الجيوب وجيوب التمام حول دائرة الوحدة في الرياضيات ، المتطابقات المثلثية أو المطابقات المثلثية أو المعادلات المثلثية هي مجموعة من المساواة تتألف من دوال مثلثية . وتعتبر المتطابقات مفيدة جدًا في تبسيط أو التحويل بين الدوال الرياضية. كما أن لها دورا كبيرا في حل المعادلات الرياضية خاصة في معكوس الدالة (كصيغة كاردان ) والتكامل (كتكامل مربع جيب تمام الزاوية ).
هي نوع من المعادلات التي تحتوي على قيم الدوال المثلثية (جا ، جتا ، ظا ) أو مقلوباتها بحيث تكون إحدى زوايا المعادلة مجهولة ويحل هذا النوع من المعادلات كباقي المعادلات الجبرية العادية وبطرق التحليل المعروفة.[ 1]
لتجنب الالتباس حول (sin−1 ( x ومثيلاتها هل هي مقاليب أم معاكيس ، سيتم استخدام (csc( x ومثيلاتها للمقاليب و(arcsin( x ومثيلاتها للمعكوسات وهكذا. الجدول التالي يبين بعض وحدات الزوايا والتحويل بينها
الدرجات
30
45
60
90
120
180
270
360
الراديان
π
/
6
{\displaystyle \pi /6}
π
/
4
{\displaystyle \pi /4}
π
/
3
{\displaystyle \pi /3}
π
/
2
{\displaystyle \pi /2}
2
π
/
3
{\displaystyle 2\pi /3}
π
{\displaystyle \pi }
3
π
/
2
{\displaystyle 3\pi /2}
2
π
{\displaystyle 2\pi }
غراد
33 ⅓
50
66 ⅔
100
133 ⅓
200
300
400
متطابقة فيثاغورس المثلثية
sin
2
θ
+
cos
2
θ
=
1
{\displaystyle \sin ^{2}\theta +\cos ^{2}\theta =1\,}
متطابقة النسبة
tan
θ
=
sin
θ
cos
θ
{\displaystyle \tan \theta ={\frac {\sin \theta }{\cos \theta }}}
[ عدل ] من دائرة الوحدة يمكن الحصول على المتطابقات التالية..
تنجم عن عملية عكس الزوايا انعكاسات في المتطابقات المثلثية كما في الجدول التالي.
انعكاس في
θ
=
0
{\displaystyle \theta =0}
انعكاس في
θ
=
π
/
2
{\displaystyle \theta =\pi /2}
انعكاس في
θ
=
π
{\displaystyle \theta =\pi }
sin
(
−
θ
)
=
−
sin
θ
cos
(
−
θ
)
=
+
cos
θ
tan
(
−
θ
)
=
−
tan
θ
csc
(
−
θ
)
=
−
csc
θ
sec
(
−
θ
)
=
+
sec
θ
cot
(
−
θ
)
=
−
cot
θ
{\displaystyle {\begin{aligned}\sin(-\theta )&=-\sin \theta \\\cos(-\theta )&=+\cos \theta \\\tan(-\theta )&=-\tan \theta \\\csc(-\theta )&=-\csc \theta \\\sec(-\theta )&=+\sec \theta \\\cot(-\theta )&=-\cot \theta \end{aligned}}}
sin
(
π
2
−
θ
)
=
+
cos
θ
cos
(
π
2
−
θ
)
=
+
sin
θ
tan
(
π
2
−
θ
)
=
+
cot
θ
csc
(
π
2
−
θ
)
=
+
sec
θ
sec
(
π
2
−
θ
)
=
+
csc
θ
cot
(
π
2
−
θ
)
=
+
tan
θ
{\displaystyle {\begin{aligned}\sin({\tfrac {\pi }{2}}-\theta )&=+\cos \theta \\\cos({\tfrac {\pi }{2}}-\theta )&=+\sin \theta \\\tan({\tfrac {\pi }{2}}-\theta )&=+\cot \theta \\\csc({\tfrac {\pi }{2}}-\theta )&=+\sec \theta \\\sec({\tfrac {\pi }{2}}-\theta )&=+\csc \theta \\\cot({\tfrac {\pi }{2}}-\theta )&=+\tan \theta \end{aligned}}}
sin
(
π
−
θ
)
=
+
sin
θ
cos
(
π
−
θ
)
=
−
cos
θ
tan
(
π
−
θ
)
=
−
tan
θ
csc
(
π
−
θ
)
=
+
csc
θ
sec
(
π
−
θ
)
=
−
sec
θ
cot
(
π
−
θ
)
=
−
cot
θ
{\displaystyle {\begin{aligned}\sin(\pi -\theta )&=+\sin \theta \\\cos(\pi -\theta )&=-\cos \theta \\\tan(\pi -\theta )&=-\tan \theta \\\csc(\pi -\theta )&=+\csc \theta \\\sec(\pi -\theta )&=-\sec \theta \\\cot(\pi -\theta )&=-\cot \theta \\\end{aligned}}}
θ θ θ
sin
(
θ
+
π
2
)
=
+
cos
θ
cos
(
θ
+
π
2
)
=
−
sin
θ
tan
(
θ
+
π
2
)
=
−
cot
θ
csc
(
θ
+
π
2
)
=
+
sec
θ
sec
(
θ
+
π
2
)
=
−
csc
θ
cot
(
θ
+
π
2
)
=
−
tan
θ
{\displaystyle {\begin{aligned}\sin(\theta +{\tfrac {\pi }{2}})&=+\cos \theta \\\cos(\theta +{\tfrac {\pi }{2}})&=-\sin \theta \\\tan(\theta +{\tfrac {\pi }{2}})&=-\cot \theta \\\csc(\theta +{\tfrac {\pi }{2}})&=+\sec \theta \\\sec(\theta +{\tfrac {\pi }{2}})&=-\csc \theta \\\cot(\theta +{\tfrac {\pi }{2}})&=-\tan \theta \end{aligned}}}
sin
(
θ
+
π
)
=
−
sin
θ
cos
(
θ
+
π
)
=
−
cos
θ
tan
(
θ
+
π
)
=
+
tan
θ
csc
(
θ
+
π
)
=
−
csc
θ
sec
(
θ
+
π
)
=
−
sec
θ
cot
(
θ
+
π
)
=
+
cot
θ
{\displaystyle {\begin{aligned}\sin(\theta +\pi )&=-\sin \theta \\\cos(\theta +\pi )&=-\cos \theta \\\tan(\theta +\pi )&=+\tan \theta \\\csc(\theta +\pi )&=-\csc \theta \\\sec(\theta +\pi )&=-\sec \theta \\\cot(\theta +\pi )&=+\cot \theta \\\end{aligned}}}
sin
(
θ
+
2
π
)
=
+
sin
θ
cos
(
θ
+
2
π
)
=
+
cos
θ
tan
(
θ
+
2
π
)
=
+
tan
θ
csc
(
θ
+
2
π
)
=
+
csc
θ
sec
(
θ
+
2
π
)
=
+
sec
θ
cot
(
θ
+
2
π
)
=
+
cot
θ
{\displaystyle {\begin{aligned}\sin(\theta +2\pi )&=+\sin \theta \\\cos(\theta +2\pi )&=+\cos \theta \\\tan(\theta +2\pi )&=+\tan \theta \\\csc(\theta +2\pi )&=+\csc \theta \\\sec(\theta +2\pi )&=+\sec \theta \\\cot(\theta +2\pi )&=+\cot \theta \end{aligned}}}
[ عدل ]
الجيب
sin
(
α
±
β
)
=
sin
α
cos
β
±
cos
α
sin
β
{\displaystyle \sin(\alpha \pm \beta )=\sin \alpha \cos \beta \pm \cos \alpha \sin \beta \,}
جيب التمام
cos
(
α
±
β
)
=
cos
α
cos
β
∓
sin
α
sin
β
{\displaystyle \cos(\alpha \pm \beta )=\cos \alpha \cos \beta \mp \sin \alpha \sin \beta \,}
الظل
tan
(
α
±
β
)
=
tan
α
±
tan
β
1
∓
tan
α
tan
β
{\displaystyle \tan(\alpha \pm \beta )={\frac {\tan \alpha \pm \tan \beta }{1\mp \tan \alpha \tan \beta }}}
قوس الجيب
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}}})}
قوس جيب التمام
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})}})}
قوس الظل
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)}
[
cos
α
−
sin
α
sin
α
cos
α
]
[
cos
β
−
sin
β
sin
β
cos
β
]
=
[
cos
(
α
+
β
)
−
sin
(
α
+
β
)
sin
(
α
+
β
)
cos
(
α
+
β
)
]
.
{\displaystyle \left[{\begin{matrix}\cos \alpha &-\sin \alpha \\\sin \alpha &\cos \alpha \end{matrix}}\right]\left[{\begin{matrix}\cos \beta &-\sin \beta \\\sin \beta &\cos \beta \end{matrix}}\right]=\left[{\begin{matrix}\cos(\alpha +\beta )&-\sin(\alpha +\beta )\\\sin(\alpha +\beta )&\cos(\alpha +\beta )\end{matrix}}\right].}
[ عدل ]
sin
(
∑
i
=
1
∞
θ
i
)
=
∑
o
d
d
k
≥
1
(
−
1
)
(
k
−
1
)
/
2
∑
A
⊆
{
1
,
2
,
3
,
…
}
|
A
|
=
k
(
∏
i
∈
A
sin
θ
i
∏
i
∉
A
cos
θ
i
)
{\displaystyle \sin \left(\sum _{i=1}^{\infty }\theta _{i}\right)=\sum _{\mathrm {odd} \ k\geq 1}(-1)^{(k-1)/2}\sum _{\begin{smallmatrix}A\subseteq \{\,1,2,3,\dots \,\}\\\left|A\right|=k\end{smallmatrix}}\left(\prod _{i\in A}\sin \theta _{i}\prod _{i\not \in A}\cos \theta _{i}\right)}
cos
(
∑
i
=
1
∞
θ
i
)
=
∑
e
v
e
n
k
≥
0
(
−
1
)
k
/
2
∑
A
⊆
{
1
,
2
,
3
,
…
}
|
A
|
=
k
(
∏
i
∈
A
sin
θ
i
∏
i
∉
A
cos
θ
i
)
{\displaystyle \cos \left(\sum _{i=1}^{\infty }\theta _{i}\right)=\sum _{\mathrm {even} \ k\geq 0}~(-1)^{k/2}~~\sum _{\begin{smallmatrix}A\subseteq \{\,1,2,3,\dots \,\}\\\left|A\right|=k\end{smallmatrix}}\left(\prod _{i\in A}\sin \theta _{i}\prod _{i\not \in A}\cos \theta _{i}\right)}
[ عدل ]
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 }},}
مثال:
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}}}
وهكذا
[ عدل ]
sec
(
θ
1
+
⋯
+
θ
n
)
=
sec
θ
1
⋯
sec
θ
n
e
0
−
e
2
+
e
4
−
⋯
{\displaystyle \sec(\theta _{1}+\cdots +\theta _{n})={\frac {\sec \theta _{1}\cdots \sec \theta _{n}}{e_{0}-e_{2}+e_{4}-\cdots }}}
مثلا,
sec
(
α
+
β
+
γ
)
=
sec
α
sec
β
sec
γ
1
−
tan
α
tan
β
−
tan
α
tan
γ
−
tan
β
tan
γ
.
{\displaystyle \sec(\alpha +\beta +\gamma )={\frac {\sec \alpha \sec \beta \sec \gamma }{1-\tan \alpha \tan \beta -\tan \alpha \tan \gamma -\tan \beta \tan \gamma }}.}
[ عدل ]
Tn هو متعدد الحدود لشيبيشيف من الدرجة n
cos
n
θ
=
T
n
(
cos
θ
)
{\displaystyle \cos n\theta =T_{n}(\cos \theta )\,}
صيغة دي موافر ،
i
{\displaystyle i}
وحدة تخيلية
cos
n
θ
+
i
sin
n
θ
=
(
cos
(
θ
)
+
i
sin
(
θ
)
)
n
{\displaystyle \cos n\theta +i\sin n\theta =(\cos(\theta )+i\sin(\theta ))^{n}\,}
1
+
2
cos
(
x
)
+
2
cos
(
2
x
)
+
2
cos
(
3
x
)
+
⋯
+
2
cos
(
n
x
)
=
sin
(
(
n
+
1
2
)
x
)
sin
(
x
/
2
)
.
{\displaystyle 1+2\cos(x)+2\cos(2x)+2\cos(3x)+\cdots +2\cos(nx)={\frac {\sin \left(\left(n+{\frac {1}{2}}\right)x\right)}{\sin(x/2)}}.}
[ عدل ] أنظر أيضا: صيغة فايرشتراس [الإنجليزية]
sin
(
2
θ
)
=
2
sin
θ
cos
θ
=
2
tan
θ
1
+
tan
2
θ
{\displaystyle \sin(2\theta )=2\sin \theta \cos \theta ={\frac {2\tan \theta }{1+\tan ^{2}\theta }}}
cos
(
2
θ
)
=
cos
2
θ
−
sin
2
θ
=
2
cos
2
θ
−
1
=
1
−
2
sin
2
θ
=
1
−
tan
2
θ
1
+
tan
2
θ
{\displaystyle \cos(2\theta )=\cos ^{2}\theta -\sin ^{2}\theta =2\cos ^{2}\theta -1=1-2\sin ^{2}\theta ={\frac {1-\tan ^{2}\theta }{1+\tan ^{2}\theta }}}
tan
(
2
θ
)
=
2
tan
θ
1
−
tan
2
θ
{\displaystyle \tan(2\theta )={\frac {2\tan \theta }{1-\tan ^{2}\theta }}}
cot
(
2
θ
)
=
cot
2
θ
−
1
2
cot
θ
{\displaystyle \cot(2\theta )={\frac {\cot ^{2}\theta -1}{2\cot \theta }}}
sec
(
2
θ
)
=
sec
2
θ
2
−
sec
2
θ
{\displaystyle \sec(2\theta )={\frac {\sec ^{2}\theta }{2-\sec ^{2}\theta }}}
csc
(
2
θ
)
=
sec
θ
csc
θ
2
{\displaystyle \csc(2\theta )={\frac {\sec \theta \csc \theta }{2}}}
[ عدل ]
sin
(
3
θ
)
=
3
sin
θ
−
4
sin
3
θ
=
4
sin
θ
sin
(
π
3
−
θ
)
sin
(
π
3
+
θ
)
{\displaystyle \sin(3\theta )=3\sin \theta -4\sin ^{3}\theta =4\sin \theta \sin({\frac {\pi }{3}}-\theta )\sin({\frac {\pi }{3}}+\theta )}
cos
(
3
θ
)
=
4
cos
3
θ
−
3
cos
θ
=
4
cos
θ
cos
(
π
3
−
θ
)
cos
(
π
3
+
θ
)
{\displaystyle \cos(3\theta )=4\cos ^{3}\theta -3\cos \theta =4\cos \theta \cos({\frac {\pi }{3}}-\theta )\cos({\frac {\pi }{3}}+\theta )}
tan
(
3
θ
)
=
3
tan
θ
−
tan
3
θ
1
−
3
tan
2
θ
=
tan
θ
tan
(
π
3
−
θ
)
tan
(
π
3
+
θ
)
{\displaystyle \tan(3\theta )={\frac {3\tan \theta -\tan ^{3}\theta }{1-3\tan ^{2}\theta }}=\tan \theta \tan({\frac {\pi }{3}}-\theta )\tan({\frac {\pi }{3}}+\theta )}
cot
(
3
θ
)
=
3
cot
θ
−
cot
3
θ
1
−
3
cot
2
θ
{\displaystyle \cot(3\theta )={\frac {3\cot \theta -\cot ^{3}\theta }{1-3\cot ^{2}\theta }}}
sec
(
3
θ
)
=
sec
3
θ
4
−
3
sec
2
θ
{\displaystyle \sec(3\theta )={\frac {\sec ^{3}\theta }{4-3\sec ^{2}\theta }}}
csc
(
3
θ
)
=
csc
3
θ
3
csc
2
θ
−
4
{\displaystyle \csc(3\theta )={\frac {\csc ^{3}\theta }{3\csc ^{2}\theta -4}}}
sin
θ
2
=
sgn
(
2
π
−
θ
+
4
π
⌊
θ
4
π
⌋
)
1
−
cos
θ
2
{\displaystyle {\begin{aligned}&\sin {\frac {\theta }{2}}=\operatorname {sgn} \left(2\pi -\theta +4\pi \left\lfloor {\frac {\theta }{4\pi }}\right\rfloor \right){\sqrt {\frac {1-\cos \theta }{2}}}\\\end{aligned}}}
حيث
sgn
x
=
±
1
{\displaystyle \operatorname {sgn} x=\pm 1}
⌊
f
(
x
)
⌋
{\displaystyle \lfloor f(x)\rfloor }
دالة الجزء الصحيح .
sin
2
θ
2
=
1
−
cos
θ
2
{\displaystyle \sin ^{2}{\frac {\theta }{2}}={\frac {1-\cos \theta }{2}}}
cos
θ
2
=
sgn
(
π
+
θ
+
4
π
⌊
π
−
θ
4
π
⌋
)
1
+
cos
θ
2
{\displaystyle \cos {\frac {\theta }{2}}=\operatorname {sgn} \left(\pi +\theta +4\pi \left\lfloor {\frac {\pi -\theta }{4\pi }}\right\rfloor \right){\sqrt {\frac {1+\cos \theta }{2}}}}
cos
2
θ
2
=
1
+
cos
θ
2
{\displaystyle \cos ^{2}{\frac {\theta }{2}}={\frac {1+\cos \theta }{2}}}
tan
θ
2
=
csc
θ
−
cot
θ
=
±
1
−
cos
θ
1
+
cos
θ
=
sin
θ
1
+
cos
θ
=
1
−
cos
θ
sin
θ
=
−
1
±
1
+
tan
2
θ
tan
θ
=
tan
θ
1
+
sec
θ
{\displaystyle {\begin{aligned}\tan {\frac {\theta }{2}}&=\csc \theta -\cot \theta =\pm \,{\sqrt {\frac {1-\cos \theta }{1+\cos \theta }}}={\frac {\sin \theta }{1+\cos \theta }}\\&={\frac {1-\cos \theta }{\sin \theta }}={\frac {-1\pm {\sqrt {1+\tan ^{2}\theta }}}{\tan \theta }}={\frac {\tan \theta }{1+\sec {\theta }}}\end{aligned}}}
cot
θ
2
=
csc
θ
+
cot
θ
=
±
1
+
cos
θ
1
−
cos
θ
=
sin
θ
1
−
cos
θ
=
1
+
cos
θ
sin
θ
{\displaystyle \cot {\frac {\theta }{2}}=\csc \theta +\cot \theta =\pm \,{\sqrt {\frac {1+\cos \theta }{1-\cos \theta }}}={\frac {\sin \theta }{1-\cos \theta }}={\frac {1+\cos \theta }{\sin \theta }}}
[ 2] [ 3]
أيضا:
tan
η
+
θ
2
=
sin
η
+
sin
θ
cos
η
+
cos
θ
{\displaystyle \tan {\frac {\eta +\theta }{2}}={\frac {\sin \eta +\sin \theta }{\cos \eta +\cos \theta }}}
tan
(
θ
2
+
π
4
)
=
sec
θ
+
tan
θ
{\displaystyle \tan \left({\frac {\theta }{2}}+{\frac {\pi }{4}}\right)=\sec \theta +\tan \theta }
1
−
sin
θ
1
+
sin
θ
=
|
1
−
tan
θ
2
|
|
1
+
tan
θ
2
|
{\displaystyle {\sqrt {\frac {1-\sin \theta }{1+\sin \theta }}}={\frac {|1-\tan {\frac {\theta }{2}}|}{|1+\tan {\frac {\theta }{2}}|}}}
[ عدل ]
sin
n
θ
=
∑
k
=
0
n
(
n
k
)
cos
k
θ
sin
n
−
k
θ
sin
(
1
2
(
n
−
k
)
π
)
{\displaystyle \sin n\theta =\sum _{k=0}^{n}{\binom {n}{k}}\cos ^{k}\theta \,\sin ^{n-k}\theta \,\sin \left({\frac {1}{2}}(n-k)\pi \right)}
cos
n
θ
=
∑
k
=
0
n
(
n
k
)
cos
k
θ
sin
n
−
k
θ
cos
(
1
2
(
n
−
k
)
π
)
{\displaystyle \cos n\theta =\sum _{k=0}^{n}{\binom {n}{k}}\cos ^{k}\theta \,\sin ^{n-k}\theta \,\cos \left({\frac {1}{2}}(n-k)\pi \right)}
tan
(
n
+
1
)
θ
=
tan
n
θ
+
tan
θ
1
−
tan
n
θ
tan
θ
.
{\displaystyle \tan \,(n{+}1)\theta ={\frac {\tan n\theta +\tan \theta }{1-\tan n\theta \,\tan \theta }}.}
cot
(
n
+
1
)
θ
=
cot
n
θ
cot
θ
−
1
cot
n
θ
+
cot
θ
.
{\displaystyle \cot \,(n{+}1)\theta ={\frac {\cot n\theta \,\cot \theta -1}{\cot n\theta +\cot \theta }}.}
tan
(
α
+
β
2
)
=
sin
α
+
sin
β
cos
α
+
cos
β
=
−
cos
α
−
cos
β
sin
α
−
sin
β
{\displaystyle \tan \left({\frac {\alpha +\beta }{2}}\right)={\frac {\sin \alpha +\sin \beta }{\cos \alpha +\cos \beta }}=-\,{\frac {\cos \alpha -\cos \beta }{\sin \alpha -\sin \beta }}}
[ عدل ]
cos
(
θ
2
)
⋅
cos
(
θ
4
)
⋅
cos
(
θ
8
)
⋯
=
∏
n
=
1
∞
cos
(
θ
2
n
)
=
sin
(
θ
)
θ
=
sinc
θ
.
{\displaystyle \cos \left({\theta \over 2}\right)\cdot \cos \left({\theta \over 4}\right)\cdot \cos \left({\theta \over 8}\right)\cdots =\prod _{n=1}^{\infty }\cos \left({\theta \over 2^{n}}\right)={\sin(\theta ) \over \theta }=\operatorname {sinc} \,\theta .}
حيث تشير sinc إلى دالة الجيب الجوهري
وهي تكافئ
sinc
(
x
)
=
sin
(
x
)
x
{\displaystyle \operatorname {sinc} (x)={\frac {\sin(x)}{x}}}
جيب
جيب التمام
أخرى
sin
2
θ
=
1
−
cos
2
θ
2
{\displaystyle \sin ^{2}\theta ={\frac {1-\cos 2\theta }{2}}}
cos
2
θ
=
1
+
cos
2
θ
2
{\displaystyle \cos ^{2}\theta ={\frac {1+\cos 2\theta }{2}}}
sin
2
θ
cos
2
θ
=
1
−
cos
4
θ
8
{\displaystyle \sin ^{2}\theta \cos ^{2}\theta ={\frac {1-\cos 4\theta }{8}}}
sin
3
θ
=
3
sin
θ
−
sin
3
θ
4
{\displaystyle \sin ^{3}\theta ={\frac {3\sin \theta -\sin 3\theta }{4}}}
cos
3
θ
=
3
cos
θ
+
cos
3
θ
4
{\displaystyle \cos ^{3}\theta ={\frac {3\cos \theta +\cos 3\theta }{4}}}
sin
3
θ
cos
3
θ
=
3
sin
2
θ
−
sin
6
θ
32
{\displaystyle \sin ^{3}\theta \cos ^{3}\theta ={\frac {3\sin 2\theta -\sin 6\theta }{32}}}
sin
4
θ
=
3
−
4
cos
2
θ
+
cos
4
θ
8
{\displaystyle \sin ^{4}\theta ={\frac {3-4\cos 2\theta +\cos 4\theta }{8}}}
cos
4
θ
=
3
+
4
cos
2
θ
+
cos
4
θ
8
{\displaystyle \cos ^{4}\theta ={\frac {3+4\cos 2\theta +\cos 4\theta }{8}}}
sin
4
θ
cos
4
θ
=
3
−
4
cos
4
θ
+
cos
8
θ
128
{\displaystyle \sin ^{4}\theta \cos ^{4}\theta ={\frac {3-4\cos 4\theta +\cos 8\theta }{128}}}
sin
5
θ
=
10
sin
θ
−
5
sin
3
θ
+
sin
5
θ
16
{\displaystyle \sin ^{5}\theta ={\frac {10\sin \theta -5\sin 3\theta +\sin 5\theta }{16}}}
cos
5
θ
=
10
cos
θ
+
5
cos
3
θ
+
cos
5
θ
16
{\displaystyle \cos ^{5}\theta ={\frac {10\cos \theta +5\cos 3\theta +\cos 5\theta }{16}}}
sin
5
θ
cos
5
θ
=
10
sin
2
θ
−
5
sin
6
θ
+
sin
10
θ
512
{\displaystyle \sin ^{5}\theta \cos ^{5}\theta ={\frac {10\sin 2\theta -5\sin 6\theta +\sin 10\theta }{512}}}
جيب التمام
جيب
إذا كان n فردي
cos
n
θ
=
2
2
n
∑
k
=
0
n
−
1
2
(
n
k
)
cos
(
(
n
−
2
k
)
θ
)
{\displaystyle \cos ^{n}\theta ={\frac {2}{2^{n}}}\sum _{k=0}^{\frac {n-1}{2}}{\binom {n}{k}}\cos {((n-2k)\theta )}}
sin
n
θ
=
2
2
n
∑
k
=
0
n
−
1
2
(
−
1
)
(
n
−
1
2
−
k
)
(
n
k
)
sin
(
(
n
−
2
k
)
θ
)
{\displaystyle \sin ^{n}\theta ={\frac {2}{2^{n}}}\sum _{k=0}^{\frac {n-1}{2}}(-1)^{({\frac {n-1}{2}}-k)}{\binom {n}{k}}\sin {((n-2k)\theta )}}
إذا كان n زوجي
cos
n
θ
=
1
2
n
(
n
n
2
)
+
2
2
n
∑
k
=
0
n
2
−
1
(
n
k
)
cos
(
(
n
−
2
k
)
θ
)
{\displaystyle \cos ^{n}\theta ={\frac {1}{2^{n}}}{\binom {n}{\frac {n}{2}}}+{\frac {2}{2^{n}}}\sum _{k=0}^{{\frac {n}{2}}-1}{\binom {n}{k}}\cos {((n-2k)\theta )}}
sin
n
θ
=
1
2
n
(
n
n
2
)
+
2
2
n
∑
k
=
0
n
2
−
1
(
−
1
)
(
n
2
−
k
)
(
n
k
)
cos
(
(
n
−
2
k
)
θ
)
{\displaystyle \sin ^{n}\theta ={\frac {1}{2^{n}}}{\binom {n}{\frac {n}{2}}}+{\frac {2}{2^{n}}}\sum _{k=0}^{{\frac {n}{2}}-1}(-1)^{({\frac {n}{2}}-k)}{\binom {n}{k}}\cos {((n-2k)\theta )}}
[ عدل ]
من الجداء إلى المجموع
cos
θ
cos
φ
=
cos
(
θ
−
φ
)
+
cos
(
θ
+
φ
)
2
{\displaystyle \cos \theta \cos \varphi ={\cos(\theta -\varphi )+\cos(\theta +\varphi ) \over 2}}
sin
θ
sin
φ
=
cos
(
θ
−
φ
)
−
cos
(
θ
+
φ
)
2
{\displaystyle \sin \theta \sin \varphi ={\cos(\theta -\varphi )-\cos(\theta +\varphi ) \over 2}}
sin
θ
cos
φ
=
sin
(
θ
+
φ
)
+
sin
(
θ
−
φ
)
2
{\displaystyle \sin \theta \cos \varphi ={\sin(\theta +\varphi )+\sin(\theta -\varphi ) \over 2}}
cos
θ
sin
φ
=
sin
(
θ
+
φ
)
−
sin
(
θ
−
φ
)
2
{\displaystyle \cos \theta \sin \varphi ={\sin(\theta +\varphi )-\sin(\theta -\varphi ) \over 2}}
من المجموع/الفرق إلى الجداء
sin
θ
±
sin
φ
=
2
sin
(
θ
±
φ
2
)
cos
(
θ
∓
φ
2
)
{\displaystyle \sin \theta \pm \sin \varphi =2\sin \left({\frac {\theta \pm \varphi }{2}}\right)\cos \left({\frac {\theta \mp \varphi }{2}}\right)}
cos
θ
+
cos
φ
=
2
cos
(
θ
+
φ
2
)
cos
(
θ
−
φ
2
)
{\displaystyle \cos \theta +\cos \varphi =2\cos \left({\frac {\theta +\varphi }{2}}\right)\cos \left({\frac {\theta -\varphi }{2}}\right)}
cos
θ
−
cos
φ
=
−
2
sin
(
θ
+
φ
2
)
sin
(
θ
−
φ
2
)
{\displaystyle \cos \theta -\cos \varphi =-2\sin \left({\theta +\varphi \over 2}\right)\sin \left({\theta -\varphi \over 2}\right)}
[ عدل ] إذا كانت
x
+
y
+
z
=
π
{\displaystyle x+y+z=\pi }
tan
(
x
)
+
tan
(
y
)
+
tan
(
z
)
=
tan
(
x
)
tan
(
y
)
tan
(
z
)
{\displaystyle \tan(x)+\tan(y)+\tan(z)=\tan(x)\tan(y)\tan(z)}
و
sin
(
2
x
)
+
sin
(
2
y
)
+
sin
(
2
z
)
=
4
sin
(
x
)
sin
(
y
)
sin
(
z
)
{\displaystyle \sin(2x)+\sin(2y)+\sin(2z)=4\sin(x)\sin(y)\sin(z)}
إذا كانت
w
+
x
+
y
+
z
=
π
{\displaystyle w+x+y+z=\pi }
sin
(
w
+
x
)
sin
(
x
+
y
)
=
sin
(
x
+
y
)
sin
(
y
+
z
)
=
sin
(
y
+
z
)
sin
(
z
+
w
)
=
sin
(
z
+
w
)
sin
(
w
+
x
)
=
sin
(
w
)
sin
(
y
)
+
sin
(
x
)
sin
(
z
)
.
{\displaystyle {\begin{aligned}\sin(w+x)\sin(x+y)&=\sin(x+y)\sin(y+z)\\&{}=\sin(y+z)\sin(z+w)\\&{}=\sin(z+w)\sin(w+x)=\sin(w)\sin(y)+\sin(x)\sin(z).\end{aligned}}}
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 )\,}
حيث:
φ
=
{
arcsin
(
b
a
2
+
b
2
)
if
a
≥
0
,
π
−
arcsin
(
b
a
2
+
b
2
)
if
a
<
0
,
{\displaystyle \varphi ={\begin{cases}\arcsin \left({\frac {b}{\sqrt {a^{2}+b^{2}}}}\right)&{\text{if }}a\geq 0,\\\pi -\arcsin \left({\frac {b}{\sqrt {a^{2}+b^{2}}}}\right)&{\text{if }}a<0,\end{cases}}}
أو:
φ
=
arctan
(
b
a
)
+
{
0
if
a
≥
0
,
π
if
a
<
0.
{\displaystyle \varphi =\arctan \left({\frac {b}{a}}\right)+{\begin{cases}0&{\text{if }}a\geq 0,\\\pi &{\text{if }}a<0.\end{cases}}}
a
sin
x
+
b
sin
(
x
+
α
)
=
c
sin
(
x
+
β
)
{\displaystyle a\sin x+b\sin(x+\alpha )=c\sin(x+\beta )\,}
حيث:
c
=
a
2
+
b
2
+
2
a
b
cos
α
,
{\displaystyle c={\sqrt {a^{2}+b^{2}+2ab\cos \alpha }},\,}
و:
β
=
arctan
(
b
sin
α
a
+
b
cos
α
)
+
{
0
if
a
+
b
cos
α
≥
0
,
π
if
a
+
b
cos
α
<
0.
{\displaystyle \beta =\arctan \left({\frac {b\sin \alpha }{a+b\cos \alpha }}\right)+{\begin{cases}0&{\text{if }}a+b\cos \alpha \geq 0,\\\pi &{\text{if }}a+b\cos \alpha <0.\end{cases}}}
[ عدل ]
sin
φ
+
sin
(
φ
+
α
)
+
sin
(
φ
+
2
α
)
+
⋯
+
sin
(
φ
+
n
α
)
=
sin
(
(
n
+
1
)
α
2
)
⋅
sin
(
φ
+
n
α
2
)
sin
α
2
.
{\displaystyle \sin {\varphi }+\sin {(\varphi +\alpha )}+\sin {(\varphi +2\alpha )}+\cdots +\sin {(\varphi +n\alpha )}={\frac {\sin {\left({\frac {(n+1)\alpha }{2}}\right)}\cdot \sin {(\varphi +{\frac {n\alpha }{2}})}}{\sin {\frac {\alpha }{2}}}}.}
cos
φ
+
cos
(
φ
+
α
)
+
cos
(
φ
+
2
α
)
+
⋯
+
cos
(
φ
+
n
α
)
=
sin
(
(
n
+
1
)
α
2
)
⋅
cos
(
φ
+
n
α
2
)
sin
α
2
.
{\displaystyle \cos {\varphi }+\cos {(\varphi +\alpha )}+\cos {(\varphi +2\alpha )}+\cdots +\cos {(\varphi +n\alpha )}={\frac {\sin {\left({\frac {(n+1)\alpha }{2}}\right)}\cdot \cos {(\varphi +{\frac {n\alpha }{2}})}}{\sin {\frac {\alpha }{2}}}}.}
a
cos
(
x
)
+
b
sin
(
x
)
=
a
2
+
b
2
cos
(
x
−
atan2
(
b
,
a
)
)
{\displaystyle a\cos(x)+b\sin(x)={\sqrt {a^{2}+b^{2}}}\cos(x-\operatorname {atan2} \,(b,a))\;}
tan
(
x
)
+
sec
(
x
)
=
tan
(
x
2
+
π
4
)
.
{\displaystyle \tan(x)+\sec(x)=\tan \left({x \over 2}+{\pi \over 4}\right).}
cot
(
x
)
cot
(
y
)
+
cot
(
y
)
cot
(
z
)
+
cot
(
z
)
cot
(
x
)
=
1.
{\displaystyle \cot(x)\cot(y)+\cot(y)\cot(z)+\cot(z)\cot(x)=1.\,}
[ عدل ]
f
(
x
)
=
(
cos
α
)
x
−
sin
α
(
sin
α
)
x
+
cos
α
,
{\displaystyle f(x)={\frac {(\cos \alpha )x-\sin \alpha }{(\sin \alpha )x+\cos \alpha }},}
وبالمثل:
g
(
x
)
=
(
cos
β
)
x
−
sin
β
(
sin
β
)
x
+
cos
β
,
{\displaystyle g(x)={\frac {(\cos \beta )x-\sin \beta }{(\sin \beta )x+\cos \beta }},}
وعليه:
f
(
g
(
x
)
)
=
g
(
f
(
x
)
)
=
(
cos
(
α
+
β
)
)
x
−
sin
(
α
+
β
)
(
sin
(
α
+
β
)
)
x
+
cos
(
α
+
β
)
.
{\displaystyle f(g(x))=g(f(x))={\frac {(\cos(\alpha +\beta ))x-\sin(\alpha +\beta )}{(\sin(\alpha +\beta ))x+\cos(\alpha +\beta )}}.}
f
α
∘
f
β
=
f
α
+
β
.
{\displaystyle f_{\alpha }\circ f_{\beta }=f_{\alpha +\beta }.\,}
[ عدل ]
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
,
if
x
>
0
−
π
/
2
,
if
x
<
0
{\displaystyle \arctan(x)+\arctan(1/x)=\left\{{\begin{matrix}\pi /2,&{\mbox{if }}x>0\\-\pi /2,&{\mbox{if }}x<0\end{matrix}}\right.}
[ عدل ]
sin
[
arccos
(
x
)
]
=
1
−
x
2
{\displaystyle \sin[\arccos(x)]={\sqrt {1-x^{2}}}\,}
tan
[
arcsin
(
x
)
]
=
x
1
−
x
2
{\displaystyle \tan[\arcsin(x)]={\frac {x}{\sqrt {1-x^{2}}}}}
sin
[
arctan
(
x
)
]
=
x
1
+
x
2
{\displaystyle \sin[\arctan(x)]={\frac {x}{\sqrt {1+x^{2}}}}}
tan
[
arccos
(
x
)
]
=
1
−
x
2
x
{\displaystyle \tan[\arccos(x)]={\frac {\sqrt {1-x^{2}}}{x}}}
cos
[
arctan
(
x
)
]
=
1
1
+
x
2
{\displaystyle \cos[\arctan(x)]={\frac {1}{\sqrt {1+x^{2}}}}}
cot
[
arcsin
(
x
)
]
=
1
−
x
2
x
{\displaystyle \cot[\arcsin(x)]={\frac {\sqrt {1-x^{2}}}{x}}}
cos
[
arcsin
(
x
)
]
=
1
−
x
2
{\displaystyle \cos[\arcsin(x)]={\sqrt {1-x^{2}}}\,}
cot
[
arccos
(
x
)
]
=
x
1
−
x
2
{\displaystyle \cot[\arccos(x)]={\frac {x}{\sqrt {1-x^{2}}}}}
e
i
x
=
cos
(
x
)
+
i
sin
(
x
)
{\displaystyle e^{ix}=\cos(x)+i\sin(x)\,}
صيغة أويلر ),
e
−
i
x
=
cos
(
−
x
)
+
i
sin
(
−
x
)
=
cos
(
x
)
−
i
sin
(
x
)
{\displaystyle e^{-ix}=\cos(-x)+i\sin(-x)=\cos(x)-i\sin(x)\,}
e
i
π
=
−
1
{\displaystyle e^{i\pi }=-1\,}
cos
(
x
)
=
e
i
x
+
e
−
i
x
2
{\displaystyle \cos(x)={\frac {e^{ix}+e^{-ix}}{2}}\;}
sin
(
x
)
=
e
i
x
−
e
−
i
x
2
i
{\displaystyle \sin(x)={\frac {e^{ix}-e^{-ix}}{2i}}\;}
tan
(
x
)
=
e
i
x
−
e
−
i
x
i
(
e
i
x
+
e
−
i
x
)
=
sin
(
x
)
cos
(
x
)
{\displaystyle \tan(x)={\frac {e^{ix}-e^{-ix}}{i({e^{ix}+e^{-ix}})}}\;={\frac {\sin(x)}{\cos(x)}}}
حيث
i
2
=
−
1
{\displaystyle i^{2}=-1}
[ عدل ]
[ عدل ]
cos
20
∘
⋅
cos
40
∘
⋅
cos
80
∘
=
1
8
{\displaystyle \cos 20^{\circ }\cdot \cos 40^{\circ }\cdot \cos 80^{\circ }={\frac {1}{8}}}
∏
j
=
0
k
−
1
cos
(
2
j
x
)
=
sin
(
2
k
x
)
2
k
sin
(
x
)
.
{\displaystyle \prod _{j=0}^{k-1}\cos(2^{j}x)={\frac {\sin(2^{k}x)}{2^{k}\sin(x)}}.}
cos
π
7
cos
2
π
7
cos
3
π
7
=
1
8
,
{\displaystyle \cos {\frac {\pi }{7}}\cos {\frac {2\pi }{7}}\cos {\frac {3\pi }{7}}={\frac {1}{8}},}
sin
20
∘
⋅
sin
40
∘
⋅
sin
80
∘
=
3
8
.
{\displaystyle \sin 20^{\circ }\cdot \sin 40^{\circ }\cdot \sin 80^{\circ }={\frac {\sqrt {3}}{8}}.}
cos
24
∘
+
cos
48
∘
+
cos
96
∘
+
cos
168
∘
=
1
2
.
{\displaystyle \cos 24^{\circ }+\cos 48^{\circ }+\cos 96^{\circ }+\cos 168^{\circ }={\frac {1}{2}}.}
cos
(
2
π
21
)
+
cos
(
2
⋅
2
π
21
)
+
cos
(
4
⋅
2
π
21
)
{\displaystyle \cos \left({\frac {2\pi }{21}}\right)\,+\,\cos \left(2\cdot {\frac {2\pi }{21}}\right)\,+\,\cos \left(4\cdot {\frac {2\pi }{21}}\right)}
+
cos
(
5
⋅
2
π
21
)
+
cos
(
8
⋅
2
π
21
)
+
cos
(
10
⋅
2
π
21
)
=
1
2
.
{\displaystyle \,+\,\cos \left(5\cdot {\frac {2\pi }{21}}\right)\,+\,\cos \left(8\cdot {\frac {2\pi }{21}}\right)\,+\,\cos \left(10\cdot {\frac {2\pi }{21}}\right)={\frac {1}{2}}.}
π
4
=
4
arctan
1
5
−
arctan
1
239
{\displaystyle {\frac {\pi }{4}}=4\arctan {\frac {1}{5}}-\arctan {\frac {1}{239}}}
π
4
=
5
arctan
1
7
+
2
arctan
3
79
.
{\displaystyle {\frac {\pi }{4}}=5\arctan {\frac {1}{7}}+2\arctan {\frac {3}{79}}.}
[ عدل ]
sin
0
=
sin
0
∘
=
0
/
2
=
cos
90
∘
=
cos
(
π
2
)
sin
(
π
6
)
=
sin
30
∘
=
1
/
2
=
cos
60
∘
=
cos
(
π
3
)
sin
(
π
4
)
=
sin
45
∘
=
2
/
2
=
cos
45
∘
=
cos
(
π
4
)
sin
(
π
3
)
=
sin
60
∘
=
3
/
2
=
cos
30
∘
=
cos
(
π
6
)
sin
(
π
2
)
=
sin
90
∘
=
4
/
2
=
cos
0
∘
=
cos
0
{\displaystyle {\begin{matrix}\sin 0&=&\sin 0^{\circ }&=&{\sqrt {0}}/2&=&\cos 90^{\circ }&=&\cos \left({\frac {\pi }{2}}\right)\\\\\sin \left({\frac {\pi }{6}}\right)&=&\sin 30^{\circ }&=&{\sqrt {1}}/2&=&\cos 60^{\circ }&=&\cos \left({\frac {\pi }{3}}\right)\\\\\sin \left({\frac {\pi }{4}}\right)&=&\sin 45^{\circ }&=&{\sqrt {2}}/2&=&\cos 45^{\circ }&=&\cos \left({\frac {\pi }{4}}\right)\\\\\sin \left({\frac {\pi }{3}}\right)&=&\sin 60^{\circ }&=&{\sqrt {3}}/2&=&\cos 30^{\circ }&=&\cos \left({\frac {\pi }{6}}\right)\\\\\sin \left({\frac {\pi }{2}}\right)&=&\sin 90^{\circ }&=&{\sqrt {4}}/2&=&\cos 0^{\circ }&=&\cos 0\end{matrix}}}
sin
π
7
=
7
6
−
7
189
∑
j
=
0
∞
(
3
j
+
1
)
!
189
j
j
!
(
2
j
+
2
)
!
{\displaystyle \sin {\frac {\pi }{7}}={\frac {\sqrt {7}}{6}}-{\frac {\sqrt {7}}{189}}\sum _{j=0}^{\infty }{\frac {(3j+1)!}{189^{j}j!\,(2j+2)!}}\!}
sin
π
18
=
1
6
∑
j
=
0
∞
(
3
j
)
!
27
j
j
!
(
2
j
+
1
)
!
{\displaystyle \sin {\frac {\pi }{18}}={\frac {1}{6}}\sum _{j=0}^{\infty }{\frac {(3j)!}{27^{j}j!\,(2j+1)!}}\!}
بـالنسبة الذهبية φ:
cos
(
π
5
)
=
cos
36
∘
=
5
+
1
4
=
φ
/
2
{\displaystyle \cos \left({\frac {\pi }{5}}\right)=\cos 36^{\circ }={{\sqrt {5}}+1 \over 4}=\varphi /2}
sin
(
π
10
)
=
sin
18
∘
=
5
−
1
4
=
φ
−
1
2
=
1
2
φ
{\displaystyle \sin \left({\frac {\pi }{10}}\right)=\sin 18^{\circ }={{\sqrt {5}}-1 \over 4}={\varphi -1 \over 2}={1 \over 2\varphi }}
في حساب التفاضل والتكامل ، تتطلب العلاقات المذكورة أدناه قياس الزوايا بالتقدير الدائري (راديان)؛ ستصبح العلاقات أكثر تعقيدًا إذا تم قياس الزوايا
بوحدة أخرى مثل الدرجات. إذا كانت الدوال المثلثية معرفة بدلالة الهندسة، إلى جانب تعريفات طول القوس والمساحة ، يمكن إيجاد مشتقاتها من خلال التحقق من نهايتين . الأولى هي:
lim
x
→
0
sin
x
x
=
1
,
{\displaystyle \lim _{x\rightarrow 0}{\frac {\sin x}{x}}=1,}
محققة باستخدام دائرة الوحدة ومبرهنة الساندويتش . النهاية الثانية هي:
lim
x
→
0
1
−
cos
x
x
=
0
{\displaystyle \lim _{x\rightarrow 0}{\frac {1-\cos x}{x}}=0}
محققة باستخدام هذه المتطابقة tan x / 2 1 − cos x / sin x . بعد تحديد هتين النهايتين، يمكن للمرء استخدام تعريف النهاية للمشتقات ومبرهنات الجمع لإظهار أن (sin x )′ = cos x و (cos x )′ = −sin x . إذا كانت دالتي الجيب وجيب التمام معرفة بمتسلسلة تايلور الخاصة بهم، فيمكن إيجاد المشتقات عن طريق اشتقاق متسلسلة القوى حدًا بحد.
d
d
x
sin
x
=
cos
x
{\displaystyle {\frac {d}{dx}}\sin x=\cos x}
يمكن اشتقاق باقي الدوال المثلثية باستخدام المتطابقات أعلاه وقواعد التفاضل:
d
d
x
sin
x
=
cos
x
,
d
d
x
arcsin
x
=
1
1
−
x
2
d
d
x
cos
x
=
−
sin
x
,
d
d
x
arccos
x
=
−
1
1
−
x
2
d
d
x
tan
x
=
sec
2
x
,
d
d
x
arctan
x
=
1
1
+
x
2
d
d
x
cot
x
=
−
csc
2
x
,
d
d
x
arccot
x
=
−
1
1
+
x
2
d
d
x
sec
x
=
tan
x
sec
x
,
d
d
x
arcsec
x
=
1
|
x
|
x
2
−
1
d
d
x
csc
x
=
−
csc
x
cot
x
,
d
d
x
arccsc
x
=
−
1
|
x
|
x
2
−
1
{\displaystyle {\begin{aligned}{d \over dx}\sin x&=\cos x,&{d \over dx}\arcsin x&={1 \over {\sqrt {1-x^{2}}}}\\\\{d \over dx}\cos x&=-\sin x,&{d \over dx}\arccos x&={-1 \over {\sqrt {1-x^{2}}}}\\\\{d \over dx}\tan x&=\sec ^{2}x,&{d \over dx}\arctan x&={1 \over 1+x^{2}}\\\\{d \over dx}\cot x&=-\csc ^{2}x,&{d \over dx}\operatorname {arccot} x&={-1 \over 1+x^{2}}\\\\{d \over dx}\sec x&=\tan x\sec x,&{d \over dx}\operatorname {arcsec} x&={1 \over |x|{\sqrt {x^{2}-1}}}\\\\{d \over dx}\csc x&=-\csc x\cot x,&{d \over dx}\operatorname {arccsc} x&={-1 \over |x|{\sqrt {x^{2}-1}}}\end{aligned}}\ }
يمكن إيجاد المتطابقات التكاملية في قائمة تكاملات الدوال المثلثية . بعض الأشكال العامة مسرودة أدناه:
∫
d
u
a
2
−
u
2
=
sin
−
1
(
u
a
)
+
C
{\displaystyle \int {\frac {du}{\sqrt {a^{2}-u^{2}}}}=\sin ^{-1}\left({\frac {u}{a}}\right)+C}
∫
d
u
a
2
+
u
2
=
1
a
tan
−
1
(
u
a
)
+
C
{\displaystyle \int {\frac {du}{a^{2}+u^{2}}}={\frac {1}{a}}\tan ^{-1}\left({\frac {u}{a}}\right)+C}
∫
d
u
u
u
2
−
a
2
=
1
a
sec
−
1
|
u
a
|
+
C
{\displaystyle \int {\frac {du}{u{\sqrt {u^{2}-a^{2}}}}}={\frac {1}{a}}\sec ^{-1}\left|{\frac {u}{a}}\right|+C}
الدالة
الدالة المعكوسة
sin
θ
=
e
i
θ
−
e
−
i
θ
2
i
{\displaystyle \sin \theta ={\frac {e^{i\theta }-e^{-i\theta }}{2i}}\,}
arcsin
x
=
−
i
ln
(
i
x
+
1
−
x
2
)
{\displaystyle \arcsin x=-i\ln \left(ix+{\sqrt {1-x^{2}}}\right)\,}
cos
θ
=
e
i
θ
+
e
−
i
θ
2
{\displaystyle \cos \theta ={\frac {e^{i\theta }+e^{-i\theta }}{2}}\,}
arccos
x
=
−
i
ln
(
x
+
x
2
−
1
)
{\displaystyle \arccos x=-i\ln \left(x+{\sqrt {x^{2}-1}}\right)\,}
tan
θ
=
e
i
θ
−
e
−
i
θ
i
(
e
i
θ
+
e
−
i
θ
)
{\displaystyle \tan \theta ={\frac {e^{i\theta }-e^{-i\theta }}{i(e^{i\theta }+e^{-i\theta })}}\,}
arctan
x
=
i
2
ln
(
i
+
x
i
−
x
)
{\displaystyle \arctan x={\frac {i}{2}}\ln \left({\frac {i+x}{i-x}}\right)\,}
csc
θ
=
2
i
e
i
θ
−
e
−
i
θ
{\displaystyle \csc \theta ={\frac {2i}{e^{i\theta }-e^{-i\theta }}}\,}
arccsc
x
=
−
i
ln
(
i
x
+
1
−
1
x
2
)
{\displaystyle \operatorname {arccsc} x=-i\ln \left({\tfrac {i}{x}}+{\sqrt {1-{\tfrac {1}{x^{2}}}}}\right)\,}
sec
θ
=
2
e
i
θ
+
e
−
i
θ
{\displaystyle \sec \theta ={\frac {2}{e^{i\theta }+e^{-i\theta }}}\,}
arcsec
x
=
−
i
ln
(
1
x
+
1
−
i
x
2
)
{\displaystyle \operatorname {arcsec} x=-i\ln \left({\tfrac {1}{x}}+{\sqrt {1-{\tfrac {i}{x^{2}}}}}\right)\,}
cot
θ
=
i
(
e
i
θ
+
e
−
i
θ
)
e
i
θ
−
e
−
i
θ
{\displaystyle \cot \theta ={\frac {i(e^{i\theta }+e^{-i\theta })}{e^{i\theta }-e^{-i\theta }}}\,}
arccot
x
=
i
2
ln
(
x
−
i
x
+
i
)
{\displaystyle \operatorname {arccot} x={\frac {i}{2}}\ln \left({\frac {x-i}{x+i}}\right)\,}
cis
θ
=
e
i
θ
{\displaystyle \operatorname {cis} \,\theta =e^{i\theta }\,}
arccis
x
=
ln
x
i
{\displaystyle \operatorname {arccis} \,x={\frac {\ln x}{i}}\,}
1
+
2
cos
(
x
)
+
2
cos
(
2
x
)
+
2
cos
(
3
x
)
+
⋯
+
2
cos
(
n
x
)
=
sin
[
(
n
+
1
2
)
x
]
sin
(
x
2
)
.
{\displaystyle 1+2\cos(x)+2\cos(2x)+2\cos(3x)+\cdots +2\cos(nx)={\frac {\sin \left[\left(n+{\frac {1}{2}}\right)x\right\rbrack }{\sin \left({\frac {x}{2}}\right)}}.}
[ عدل ] إذا وضعنا
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}}}}
cos
(
x
)
=
1
−
t
2
1
+
t
2
{\displaystyle \cos(x)={\frac {1-t^{2}}{1+t^{2}}}}
e
i
x
=
1
+
i
t
1
−
i
t
{\displaystyle e^{ix}={\frac {1+it}{1-it}}}
بوابة رياضيات
![{\displaystyle \left[{\begin{matrix}\cos \alpha &-\sin \alpha \\\sin \alpha &\cos \alpha \end{matrix}}\right]\left[{\begin{matrix}\cos \beta &-\sin \beta \\\sin \beta &\cos \beta \end{matrix}}\right]=\left[{\begin{matrix}\cos(\alpha +\beta )&-\sin(\alpha +\beta )\\\sin(\alpha +\beta )&\cos(\alpha +\beta )\end{matrix}}\right].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1466009a9e82d549369a00717e6e796074344559)
![{\displaystyle \sin[\arccos(x)]={\sqrt {1-x^{2}}}\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/89020e4173e7c25657019ade3629e832bb572ff7)
![{\displaystyle \tan[\arcsin(x)]={\frac {x}{\sqrt {1-x^{2}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/35f9ce679263ce5306237e3342d22abeb3d9d346)
![{\displaystyle \sin[\arctan(x)]={\frac {x}{\sqrt {1+x^{2}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7d3ed8cedb4c73b01b65314a15152647c9000086)
![{\displaystyle \tan[\arccos(x)]={\frac {\sqrt {1-x^{2}}}{x}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/22e131c5707de0e5ab51c45a4c518f8b137ffb99)
![{\displaystyle \cos[\arctan(x)]={\frac {1}{\sqrt {1+x^{2}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d2e67d269ac585855446bba315fc260582bc38be)
![{\displaystyle \cot[\arcsin(x)]={\frac {\sqrt {1-x^{2}}}{x}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/03accb0c26aee5933b8bab9603a6d213c5a9b196)
![{\displaystyle \cos[\arcsin(x)]={\sqrt {1-x^{2}}}\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6123652c97881e3910234616ea5ee831deae4a73)
![{\displaystyle \cot[\arccos(x)]={\frac {x}{\sqrt {1-x^{2}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6b3fb51eba767321e7afcc50e38aa936d845c0af)
