(
x
)
arccsc
(
x
)
=
π
2
−
arcsec
(
x
)
{\displaystyle {\begin{aligned}\arccos(x)&={\frac {\pi }{2}}-\arcsin(x)\\[0.5em]\operatorname {arccot}(x)&={\frac {\pi }{2}}-\arctan(x)\\[0.5em]\operatorname {arccsc}(x)&={\frac {\pi }{2}}-\operatorname {arcsec}(x)\end{aligned}}}
مداخلها عبارة عن مقابل متغيرها:
arcsin
(
−
x
)
=
−
arcsin
(
x
)
arccos
(
−
x
)
=
π
−
arccos
(
x
)
arctan
(
−
x
)
=
−
arctan
(
x
)
arccot
(
−
x
)
=
π
−
arccot
(
x
)
arcsec
(
−
x
)
=
π
−
arcsec
(
x
)
arccsc
(
−
x
)
=
−
arccsc
(
x
)
{\displaystyle {\begin{aligned}\arcsin(-x)&=-\arcsin(x)\\\arccos(-x)&=\pi -\arccos(x)\\\arctan(-x)&=-\arctan(x)\\\operatorname {arccot}(-x)&=\pi -\operatorname {arccot}(x)\\\operatorname {arcsec}(-x)&=\pi -\operatorname {arcsec}(x)\\\operatorname {arccsc}(-x)&=-\operatorname {arccsc}(x)\end{aligned}}}
مداخلها عبارة عن مقلوب متغيرها:
arccos
(
1
x
)
=
arcsec
(
x
)
arcsin
(
1
x
)
=
arccsc
(
x
)
arctan
(
1
x
)
=
π
2
−
arctan
(
x
)
=
arccot
(
x
)
,
if
x
>
0
arctan
(
1
x
)
=
−
π
2
−
arctan
(
x
)
=
arccot
(
x
)
−
π
,
if
x
<
0
arccot
(
1
x
)
=
π
2
−
arccot
(
x
)
=
arctan
(
x
)
,
if
x
>
0
arccot
(
1
x
)
=
3
π
2
−
arccot
(
x
)
=
π
+
arctan
(
x
)
,
if
x
<
0
arcsec
(
1
x
)
=
arccos
(
x
)
arccsc
(
1
x
)
=
arcsin
(
x
)
{\displaystyle {\begin{aligned}\arccos \left({\frac {1}{x}}\right)&=\operatorname {arcsec}(x)\\[0.3em]\arcsin \left({\frac {1}{x}}\right)&=\operatorname {arccsc}(x)\\[0.3em]\arctan \left({\frac {1}{x}}\right)&={\frac {\pi }{2}}-\arctan(x)=\operatorname {arccot}(x)\,,{\text{ if }}x>0\\[0.3em]\arctan \left({\frac {1}{x}}\right)&=-{\frac {\pi }{2}}-\arctan(x)=\operatorname {arccot}(x)-\pi \,,{\text{ if }}x<0\\[0.3em]\operatorname {arccot} \left({\frac {1}{x}}\right)&={\frac {\pi }{2}}-\operatorname {arccot}(x)=\arctan(x)\,,{\text{ if }}x>0\\[0.3em]\operatorname {arccot} \left({\frac {1}{x}}\right)&={\frac {3\pi }{2}}-\operatorname {arccot}(x)=\pi +\arctan(x)\,,{\text{ if }}x<0\\[0.3em]\operatorname {arcsec} \left({\frac {1}{x}}\right)&=\arccos(x)\\[0.3em]\operatorname {arccsc} \left({\frac {1}{x}}\right)&=\arcsin(x)\end{aligned}}}
عدل
arcsin
α
±
arcsin
β
=
arcsin
(
α
1
−
β
2
±
β
1
−
α
2
)
{\displaystyle \arcsin \alpha \pm \arcsin \beta =\arcsin(\alpha {\sqrt {1-\beta ^{2}}}\pm \beta {\sqrt {1-\alpha ^{2}}})}
arccos
α
±
arccos
β
=
arccos
(
α
β
∓
(
1
−
α
2
)
(
1
−
β
2
)
)
{\displaystyle \arccos \alpha \pm \arccos \beta =\arccos(\alpha \beta \mp {\sqrt {(1-\alpha ^{2})(1-\beta ^{2})}})}
arctan
α
±
arctan
β
=
arctan
(
α
±
β
1
∓
α
β
)
{\displaystyle \arctan \alpha \pm \arctan \beta =\arctan \left({\frac {\alpha \pm \beta }{1\mp \alpha \beta }}\right)}
arcsin
(
x
)
+
arccos
(
x
)
=
π
2
{\displaystyle \arcsin(x)+\arccos(x)={\pi \over 2}\;}
arctan
(
x
)
+
arccot
(
x
)
=
π
2
.
{\displaystyle \arctan(x)+\operatorname {arccot}(x)={\pi \over 2}.\;}
arctan
(
x
)
+
arctan
(
1
x
)
=
{
π
2
,
if
x
>
0
−
π
2
,
if
x
<
0
{\displaystyle \arctan(x)+\arctan \left({1 \over x}\right)=\left\{{\begin{matrix}{\pi \over 2},&{\mbox{if }}x>0\\-{\pi \over 2},&{\mbox{if }}x<0\end{matrix}}\right.}
arccos
(
x
)
+
arccos
(
−
x
)
=
π
.
{\displaystyle \arccos(x)+\arccos(-x)=\pi .\;}
arccos
(
x
)
=
arcsin
(
1
−
x
2
)
,
if
0
≤
x
≤
1
arccos
(
x
)
=
1
2
arccos
(
2
x
2
−
1
)
,
if
0
≤
x
≤
1
arcsin
(
x
)
=
1
2
arccos
(
1
−
2
x
2
)
,
if
0
≤
x
≤
1
arcsin
(
x
)
=
arctan
(
x
1
−
x
2
)
arctan
(
x
)
=
arcsin
(
x
1
+
x
2
)
{\displaystyle {\begin{aligned}\arccos(x)&=\arcsin \left({\sqrt {1-x^{2}}}\right)\,,{\text{ if }}0\leq x\leq 1\\\arccos(x)&={\frac {1}{2}}\arccos \left(2x^{2}-1\right)\,,{\text{ if }}0\leq x\leq 1\\\arcsin(x)&={\frac {1}{2}}\arccos \left(1-2x^{2}\right)\,,{\text{ if }}0\leq x\leq 1\\\arcsin(x)&=\arctan \left({\frac {x}{\sqrt {1-x^{2}}}}\right)\\\arctan(x)&=\arcsin \left({\frac {x}{\sqrt {1+x^{2}}}}\right)\end{aligned}}}
عدل
عدل
تُبين فيما يلي، اشتقاقات الدوال المثلثية العكسية بالنسبة لقيم عقدية أو حقيقية للمتغير x:
d
d
x
arcsin
x
=
1
1
−
x
2
d
d
x
arccos
x
=
−
1
1
−
x
2
d
d
x
arctan
x
=
1
1
+
x
2
d
d
x
arccot
x
=
−
1
1
+
x
2
d
d
x
arcsec
x
=
1
x
x
2
−
1
d
d
x
arccsc
x
=
−
1
x
x
2
−
1
{\displaystyle {\begin{aligned}{\frac {d}{dx}}\arcsin x&{}={\frac {1}{\sqrt {1-x^{2}}}}\\{\frac {d}{dx}}\arccos x&{}={\frac {-1}{\sqrt {1-x^{2}}}}\\{\frac {d}{dx}}\arctan x&{}={\frac {1}{1+x^{2}}}\\{\frac {d}{dx}}\operatorname {arccot} x&{}={\frac {-1}{1+x^{2}}}\\{\frac {d}{dx}}\operatorname {arcsec} x&{}={\frac {1}{x\,{\sqrt {x^{2}-1}}}}\\{\frac {d}{dx}}\operatorname {arccsc} x&{}={\frac {-1}{x\,{\sqrt {x^{2}-1}}}}\end{aligned}}}
=
المتساويتان التاليتان صالحتان فقط عندما يكون العدد x حقيقيا:
d
d
x
arcsec
x
=
1
|
x
|
x
2
−
1
;
|
x
|
>
1
d
d
x
arccsc
x
=
−
1
|
x
|
x
2
−
1
;
|
x
|
>
1
{\displaystyle {\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}}}
على سبيل المثال، إذا توفر
θ
=
arcsin
x
{\displaystyle \theta =\arcsin x\!}
d
arcsin
x
d
x
=
d
θ
d
sin
θ
=
d
θ
cos
θ
d
θ
=
1
cos
θ
=
1
1
−
sin
2
θ
=
1
1
−
x
2
{\displaystyle {\frac {d\arcsin x}{dx}}={\frac {d\theta }{d\sin \theta }}={\frac {d\theta }{\cos \theta d\theta }}={\frac {1}{\cos \theta }}={\frac {1}{\sqrt {1-\sin ^{2}\theta }}}={\frac {1}{\sqrt {1-x^{2}}}}}
عدل
باستخدام التكامل بالتجزئة ، نجد أن:
∫
arcsin
(
x
)
d
x
=
x
arcsin
(
x
)
+
1
−
x
2
+
C
∫
arccos
(
x
)
d
x
=
x
arccos
(
x
)
−
1
−
x
2
+
C
∫
arctan
(
x
)
d
x
=
x
arctan
(
x
)
−
1
2
ln
(
1
+
x
2
)
+
C
∫
arccot
(
x
)
d
x
=
x
arccot
(
x
)
+
1
2
ln
(
1
+
x
2
)
+
C
∫
arcsec
(
x
)
d
x
=
x
arcsec
(
x
)
−
ln
(
x
+
x
2
−
1
)
+
C
∫
arccsc
(
x
)
d
x
=
x
arccsc
(
x
)
+
ln
(
x
+
x
2
−
1
)
+
C
{\displaystyle {\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 \arctan(x)\,dx&{}=x\,\arctan(x)-{\frac {1}{2}}\ln \left(1+x^{2}\right)+C\\\int \operatorname {arccot}(x)\,dx&{}=x\,\operatorname {arccot}(x)+{\frac {1}{2}}\ln \left(1+x^{2}\right)+C\\\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}}}
عدل
يمكننا تعبير عن بعض د.م.ع. بواسطة متسلسلة ماكلورين :
arcsin
(
x
)
=
x
+
(
1
2
)
x
3
3
+
(
1
⋅
3
2
⋅
4
)
x
5
5
+
(
1
⋅
3
⋅
5
2
⋅
4
⋅
6
)
x
7
7
+
⋯
=
∑
n
=
0
∞
(
2
n
−
1
)
!
!
(
2
n
)
!
!
x
2
n
+
1
2
n
+
1
=
∑
n
=
0
∞
(
2
n
)
!
(
2
n
n
!
)
2
x
2
n
+
1
2
n
+
1
{\displaystyle {\begin{aligned}\arcsin(x)&=x+\left({\frac {1}{2}}\right){\frac {x^{3}}{3}}+\left({\frac {1\cdot 3}{2\cdot 4}}\right){\frac {x^{5}}{5}}+\left({\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6}}\right){\frac {x^{7}}{7}}+\cdots \\[5pt]&=\sum _{n=0}^{\infty }{\frac {(2n-1)!!}{(2n)!!}}{\frac {x^{2n+1}}{2n+1}}\\[5pt]&=\sum _{n=0}^{\infty }{\frac {(2n)!}{(2^{n}n!)^{2}}}{\frac {x^{2n+1}}{2n+1}}\,\end{aligned}}}
arctan
(
x
)
=
x
−
x
3
3
+
x
5
5
−
x
7
7
+
⋯
=
∑
n
=
0
∞
(
−
1
)
n
x
2
n
+
1
2
n
+
1
{\displaystyle \arctan(x)=x-{\frac {x^{3}}{3}}+{\frac {x^{5}}{5}}-{\frac {x^{7}}{7}}+\cdots =\sum _{n=0}^{\infty }{\frac {(-1)^{n}x^{2n+1}}{2n+1}}\,}
arccos
(
x
)
=
π
2
−
arcsin
(
x
)
=
π
2
−
x
−
(
1
2
)
x
3
3
−
(
1
⋅
3
2
⋅
4
)
x
5
5
−
(
1
⋅
3
⋅
5
2
⋅
4
⋅
6
)
x
7
7
−
⋯
=
π
2
−
∑
n
=
0
∞
(
2
n
−
1
)
!
!
(
2
n
)
!
!
x
2
n
+
1
2
n
+
1
=
π
2
−
∑
n
=
0
∞
(
2
n
)
!
(
2
n
n
!
)
2
x
2
n
+
1
2
n
+
1
{\displaystyle {\begin{aligned}\arccos(x)&={\frac {\pi }{2}}-\arcsin(x)={\frac {\pi }{2}}-x-\left({\frac {1}{2}}\right){\frac {x^{3}}{3}}-\left({\frac {1\cdot 3}{2\cdot 4}}\right){\frac {x^{5}}{5}}-\left({\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6}}\right){\frac {x^{7}}{7}}-\cdots \\[5pt]&={\frac {\pi }{2}}-\sum _{n=0}^{\infty }{\frac {(2n-1)!!}{(2n)!!}}{\frac {x^{2n+1}}{2n+1}}\\[5pt]&={\frac {\pi }{2}}-\sum _{n=0}^{\infty }{\frac {(2n)!}{(2^{n}n!)^{2}}}{\frac {x^{2n+1}}{2n+1}}\,\end{aligned}}}
حيث تشير n!! إلى عاملي ثنائي (ميز عن «عاملي مرتين» (n!)! ).
عدل
فيما يلي، كسران مستمران معممان يمثلان دالة الظل العكسية. قد يستعملان تعويضا لمتسلسلة القوى للتعبير عن دالة الظل العكسية.
arctan
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⋱
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{\displaystyle \arctan z={\cfrac {z}{1+{\cfrac {(1z)^{2}}{3-1z^{2}+{\cfrac {(3z)^{2}}{5-3z^{2}+{\cfrac {(5z)^{2}}{7-5z^{2}+{\cfrac {(7z)^{2}}{9-7z^{2}+\ddots }}}}}}}}}}={\cfrac {z}{1+{\cfrac {(1z)^{2}}{3+{\cfrac {(2z)^{2}}{5+{\cfrac {(3z)^{2}}{7+{\cfrac {(4z)^{2}}{9+\ddots \,}}}}}}}}}}\,}
عدل
قد يتم التعبير عن هذه الدوال أيضًا باستخدام اللوغاريتمات العقدية . هذا يمَدِّد مجالاتهما إلى المستوي العقدي (المركّب) بطريقة طبيعية. تشبه هذه التعبيرات العبارات اللوغاريتمية للدوال الزائدية العكسية .
arcsin
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ln
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arccsc
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arccos
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arcsin
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arcsec
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arctan
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ln
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ln
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arccot
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arccot
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ln
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arctan
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arcsec
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arccsc
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arccos
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arccsc
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arcsin
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{\displaystyle {\begin{aligned}\arcsin(z)&{}=-i\ln \left(iz+{\sqrt {1-z^{2}}}\right)&{}=\operatorname {arccsc} \left({\frac {1}{z}}\right)\\[10pt]\arccos(z)&{}=-i\ln \left(z+{\sqrt {z^{2}-1}}\right)={\frac {\pi }{2}}\,+i\ln \left(iz+{\sqrt {1-z^{2}}}\right)={\frac {\pi }{2}}-\arcsin(z)&{}=\operatorname {arcsec} \left({\frac {1}{z}}\right)\\[10pt]\arctan(z)&{}={\frac {i}{2}}\ln \left({\frac {i+z}{i-z}}\right)={\frac {i}{2}}\left[\ln(1-iz)-\ln(1+iz)\right]&{}=\operatorname {arccot} \left({\frac {1}{z}}\right)\\[10pt]\operatorname {arccot}(z)&{}={\frac {i}{2}}\ln \left({\frac {z-i}{z+i}}\right)={\frac {i}{2}}\left[\ln \left(1-{\frac {i}{z}}\right)-\ln \left(1+{\frac {i}{z}}\right)\right]&{}=\arctan \left({\frac {1}{z}}\right)\\[10pt]\operatorname {arcsec}(z)&{}=-i\ln \left({\sqrt {{\frac {1}{z^{2}}}-1}}+{\frac {1}{z}}\right)=i\,\ln \left({\sqrt {1-{\frac {1}{z^{2}}}}}+{\frac {i}{z}}\right)+{\frac {\pi }{2}}={\frac {\pi }{2}}-\operatorname {arccsc}(z)&{}=\arccos \left({\frac {1}{z}}\right)\\[10pt]\operatorname {arccsc}(z)&{}=-i\ln \left({\sqrt {1-{\frac {1}{z^{2}}}}}+{\frac {i}{z}}\right)&{}=\arcsin \left({\frac {1}{z}}\right)\end{aligned}}}
عدل
التمثيلات البيانية للدوال في المَعْلَم الديكارتي.
ت.ب لدالتي قوس الجيب (بالأحمر) وقوس جيب التمام (بالأزرق)
ت.ب لدالتي قوس الظل (بالأحمر) وقوس ظل التمام (بالأزرق)
ت.ب لدالتي قوس القاطع (بالأحمر) وقوس قاطع التمام (بالأزرق)
^ "معلومات عن الدوال المثلثية العكسية على موقع mathworld.wolfram.com" . mathworld.wolfram.com. مؤرشف من الأصل في 2019-03-25.^ ميشال إبراهيم ورامي أبو سليمان وفادي (1 يناير 2007). Dictionaire des termes scientifiques (Anglais/Français/Arabe): قاموس المصطلحات العلمية - انكليزي/فرنسي/عربي ISBN :978-2-7451-5445-3 الأصل في 2020-02-20. ^ Graham Hall et Fred Goodrich Frink, chap. II « The Acute Angle (14) Inverse trigonometric functions », dans Trigonometry, Ann Arbor, Michigan, USA, Henry Holt and Company / Norwood Press / J. S. Cushing Co. - Berwick & Smith Co., Norwood, Massachusetts, USA, janvier 1909 ,I: Plane Trigonometry, p. 15.. نسخة محفوظة 5 يوليو 2019 على موقع واي باك مشين .
![{\displaystyle {\begin{aligned}\arccos(x)&={\frac {\pi }{2}}-\arcsin(x)\\[0.5em]\operatorname {arccot}(x)&={\frac {\pi }{2}}-\arctan(x)\\[0.5em]\operatorname {arccsc}(x)&={\frac {\pi }{2}}-\operatorname {arcsec}(x)\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ec43798232f580abb074cf15f3d77692edd36af0)
![{\displaystyle {\begin{aligned}\arccos \left({\frac {1}{x}}\right)&=\operatorname {arcsec}(x)\\[0.3em]\arcsin \left({\frac {1}{x}}\right)&=\operatorname {arccsc}(x)\\[0.3em]\arctan \left({\frac {1}{x}}\right)&={\frac {\pi }{2}}-\arctan(x)=\operatorname {arccot}(x)\,,{\text{ if }}x>0\\[0.3em]\arctan \left({\frac {1}{x}}\right)&=-{\frac {\pi }{2}}-\arctan(x)=\operatorname {arccot}(x)-\pi \,,{\text{ if }}x<0\\[0.3em]\operatorname {arccot} \left({\frac {1}{x}}\right)&={\frac {\pi }{2}}-\operatorname {arccot}(x)=\arctan(x)\,,{\text{ if }}x>0\\[0.3em]\operatorname {arccot} \left({\frac {1}{x}}\right)&={\frac {3\pi }{2}}-\operatorname {arccot}(x)=\pi +\arctan(x)\,,{\text{ if }}x<0\\[0.3em]\operatorname {arcsec} \left({\frac {1}{x}}\right)&=\arccos(x)\\[0.3em]\operatorname {arccsc} \left({\frac {1}{x}}\right)&=\arcsin(x)\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ecb0a376d9148e3cf65ea6e6ed5fb37752a581c7)
![{\displaystyle {\begin{aligned}\arcsin(x)&=x+\left({\frac {1}{2}}\right){\frac {x^{3}}{3}}+\left({\frac {1\cdot 3}{2\cdot 4}}\right){\frac {x^{5}}{5}}+\left({\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6}}\right){\frac {x^{7}}{7}}+\cdots \\[5pt]&=\sum _{n=0}^{\infty }{\frac {(2n-1)!!}{(2n)!!}}{\frac {x^{2n+1}}{2n+1}}\\[5pt]&=\sum _{n=0}^{\infty }{\frac {(2n)!}{(2^{n}n!)^{2}}}{\frac {x^{2n+1}}{2n+1}}\,\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/92b62bf5b37765271c47e9788914ff0efdc965a2)
![{\displaystyle {\begin{aligned}\arccos(x)&={\frac {\pi }{2}}-\arcsin(x)={\frac {\pi }{2}}-x-\left({\frac {1}{2}}\right){\frac {x^{3}}{3}}-\left({\frac {1\cdot 3}{2\cdot 4}}\right){\frac {x^{5}}{5}}-\left({\frac {1\cdot 3\cdot 5}{2\cdot 4\cdot 6}}\right){\frac {x^{7}}{7}}-\cdots \\[5pt]&={\frac {\pi }{2}}-\sum _{n=0}^{\infty }{\frac {(2n-1)!!}{(2n)!!}}{\frac {x^{2n+1}}{2n+1}}\\[5pt]&={\frac {\pi }{2}}-\sum _{n=0}^{\infty }{\frac {(2n)!}{(2^{n}n!)^{2}}}{\frac {x^{2n+1}}{2n+1}}\,\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8ffbad34ed3a8e64858c108533407e7a0a0551c6)
![{\displaystyle {\begin{aligned}\arcsin(z)&{}=-i\ln \left(iz+{\sqrt {1-z^{2}}}\right)&{}=\operatorname {arccsc} \left({\frac {1}{z}}\right)\\[10pt]\arccos(z)&{}=-i\ln \left(z+{\sqrt {z^{2}-1}}\right)={\frac {\pi }{2}}\,+i\ln \left(iz+{\sqrt {1-z^{2}}}\right)={\frac {\pi }{2}}-\arcsin(z)&{}=\operatorname {arcsec} \left({\frac {1}{z}}\right)\\[10pt]\arctan(z)&{}={\frac {i}{2}}\ln \left({\frac {i+z}{i-z}}\right)={\frac {i}{2}}\left[\ln(1-iz)-\ln(1+iz)\right]&{}=\operatorname {arccot} \left({\frac {1}{z}}\right)\\[10pt]\operatorname {arccot}(z)&{}={\frac {i}{2}}\ln \left({\frac {z-i}{z+i}}\right)={\frac {i}{2}}\left[\ln \left(1-{\frac {i}{z}}\right)-\ln \left(1+{\frac {i}{z}}\right)\right]&{}=\arctan \left({\frac {1}{z}}\right)\\[10pt]\operatorname {arcsec}(z)&{}=-i\ln \left({\sqrt {{\frac {1}{z^{2}}}-1}}+{\frac {1}{z}}\right)=i\,\ln \left({\sqrt {1-{\frac {1}{z^{2}}}}}+{\frac {i}{z}}\right)+{\frac {\pi }{2}}={\frac {\pi }{2}}-\operatorname {arccsc}(z)&{}=\arccos \left({\frac {1}{z}}\right)\\[10pt]\operatorname {arccsc}(z)&{}=-i\ln \left({\sqrt {1-{\frac {1}{z^{2}}}}}+{\frac {i}{z}}\right)&{}=\arcsin \left({\frac {1}{z}}\right)\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7621fcfe397c23cf51412c21812c2be167565eb0)
ت.ب لدالتي قوس الجيب (بالأحمر) وقوس جيب التمام (بالأزرق)
ت.ب لدالتي قوس الظل (بالأحمر) وقوس ظل التمام (بالأزرق)
ت.ب لدالتي قوس القاطع (بالأحمر) وقوس قاطع التمام (بالأزرق)
