من ويكيبيديا، الموسوعة الحرة
العلاقات بين الدوال المثلثية والدوال المثلثية العكسية[ عدل | عدل المصدر ]
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}}}
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}}}
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}}}
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\!}
، فإنه يُحصل على ما يلي:
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}}}}}
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}}}
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!)! ).
فيما يلي، كسران مستمران معممان يمثلان دالة الظل العكسية. قد يستعملان تعويضا لمتسلسلة القوى للتعبير عن دالة الظل العكسية.
arctan
z
=
z
1
+
(
1
z
)
2
3
−
1
z
2
+
(
3
z
)
2
5
−
3
z
2
+
(
5
z
)
2
7
−
5
z
2
+
(
7
z
)
2
9
−
7
z
2
+
⋱
=
z
1
+
(
1
z
)
2
3
+
(
2
z
)
2
5
+
(
3
z
)
2
7
+
(
4
z
)
2
9
+
⋱
{\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 \,}}}}}}}}}}\,}
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
(
z
)
=
−
i
ln
(
i
z
+
1
−
z
2
)
=
arccsc
(
1
z
)
arccos
(
z
)
=
−
i
ln
(
z
+
z
2
−
1
)
=
π
2
+
i
ln
(
i
z
+
1
−
z
2
)
=
π
2
−
arcsin
(
z
)
=
arcsec
(
1
z
)
arctan
(
z
)
=
i
2
ln
(
i
+
z
i
−
z
)
=
i
2
[
ln
(
1
−
i
z
)
−
ln
(
1
+
i
z
)
]
=
arccot
(
1
z
)
arccot
(
z
)
=
i
2
ln
(
z
−
i
z
+
i
)
=
i
2
[
ln
(
1
−
i
z
)
−
ln
(
1
+
i
z
)
]
=
arctan
(
1
z
)
arcsec
(
z
)
=
−
i
ln
(
1
z
2
−
1
+
1
z
)
=
i
ln
(
1
−
1
z
2
+
i
z
)
+
π
2
=
π
2
−
arccsc
(
z
)
=
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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}}}
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 . Dar Al Kotob Al Ilmiyah دار الكتب العلمية. ISBN :978-2-7451-5445-3 . مؤرشف من 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