x
n
{\displaystyle {\sqrt[{n}]{x}}}
Ein Dreieck mit den üblichen Bezeichnungen
Die folgende Liste enthält die meisten bekannten Formeln aus der Trigonometrie in der Ebene . Die meisten dieser Beziehungen verwenden trigonometrische Funktionen .
Dabei werden die folgenden Bezeichnungen verwendet: Das Dreieck
A
B
C
{\displaystyle ABC}
habe die Seiten
a
=
B
C
{\displaystyle a=BC}
,
b
=
C
A
{\displaystyle b=CA}
und
c
=
A
B
{\displaystyle c=AB}
, die Winkel
α
{\displaystyle \alpha }
,
β
{\displaystyle \beta }
und
γ
{\displaystyle \gamma }
bei den Ecken
A
{\displaystyle A}
,
B
{\displaystyle B}
und
C
{\displaystyle C}
. Ferner seien
r
{\displaystyle r}
der Umkreisradius ,
ρ
{\displaystyle \rho }
der Inkreisradius und
ρ
a
{\displaystyle \rho _{a}}
,
ρ
b
{\displaystyle \rho _{b}}
und
ρ
c
{\displaystyle \rho _{c}}
die Ankreisradien (und zwar die Radien der Ankreise, die den Ecken
A
{\displaystyle A}
,
B
{\displaystyle B}
bzw.
C
{\displaystyle C}
gegenüberliegen) des Dreiecks
A
B
C
{\displaystyle ABC}
. Die Variable
s
{\displaystyle s}
steht für den halben Umfang des Dreiecks
A
B
C
{\displaystyle ABC}
:
s
=
a
+
b
+
c
2
{\displaystyle s={\frac {a+b+c}{2}}}
.
Schließlich wird die Fläche des Dreiecks
A
B
C
{\displaystyle ABC}
mit
F
{\displaystyle F}
bezeichnet. Alle anderen Bezeichnungen werden jeweils in den entsprechenden Abschnitten, in denen sie vorkommen, erläutert.
Es ist zu beachten, dass hier die Bezeichnungen für den Umkreisradius
r
{\displaystyle r}
, den Inkreisradius
ρ
{\displaystyle \rho }
und die drei Ankreisradien
ρ
a
{\displaystyle \rho _{a}}
,
ρ
b
{\displaystyle \rho _{b}}
,
ρ
c
{\displaystyle \rho _{c}}
benutzt werden. Oft werden davon abweichend für dieselben Größen auch die Bezeichnungen
R
{\displaystyle R}
,
r
{\displaystyle r}
,
r
a
{\displaystyle r_{a}}
,
r
b
{\displaystyle r_{b}}
,
r
c
{\displaystyle r_{c}}
verwendet.
α
+
β
+
γ
=
180
∘
{\displaystyle \alpha +\beta +\gamma =180^{\circ }}
Formel 1:
a
sin
α
=
b
sin
β
=
c
sin
γ
=
2
r
=
a
b
c
2
F
{\displaystyle {\frac {a}{\sin \alpha }}={\frac {b}{\sin \beta }}={\frac {c}{\sin \gamma }}=2r={\frac {abc}{2F}}}
Formel 2:
wenn
α
=
90
∘
{\displaystyle \alpha =90^{\circ }}
sin
β
=
b
a
{\displaystyle \sin \beta ={\frac {b}{a}}}
sin
γ
=
c
a
{\displaystyle \sin \gamma ={\frac {c}{a}}}
wenn
β
=
90
∘
{\displaystyle \beta =90^{\circ }}
sin
α
=
a
b
{\displaystyle \sin \alpha ={\frac {a}{b}}}
sin
γ
=
c
b
{\displaystyle \sin \gamma ={\frac {c}{b}}}
wenn
γ
=
90
∘
{\displaystyle \gamma =90^{\circ }}
sin
α
=
a
c
{\displaystyle \sin \alpha ={\frac {a}{c}}}
sin
β
=
b
c
{\displaystyle \sin \beta ={\frac {b}{c}}}
Formel 1:
a
2
=
b
2
+
c
2
−
2
b
c
cos
α
{\displaystyle a^{2}=b^{2}+c^{2}-2bc\ \cos \alpha }
b
2
=
c
2
+
a
2
−
2
c
a
cos
β
{\displaystyle b^{2}=c^{2}+a^{2}-2ca\ \cos \beta }
c
2
=
a
2
+
b
2
−
2
a
b
cos
γ
{\displaystyle c^{2}=a^{2}+b^{2}-2ab\ \cos \gamma }
Formel 2:
wenn
α
=
90
∘
{\displaystyle \alpha =90^{\circ }}
cos
β
=
c
a
{\displaystyle \cos \beta ={\frac {c}{a}}}
cos
γ
=
b
a
{\displaystyle \cos \gamma ={\frac {b}{a}}}
wenn
β
=
90
∘
{\displaystyle \beta =90^{\circ }}
cos
α
=
c
b
{\displaystyle \cos \alpha ={\frac {c}{b}}}
cos
γ
=
a
b
{\displaystyle \cos \gamma ={\frac {a}{b}}}
wenn
γ
=
90
∘
{\displaystyle \gamma =90^{\circ }}
a
2
+
b
2
=
c
2
{\displaystyle a^{2}+b^{2}=c^{2}}
(Satz des Pythagoras )
cos
α
=
b
c
{\displaystyle \cos \alpha ={\frac {b}{c}}}
cos
β
=
a
c
{\displaystyle \cos \beta ={\frac {a}{c}}}
a
=
b
cos
γ
+
c
cos
β
{\displaystyle a=b\,\cos \gamma +c\,\cos \beta }
b
=
c
cos
α
+
a
cos
γ
{\displaystyle b=c\,\cos \alpha +a\,\cos \gamma }
c
=
a
cos
β
+
b
cos
α
{\displaystyle c=a\,\cos \beta +b\,\cos \alpha }
b
+
c
a
=
cos
β
−
γ
2
sin
α
2
,
c
+
a
b
=
cos
γ
−
α
2
sin
β
2
,
a
+
b
c
=
cos
α
−
β
2
sin
γ
2
{\displaystyle {\frac {b+c}{a}}={\frac {\cos {\frac {\beta -\gamma }{2}}}{\sin {\frac {\alpha }{2}}}},\quad {\frac {c+a}{b}}={\frac {\cos {\frac {\gamma -\alpha }{2}}}{\sin {\frac {\beta }{2}}}},\quad {\frac {a+b}{c}}={\frac {\cos {\frac {\alpha -\beta }{2}}}{\sin {\frac {\gamma }{2}}}}}
b
−
c
a
=
sin
β
−
γ
2
cos
α
2
,
c
−
a
b
=
sin
γ
−
α
2
cos
β
2
,
a
−
b
c
=
sin
α
−
β
2
cos
γ
2
{\displaystyle {\frac {b-c}{a}}={\frac {\sin {\frac {\beta -\gamma }{2}}}{\cos {\frac {\alpha }{2}}}},\quad {\frac {c-a}{b}}={\frac {\sin {\frac {\gamma -\alpha }{2}}}{\cos {\frac {\beta }{2}}}},\quad {\frac {a-b}{c}}={\frac {\sin {\frac {\alpha -\beta }{2}}}{\cos {\frac {\gamma }{2}}}}}
Formel 1:
b
+
c
b
−
c
=
tan
β
+
γ
2
tan
β
−
γ
2
=
cot
α
2
tan
β
−
γ
2
{\displaystyle {\frac {b+c}{b-c}}={\frac {\tan {\frac {\beta +\gamma }{2}}}{\tan {\frac {\beta -\gamma }{2}}}}={\frac {\cot {\frac {\alpha }{2}}}{\tan {\frac {\beta -\gamma }{2}}}}}
Analoge Formeln gelten für
a
+
b
a
−
b
{\displaystyle {\frac {a+b}{a-b}}}
und
c
+
a
c
−
a
{\displaystyle {\frac {c+a}{c-a}}}
:
a
+
b
a
−
b
=
tan
α
+
β
2
tan
α
−
β
2
=
cot
γ
2
tan
α
−
β
2
{\displaystyle {\frac {a+b}{a-b}}={\frac {\tan {\frac {\alpha +\beta }{2}}}{\tan {\frac {\alpha -\beta }{2}}}}={\frac {\cot {\frac {\gamma }{2}}}{\tan {\frac {\alpha -\beta }{2}}}}}
c
+
a
c
−
a
=
tan
γ
+
α
2
tan
γ
−
α
2
=
cot
β
2
tan
γ
−
α
2
{\displaystyle {\frac {c+a}{c-a}}={\frac {\tan {\frac {\gamma +\alpha }{2}}}{\tan {\frac {\gamma -\alpha }{2}}}}={\frac {\cot {\frac {\beta }{2}}}{\tan {\frac {\gamma -\alpha }{2}}}}}
Wegen
tan
(
−
x
)
=
−
tan
(
x
)
{\displaystyle \tan(-x)=-\tan(x)}
bleibt eine dieser Formel gültig, wenn sowohl die Seiten als auch die zugehörigen Winkel vertauscht werden, also etwa:
a
+
c
a
−
c
=
tan
α
+
γ
2
tan
α
−
γ
2
=
cot
β
2
tan
α
−
γ
2
{\displaystyle {\frac {a+c}{a-c}}={\frac {\tan {\frac {\alpha +\gamma }{2}}}{\tan {\frac {\alpha -\gamma }{2}}}}={\frac {\cot {\frac {\beta }{2}}}{\tan {\frac {\alpha -\gamma }{2}}}}}
Formel 2:
wenn
α
=
90
∘
{\displaystyle \alpha =90^{\circ }}
tan
β
=
b
c
{\displaystyle \tan \beta ={\frac {b}{c}}}
tan
γ
=
c
b
{\displaystyle \tan \gamma ={\frac {c}{b}}}
wenn
β
=
90
∘
{\displaystyle \beta =90^{\circ }}
tan
α
=
a
c
{\displaystyle \tan \alpha ={\frac {a}{c}}}
tan
γ
=
c
a
{\displaystyle \tan \gamma ={\frac {c}{a}}}
wenn
γ
=
90
∘
{\displaystyle \gamma =90^{\circ }}
tan
α
=
a
b
{\displaystyle \tan \alpha ={\frac {a}{b}}}
tan
β
=
b
a
{\displaystyle \tan \beta ={\frac {b}{a}}}
Im Folgenden bedeutet
s
{\displaystyle s}
immer die Hälfte des Umfangs des Dreiecks
A
B
C
{\displaystyle ABC}
, also
s
=
a
+
b
+
c
2
{\displaystyle s={\frac {a+b+c}{2}}}
.
s
−
a
=
b
+
c
−
a
2
{\displaystyle s-a={\frac {b+c-a}{2}}}
s
−
b
=
c
+
a
−
b
2
{\displaystyle s-b={\frac {c+a-b}{2}}}
s
−
c
=
a
+
b
−
c
2
{\displaystyle s-c={\frac {a+b-c}{2}}}
(
s
−
b
)
+
(
s
−
c
)
=
a
{\displaystyle \left(s-b\right)+\left(s-c\right)=a}
(
s
−
c
)
+
(
s
−
a
)
=
b
{\displaystyle \left(s-c\right)+\left(s-a\right)=b}
(
s
−
a
)
+
(
s
−
b
)
=
c
{\displaystyle \left(s-a\right)+\left(s-b\right)=c}
(
s
−
a
)
+
(
s
−
b
)
+
(
s
−
c
)
=
s
{\displaystyle \left(s-a\right)+\left(s-b\right)+\left(s-c\right)=s}
sin
α
2
=
(
s
−
b
)
(
s
−
c
)
b
c
{\displaystyle \sin {\frac {\alpha }{2}}={\sqrt {\frac {\left(s-b\right)\left(s-c\right)}{bc}}}}
sin
β
2
=
(
s
−
c
)
(
s
−
a
)
c
a
{\displaystyle \sin {\frac {\beta }{2}}={\sqrt {\frac {\left(s-c\right)\left(s-a\right)}{ca}}}}
sin
γ
2
=
(
s
−
a
)
(
s
−
b
)
a
b
{\displaystyle \sin {\frac {\gamma }{2}}={\sqrt {\frac {\left(s-a\right)\left(s-b\right)}{ab}}}}
cos
α
2
=
s
(
s
−
a
)
b
c
{\displaystyle \cos {\frac {\alpha }{2}}={\sqrt {\frac {s\left(s-a\right)}{bc}}}}
cos
β
2
=
s
(
s
−
b
)
c
a
{\displaystyle \cos {\frac {\beta }{2}}={\sqrt {\frac {s\left(s-b\right)}{ca}}}}
cos
γ
2
=
s
(
s
−
c
)
a
b
{\displaystyle \cos {\frac {\gamma }{2}}={\sqrt {\frac {s\left(s-c\right)}{ab}}}}
tan
α
2
=
(
s
−
b
)
(
s
−
c
)
s
(
s
−
a
)
{\displaystyle \tan {\frac {\alpha }{2}}={\sqrt {\frac {\left(s-b\right)\left(s-c\right)}{s\left(s-a\right)}}}}
tan
β
2
=
(
s
−
c
)
(
s
−
a
)
s
(
s
−
b
)
{\displaystyle \tan {\frac {\beta }{2}}={\sqrt {\frac {\left(s-c\right)\left(s-a\right)}{s\left(s-b\right)}}}}
tan
γ
2
=
(
s
−
a
)
(
s
−
b
)
s
(
s
−
c
)
{\displaystyle \tan {\frac {\gamma }{2}}={\sqrt {\frac {\left(s-a\right)\left(s-b\right)}{s\left(s-c\right)}}}}
s
=
4
r
cos
α
2
cos
β
2
cos
γ
2
{\displaystyle s=4r\cos {\frac {\alpha }{2}}\cos {\frac {\beta }{2}}\cos {\frac {\gamma }{2}}}
s
−
a
=
4
r
cos
α
2
sin
β
2
sin
γ
2
{\displaystyle s-a=4r\cos {\frac {\alpha }{2}}\sin {\frac {\beta }{2}}\sin {\frac {\gamma }{2}}}
Der Flächeninhalt des Dreiecks wird hier mit
F
{\displaystyle F}
bezeichnet (nicht, wie heute üblich, mit
A
{\displaystyle A}
, um eine Verwechselung mit der Dreiecksecke
A
{\displaystyle A}
auszuschließen):
Heronsche Formel:
F
=
s
(
s
−
a
)
(
s
−
b
)
(
s
−
c
)
=
1
4
(
a
+
b
+
c
)
(
b
+
c
−
a
)
(
c
+
a
−
b
)
(
a
+
b
−
c
)
{\displaystyle F={\sqrt {s\left(s-a\right)\left(s-b\right)\left(s-c\right)}}={\frac {1}{4}}{\sqrt {\left(a+b+c\right)\left(b+c-a\right)\left(c+a-b\right)\left(a+b-c\right)}}}
F
=
1
4
2
(
b
2
c
2
+
c
2
a
2
+
a
2
b
2
)
−
(
a
4
+
b
4
+
c
4
)
{\displaystyle F={\frac {1}{4}}{\sqrt {2\left(b^{2}c^{2}+c^{2}a^{2}+a^{2}b^{2}\right)-\left(a^{4}+b^{4}+c^{4}\right)}}}
Weitere Flächenformeln:
F
=
1
2
b
c
sin
α
=
1
2
c
a
sin
β
=
1
2
a
b
sin
γ
{\displaystyle F={\frac {1}{2}}bc\sin \alpha ={\frac {1}{2}}ca\sin \beta ={\frac {1}{2}}ab\sin \gamma }
F
=
1
2
a
h
a
=
1
2
b
h
b
=
1
2
c
h
c
{\displaystyle F={\frac {1}{2}}ah_{a}={\frac {1}{2}}bh_{b}={\frac {1}{2}}ch_{c}}
, wobei
h
a
{\displaystyle h_{a}}
,
h
b
{\displaystyle h_{b}}
und
h
c
{\displaystyle h_{c}}
die Längen der von
A
{\displaystyle A}
,
B
{\displaystyle B}
bzw.
C
{\displaystyle C}
ausgehenden Höhen des Dreiecks
A
B
C
{\displaystyle ABC}
sind.
F
=
2
r
2
sin
α
sin
β
sin
γ
{\displaystyle F=2r^{2}\sin \,\alpha \,\sin \,\beta \,\sin \,\gamma }
F
=
a
b
c
4
r
{\displaystyle F={\frac {abc}{4r}}}
F
=
ρ
s
=
ρ
a
(
s
−
a
)
=
ρ
b
(
s
−
b
)
=
ρ
c
(
s
−
c
)
{\displaystyle F=\rho s=\rho _{a}\left(s-a\right)=\rho _{b}\left(s-b\right)=\rho _{c}\left(s-c\right)}
F
=
ρ
ρ
a
ρ
b
ρ
c
{\displaystyle F={\sqrt {\rho \rho _{a}\rho _{b}\rho _{c}}}}
F
=
4
ρ
r
cos
α
2
cos
β
2
cos
γ
2
{\displaystyle F=4\rho r\cos \,{\frac {\alpha }{2}}\,\cos \,{\frac {\beta }{2}}\,\cos \,{\frac {\gamma }{2}}}
F
=
s
2
tan
α
2
tan
β
2
tan
γ
2
{\displaystyle F=s^{2}\tan \,{\frac {\alpha }{2}}\,\tan \,{\frac {\beta }{2}}\,\tan \,{\frac {\gamma }{2}}}
F
=
ρ
2
h
a
h
b
h
c
(
h
a
−
2
ρ
)
(
h
b
−
2
ρ
)
(
h
c
−
2
ρ
)
{\displaystyle F=\rho ^{2}{\sqrt {\dfrac {h_{a}\,h_{b}\,h_{c}}{(h_{a}-2\rho )(h_{b}-2\rho )(h_{c}-2\rho )}}}}
, mit
1
ρ
=
1
h
a
+
1
h
b
+
1
h
c
{\displaystyle {\dfrac {1}{\rho }}={\dfrac {1}{h_{a}}}+{\dfrac {1}{h_{b}}}+{\dfrac {1}{h_{c}}}}
F
=
r
h
a
h
b
h
c
2
{\displaystyle F={\sqrt {\dfrac {r\,h_{a}\,h_{b}\,h_{c}}{2}}}}
F
=
h
a
h
b
h
c
2
ρ
(
sin
α
+
sin
β
+
sin
γ
)
{\displaystyle F={\dfrac {\,h_{a}\,h_{b}\,h_{c}}{2\rho \,{(\sin \alpha +\sin \beta +\sin \gamma )}}}}
Erweiterter Sinussatz:
a
sin
α
=
b
sin
β
=
c
sin
γ
=
2
r
=
a
b
c
2
F
{\displaystyle {\frac {a}{\sin \alpha }}={\frac {b}{\sin \beta }}={\frac {c}{\sin \gamma }}=2r={\frac {abc}{2F}}}
a
=
2
r
sin
α
{\displaystyle a=2r\,\sin \alpha }
b
=
2
r
sin
β
{\displaystyle b=2r\,\sin \beta }
c
=
2
r
sin
γ
{\displaystyle c=2r\,\sin \gamma }
r
=
a
b
c
4
F
{\displaystyle r={\frac {abc}{4F}}}
In diesem Abschnitt werden Formeln aufgelistet, in denen der Inkreisradius
ρ
{\displaystyle \rho }
und die Ankreisradien
ρ
a
{\displaystyle \rho _{a}}
,
ρ
b
{\displaystyle \rho _{b}}
und
ρ
c
{\displaystyle \rho _{c}}
des Dreiecks
A
B
C
{\displaystyle ABC}
vorkommen.
ρ
=
(
s
−
a
)
tan
α
2
=
(
s
−
b
)
tan
β
2
=
(
s
−
c
)
tan
γ
2
{\displaystyle \rho =\left(s-a\right)\tan {\frac {\alpha }{2}}=\left(s-b\right)\tan {\frac {\beta }{2}}=\left(s-c\right)\tan {\frac {\gamma }{2}}}
ρ
=
4
r
sin
α
2
sin
β
2
sin
γ
2
=
s
tan
α
2
tan
β
2
tan
γ
2
{\displaystyle \rho =4r\sin {\frac {\alpha }{2}}\sin {\frac {\beta }{2}}\sin {\frac {\gamma }{2}}=s\tan {\frac {\alpha }{2}}\tan {\frac {\beta }{2}}\tan {\frac {\gamma }{2}}}
ρ
=
r
(
cos
α
+
cos
β
+
cos
γ
−
1
)
{\displaystyle \rho =r\left(\cos \alpha +\cos \beta +\cos \gamma -1\right)}
ρ
=
F
s
=
a
b
c
4
r
s
{\displaystyle \rho ={\frac {F}{s}}={\frac {abc}{4rs}}}
ρ
=
(
s
−
a
)
(
s
−
b
)
(
s
−
c
)
s
=
1
2
(
b
+
c
−
a
)
(
c
+
a
−
b
)
(
a
+
b
−
c
)
a
+
b
+
c
{\displaystyle \rho ={\sqrt {\frac {\left(s-a\right)\left(s-b\right)\left(s-c\right)}{s}}}={\frac {1}{2}}{\sqrt {\frac {\left(b+c-a\right)\left(c+a-b\right)\left(a+b-c\right)}{a+b+c}}}}
ρ
=
a
cot
β
2
+
cot
γ
2
=
b
cot
γ
2
+
cot
α
2
=
c
cot
α
2
+
cot
β
2
{\displaystyle \rho ={\frac {a}{\cot {\frac {\beta }{2}}+\cot {\frac {\gamma }{2}}}}={\frac {b}{\cot {\frac {\gamma }{2}}+\cot {\frac {\alpha }{2}}}}={\frac {c}{\cot {\frac {\alpha }{2}}+\cot {\frac {\beta }{2}}}}}
a
⋅
b
+
b
⋅
c
+
c
⋅
a
=
s
2
+
ρ
2
+
4
⋅
ρ
⋅
r
{\displaystyle a\cdot b+b\cdot c+c\cdot a=s^{2}+\rho ^{2}+4\cdot \rho \cdot r}
[ 1]
Wichtige Ungleichung:
2
ρ
≤
r
{\displaystyle 2\rho \leq r}
; Gleichheit tritt nur dann ein, wenn Dreieck
A
B
C
{\displaystyle ABC}
gleichseitig ist.
ρ
a
=
s
tan
α
2
=
(
s
−
b
)
cot
γ
2
=
(
s
−
c
)
cot
β
2
{\displaystyle \rho _{a}=s\tan {\frac {\alpha }{2}}=\left(s-b\right)\cot {\frac {\gamma }{2}}=\left(s-c\right)\cot {\frac {\beta }{2}}}
ρ
a
=
4
r
sin
α
2
cos
β
2
cos
γ
2
=
(
s
−
a
)
tan
α
2
cot
β
2
cot
γ
2
{\displaystyle \rho _{a}=4r\sin {\frac {\alpha }{2}}\cos {\frac {\beta }{2}}\cos {\frac {\gamma }{2}}=\left(s-a\right)\tan {\frac {\alpha }{2}}\cot {\frac {\beta }{2}}\cot {\frac {\gamma }{2}}}
ρ
a
=
r
(
−
cos
α
+
cos
β
+
cos
γ
+
1
)
{\displaystyle \rho _{a}=r\left(-\cos \alpha +\cos \beta +\cos \gamma +1\right)}
ρ
a
=
F
s
−
a
=
a
b
c
4
r
(
s
−
a
)
{\displaystyle \rho _{a}={\frac {F}{s-a}}={\frac {abc}{4r\left(s-a\right)}}}
ρ
a
=
s
(
s
−
b
)
(
s
−
c
)
s
−
a
=
1
2
(
a
+
b
+
c
)
(
c
+
a
−
b
)
(
a
+
b
−
c
)
b
+
c
−
a
{\displaystyle \rho _{a}={\sqrt {\frac {s\left(s-b\right)\left(s-c\right)}{s-a}}}={\frac {1}{2}}{\sqrt {\frac {\left(a+b+c\right)\left(c+a-b\right)\left(a+b-c\right)}{b+c-a}}}}
Die Ankreise sind gleichberechtigt: Jede Formel für
ρ
a
{\displaystyle \rho _{a}}
gilt in analoger Form für
ρ
b
{\displaystyle \rho _{b}}
und
ρ
c
{\displaystyle \rho _{c}}
.
1
ρ
=
1
ρ
a
+
1
ρ
b
+
1
ρ
c
{\displaystyle {\frac {1}{\rho }}={\frac {1}{\rho _{a}}}+{\frac {1}{\rho _{b}}}+{\frac {1}{\rho _{c}}}}
Die Längen der von
A
{\displaystyle A}
,
B
{\displaystyle B}
bzw.
C
{\displaystyle C}
ausgehenden Höhen des Dreiecks
A
B
C
{\displaystyle ABC}
werden mit
h
a
{\displaystyle h_{a}}
,
h
b
{\displaystyle h_{b}}
und
h
c
{\displaystyle h_{c}}
bezeichnet.
h
a
=
b
sin
γ
=
c
sin
β
=
2
F
a
=
2
r
sin
β
sin
γ
=
2
r
(
cos
α
+
cos
β
cos
γ
)
{\displaystyle h_{a}=b\sin \gamma =c\sin \beta ={\frac {2F}{a}}=2r\sin \beta \sin \gamma =2r\left(\cos \alpha +\cos \beta \cos \gamma \right)}
h
b
=
c
sin
α
=
a
sin
γ
=
2
F
b
=
2
r
sin
γ
sin
α
=
2
r
(
cos
β
+
cos
α
cos
γ
)
{\displaystyle h_{b}=c\sin \alpha =a\sin \gamma ={\frac {2F}{b}}=2r\sin \gamma \sin \alpha =2r\left(\cos \beta +\cos \alpha \cos \gamma \right)}
h
c
=
a
sin
β
=
b
sin
α
=
2
F
c
=
2
r
sin
α
sin
β
=
2
r
(
cos
γ
+
cos
α
cos
β
)
{\displaystyle h_{c}=a\sin \beta =b\sin \alpha ={\frac {2F}{c}}=2r\sin \alpha \sin \beta =2r\left(\cos \gamma +\cos \alpha \cos \beta \right)}
h
a
=
a
cot
β
+
cot
γ
;
h
b
=
b
cot
γ
+
cot
α
;
h
c
=
c
cot
α
+
cot
β
{\displaystyle h_{a}={\frac {a}{\cot \beta +\cot \gamma }};\;\;\;\;\;h_{b}={\frac {b}{\cot \gamma +\cot \alpha }};\;\;\;\;\;h_{c}={\frac {c}{\cot \alpha +\cot \beta }}}
F
=
1
2
a
h
a
=
1
2
b
h
b
=
1
2
c
h
c
{\displaystyle F={\frac {1}{2}}ah_{a}={\frac {1}{2}}bh_{b}={\frac {1}{2}}ch_{c}}
1
h
a
+
1
h
b
+
1
h
c
=
1
ρ
=
1
ρ
a
+
1
ρ
b
+
1
ρ
c
{\displaystyle {\frac {1}{h_{a}}}+{\frac {1}{h_{b}}}+{\frac {1}{h_{c}}}={\frac {1}{\rho }}={\frac {1}{\rho _{a}}}+{\frac {1}{\rho _{b}}}+{\frac {1}{\rho _{c}}}}
Hat das Dreieck
A
B
C
{\displaystyle ABC}
einen rechten Winkel bei
C
{\displaystyle C}
(ist also
γ
=
90
∘
{\displaystyle \gamma =90^{\circ }}
), dann gilt
h
c
=
a
b
c
{\displaystyle h_{c}={\frac {ab}{c}}}
h
a
=
b
{\displaystyle h_{a}=b}
h
b
=
a
{\displaystyle h_{b}=a}
Die Längen der von
A
{\displaystyle A}
,
B
{\displaystyle B}
bzw.
C
{\displaystyle C}
ausgehenden Seitenhalbierenden des Dreiecks
A
B
C
{\displaystyle ABC}
werden
s
a
{\displaystyle s_{a}}
,
s
b
{\displaystyle s_{b}}
und
s
c
{\displaystyle s_{c}}
genannt.
s
a
=
1
2
2
b
2
+
2
c
2
−
a
2
=
1
2
b
2
+
c
2
+
2
b
c
cos
α
=
a
2
4
+
b
c
cos
α
{\displaystyle s_{a}={\frac {1}{2}}{\sqrt {2b^{2}+2c^{2}-a^{2}}}={\frac {1}{2}}{\sqrt {b^{2}+c^{2}+2bc\cos \alpha }}={\sqrt {{\frac {a^{2}}{4}}+bc\cos \alpha }}}
s
b
=
1
2
2
c
2
+
2
a
2
−
b
2
=
1
2
c
2
+
a
2
+
2
c
a
cos
β
=
b
2
4
+
c
a
cos
β
{\displaystyle s_{b}={\frac {1}{2}}{\sqrt {2c^{2}+2a^{2}-b^{2}}}={\frac {1}{2}}{\sqrt {c^{2}+a^{2}+2ca\cos \beta }}={\sqrt {{\frac {b^{2}}{4}}+ca\cos \beta }}}
s
c
=
1
2
2
a
2
+
2
b
2
−
c
2
=
1
2
a
2
+
b
2
+
2
a
b
cos
γ
=
c
2
4
+
a
b
cos
γ
{\displaystyle s_{c}={\frac {1}{2}}{\sqrt {2a^{2}+2b^{2}-c^{2}}}={\frac {1}{2}}{\sqrt {a^{2}+b^{2}+2ab\cos \gamma }}={\sqrt {{\frac {c^{2}}{4}}+ab\cos \gamma }}}
s
a
2
+
s
b
2
+
s
c
2
=
3
4
(
a
2
+
b
2
+
c
2
)
{\displaystyle s_{a}^{2}+s_{b}^{2}+s_{c}^{2}={\frac {3}{4}}\left(a^{2}+b^{2}+c^{2}\right)}
Wir bezeichnen mit
w
α
{\displaystyle w_{\alpha }}
,
w
β
{\displaystyle w_{\beta }}
und
w
γ
{\displaystyle w_{\gamma }}
die Längen der von
A
{\displaystyle A}
,
B
{\displaystyle B}
bzw.
C
{\displaystyle C}
ausgehenden Winkelhalbierenden im Dreieck
A
B
C
{\displaystyle ABC}
.
w
α
=
2
b
c
cos
α
2
b
+
c
=
2
F
a
cos
β
−
γ
2
=
b
c
(
b
+
c
−
a
)
(
a
+
b
+
c
)
b
+
c
{\displaystyle w_{\alpha }={\frac {2bc\cos {\frac {\alpha }{2}}}{b+c}}={\frac {2F}{a\cos {\frac {\beta -\gamma }{2}}}}={\frac {\sqrt {bc(b+c-a)(a+b+c)}}{b+c}}}
w
β
=
2
c
a
cos
β
2
c
+
a
=
2
F
b
cos
γ
−
α
2
=
c
a
(
c
+
a
−
b
)
(
a
+
b
+
c
)
c
+
a
{\displaystyle w_{\beta }={\frac {2ca\cos {\frac {\beta }{2}}}{c+a}}={\frac {2F}{b\cos {\frac {\gamma -\alpha }{2}}}}={\frac {\sqrt {ca(c+a-b)(a+b+c)}}{c+a}}}
w
γ
=
2
a
b
cos
γ
2
a
+
b
=
2
F
c
cos
α
−
β
2
=
a
b
(
a
+
b
−
c
)
(
a
+
b
+
c
)
a
+
b
{\displaystyle w_{\gamma }={\frac {2ab\cos {\frac {\gamma }{2}}}{a+b}}={\frac {2F}{c\cos {\frac {\alpha -\beta }{2}}}}={\frac {\sqrt {ab(a+b-c)(a+b+c)}}{a+b}}}
Die trigonometrischen Funktionen am Einheitskreis :
C
P
¯
=
sin
b
{\displaystyle {\overline {CP}}=\sin b}
S
P
¯
=
cos
b
{\displaystyle {\overline {SP}}=\cos b}
D
T
¯
=
tan
b
{\displaystyle {\overline {DT}}=\tan b}
E
K
¯
=
cot
b
{\displaystyle {\overline {EK}}=\cot b}
O
T
¯
=
sec
b
{\displaystyle {\overline {OT}}=\operatorname {sec} \,b}
O
K
¯
=
csc
b
{\displaystyle {\overline {OK}}=\operatorname {csc} \,b}
sin
x
=
sin
(
x
+
2
n
π
)
;
n
∈
Z
{\displaystyle \sin x\quad =\quad \sin(x+2n\pi );\quad n\in \mathbb {Z} }
cos
x
=
cos
(
x
+
2
n
π
)
;
n
∈
Z
{\displaystyle \cos x\quad =\quad \cos(x+2n\pi );\quad n\in \mathbb {Z} }
tan
x
=
tan
(
x
+
n
π
)
;
n
∈
Z
{\displaystyle \tan x\quad =\quad \tan(x+n\pi );\quad n\in \mathbb {Z} }
cot
x
=
cot
(
x
+
n
π
)
;
n
∈
Z
{\displaystyle \cot x\quad =\quad \cot(x+n\pi );\quad n\in \mathbb {Z} }
Die trigonometrischen Funktionen lassen sich ineinander umwandeln oder gegenseitig darstellen. Es gelten folgende Zusammenhänge:
tan
x
=
sin
x
cos
x
{\displaystyle \tan x={\frac {\sin x}{\cos x}}}
sin
2
x
+
cos
2
x
=
1
{\displaystyle \sin ^{2}x+\cos ^{2}x=1}
(„Trigonometrischer Pythagoras “)
1
+
tan
2
x
=
1
cos
2
x
=
sec
2
x
{\displaystyle 1+\tan ^{2}x={\frac {1}{\cos ^{2}x}}=\sec ^{2}x}
1
+
cot
2
x
=
1
sin
2
x
=
csc
2
x
{\displaystyle 1+\cot ^{2}x={\frac {1}{\sin ^{2}x}}=\csc ^{2}x}
(Siehe auch den Abschnitt Phasenverschiebungen .)
Mittels dieser Gleichungen lassen sich die drei vorkommenden Funktionen durch eine der beiden anderen darstellen:
sin
x
=
1
−
cos
2
x
{\displaystyle \sin x\;=\;{\sqrt {1-\cos ^{2}x}}}
für
x
∈
[
0
,
π
[
=
[
0
∘
,
180
∘
[
{\displaystyle x\in \left[0,\pi \right[\quad =\quad [0^{\circ },180^{\circ }[}
sin
x
=
−
1
−
cos
2
x
{\displaystyle \sin x\;=\;-{\sqrt {1-\cos ^{2}x}}}
für
x
∈
[
π
,
2
π
[
=
[
180
∘
,
360
∘
[
{\displaystyle x\in \left[\pi ,2\pi \right[\quad =\quad [180^{\circ },360^{\circ }[}
sin
x
=
tan
x
1
+
tan
2
x
{\displaystyle \sin x\;=\;{\frac {\tan x}{\sqrt {1+\tan ^{2}x}}}}
für
x
∈
[
0
,
π
2
[
∪
]
3
π
2
,
2
π
[
=
[
0
∘
,
90
∘
[
∪
]
270
∘
,
360
∘
[
{\displaystyle x\in \left[0,{\frac {\pi }{2}}\right[\;\cup \;\left]{\frac {3\pi }{2}},2\pi \right[\quad =\quad [0^{\circ },90^{\circ }[\;\cup \;]270^{\circ },360^{\circ }[}
sin
x
=
−
tan
x
1
+
tan
2
x
{\displaystyle \sin x\;=\;-{\frac {\tan x}{\sqrt {1+\tan ^{2}x}}}}
für
x
∈
]
π
2
,
3
π
2
[
=
]
90
∘
,
270
∘
[
{\displaystyle x\in \left]{\frac {\pi }{2}},{\frac {3\pi }{2}}\right[\quad =\quad ]90^{\circ },270^{\circ }[}
cos
x
=
1
−
sin
2
x
{\displaystyle \cos x\;=\;{\sqrt {1-\sin ^{2}x}}}
für
x
∈
[
0
,
π
2
[
∪
[
3
π
2
,
2
π
[
=
[
0
∘
,
90
∘
[
∪
[
270
∘
,
360
∘
[
{\displaystyle x\in \left[0,{\frac {\pi }{2}}\right[\;\cup \;\left[{\frac {3\pi }{2}},2\pi \right[\quad =\quad [0^{\circ },90^{\circ }[\;\cup \;[270^{\circ },360^{\circ }[}
cos
x
=
−
1
−
sin
2
x
{\displaystyle \cos x\;=\;-{\sqrt {1-\sin ^{2}x}}}
für
x
∈
[
π
2
,
3
π
2
[
=
[
90
∘
,
270
∘
[
{\displaystyle x\in \left[{\frac {\pi }{2}},{\frac {3\pi }{2}}\right[\quad =\quad [90^{\circ },270^{\circ }[}
cos
x
=
1
1
+
tan
2
x
{\displaystyle \cos x={\frac {1}{\sqrt {1+\tan ^{2}x}}}}
für
x
∈
[
0
,
π
2
[
∪
]
3
π
2
,
2
π
[
=
[
0
∘
,
90
∘
[
∪
]
270
∘
,
360
∘
[
{\displaystyle x\in \left[0,{\frac {\pi }{2}}\right[\;\cup \;\left]{\frac {3\pi }{2}},2\pi \right[\quad =\quad [0^{\circ },90^{\circ }[\;\cup \;]270^{\circ },360^{\circ }[}
cos
x
=
−
1
1
+
tan
2
x
{\displaystyle \cos x=-{\frac {1}{\sqrt {1+\tan ^{2}x}}}}
für
x
∈
]
π
2
,
3
π
2
[
=
]
90
∘
,
270
∘
[
{\displaystyle x\in \left]{\frac {\pi }{2}},{\frac {3\pi }{2}}\right[\quad =\quad ]90^{\circ },270^{\circ }[}
tan
x
=
1
−
cos
2
x
cos
x
{\displaystyle \tan x={\frac {\sqrt {1-\cos ^{2}x}}{\cos x}}}
für
x
∈
[
0
,
π
2
[
∪
]
π
2
,
π
[
=
[
0
∘
,
90
∘
[
∪
]
90
∘
,
180
∘
[
{\displaystyle x\in \left[0,{\frac {\pi }{2}}\right[\;\cup \;\left]{\frac {\pi }{2}},\pi \right[\quad =\quad [0^{\circ },90^{\circ }[\;\cup \;]90^{\circ },180^{\circ }[}
tan
x
=
−
1
−
cos
2
x
cos
x
{\displaystyle \tan x=-{\frac {\sqrt {1-\cos ^{2}x}}{\cos x}}}
für
x
∈
[
π
,
3
π
2
[
∪
]
3
π
2
,
2
π
[
=
[
180
∘
,
270
∘
[
∪
]
270
∘
,
360
∘
[
{\displaystyle x\in \left[\pi ,{\frac {3\pi }{2}}\right[\;\cup \;\left]{\frac {3\pi }{2}},2\pi \right[\quad =\quad [180^{\circ },270^{\circ }[\;\cup \;]270^{\circ },360^{\circ }[}
tan
x
=
sin
x
1
−
sin
2
x
{\displaystyle \tan x={\frac {\sin x}{\sqrt {1-\sin ^{2}x}}}}
für
x
∈
[
0
,
π
2
[
∪
]
3
π
2
,
2
π
[
=
[
0
∘
,
90
∘
[
∪
]
270
∘
,
360
∘
[
{\displaystyle x\in \left[0,{\frac {\pi }{2}}\right[\;\cup \;\left]{\frac {3\pi }{2}},2\pi \right[\quad =\quad [0^{\circ },90^{\circ }[\;\cup \;]270^{\circ },360^{\circ }[}
tan
x
=
−
sin
x
1
−
sin
2
x
{\displaystyle \tan x=-{\frac {\sin x}{\sqrt {1-\sin ^{2}x}}}}
für
x
∈
]
π
2
,
3
π
2
[
=
]
90
∘
,
270
∘
[
{\displaystyle x\in \left]{\frac {\pi }{2}},{\frac {3\pi }{2}}\right[\quad =\quad ]90^{\circ },270^{\circ }[}
sin
x
>
0
für
x
∈
]
0
∘
,
180
∘
[
{\displaystyle \sin x>0\quad {\text{für}}\quad x\in \left]0^{\circ },180^{\circ }\right[}
sin
x
<
0
für
x
∈
]
180
∘
,
360
∘
[
{\displaystyle \sin x<0\quad {\text{für}}\quad x\in \left]180^{\circ },360^{\circ }\right[}
cos
x
>
0
für
x
∈
[
0
∘
,
90
∘
[
∪
]
270
∘
,
360
∘
]
{\displaystyle \cos x>0\quad {\text{für}}\quad x\in \left[0^{\circ },90^{\circ }\right[\cup \left]270^{\circ },360^{\circ }\right]}
cos
x
<
0
für
x
∈
]
90
∘
,
270
∘
[
{\displaystyle \cos x<0\quad {\text{für}}\quad x\in \left]90^{\circ },270^{\circ }\right[}
tan
x
>
0
für
x
∈
]
0
∘
,
90
∘
[
∪
]
180
∘
,
270
∘
[
{\displaystyle \tan x>0\quad {\text{für}}\quad x\in \left]0^{\circ },90^{\circ }\right[\cup \left]180^{\circ },270^{\circ }\right[}
tan
x
<
0
für
x
∈
]
90
∘
,
180
∘
[
∪
]
270
∘
,
360
∘
[
{\displaystyle \tan x<0\quad {\text{für}}\quad x\in \left]90^{\circ },180^{\circ }\right[\cup \left]270^{\circ },360^{\circ }\right[}
Die Vorzeichen von
cot
{\displaystyle \cot }
,
sec
{\displaystyle \sec }
und
csc
{\displaystyle \csc }
stimmen überein mit denen ihrer Kehrwertfunktionen
tan
{\displaystyle \tan }
,
cos
{\displaystyle \cos }
bzw.
sin
{\displaystyle \sin }
.
Darstellung wichtiger Funktionswerte von Kosinus (1. Klammerwert) und Sinus (2. Klammerwert) auf dem Einheitskreis
α
{\displaystyle \alpha }
α
{\displaystyle \alpha }
(rad)
sin
α
{\displaystyle \sin \alpha }
cos
α
{\displaystyle \cos \alpha }
tan
α
{\displaystyle \tan \alpha }
cot
α
{\displaystyle \cot \alpha }
0
∘
{\displaystyle 0^{\circ }}
0
{\displaystyle \,0}
0
{\displaystyle \,0}
1
{\displaystyle \,1}
0
{\displaystyle \,0}
±
∞
{\displaystyle \pm \infty }
15
∘
{\displaystyle 15^{\circ }}
π
12
{\displaystyle {\tfrac {\pi }{12}}}
1
4
(
6
−
2
)
{\displaystyle {\tfrac {1}{4}}({\sqrt {6}}-{\sqrt {2}})}
1
4
(
6
+
2
)
{\displaystyle {\tfrac {1}{4}}({\sqrt {6}}+{\sqrt {2}})}
2
−
3
{\displaystyle 2-{\sqrt {3}}}
2
+
3
{\displaystyle 2+{\sqrt {3}}}
18
∘
{\displaystyle 18^{\circ }}
π
10
{\displaystyle {\tfrac {\pi }{10}}}
1
4
(
5
−
1
)
{\displaystyle {\tfrac {1}{4}}\left({\sqrt {5}}-1\right)}
1
4
10
+
2
5
{\displaystyle {\tfrac {1}{4}}{\sqrt {10+2{\sqrt {5}}}}}
1
5
25
−
10
5
{\displaystyle {\tfrac {1}{5}}{\sqrt {25-10{\sqrt {5}}}}}
5
+
2
5
{\displaystyle {\sqrt {5+2{\sqrt {5}}}}}
30
∘
{\displaystyle 30^{\circ }}
π
6
{\displaystyle {\tfrac {\pi }{6}}}
1
2
{\displaystyle {\tfrac {1}{2}}}
1
2
3
{\displaystyle {\tfrac {1}{2}}{\sqrt {3}}}
1
3
3
{\displaystyle {\tfrac {1}{3}}{\sqrt {3}}}
3
{\displaystyle {\sqrt {3}}}
36
∘
{\displaystyle 36^{\circ }}
π
5
{\displaystyle {\tfrac {\pi }{5}}}
1
4
10
−
2
5
{\displaystyle {\tfrac {1}{4}}{\sqrt {10-2{\sqrt {5}}}}}
1
4
(
1
+
5
)
{\displaystyle {\tfrac {1}{4}}\left(1+{\sqrt {5}}\right)}
5
−
2
5
{\displaystyle {\sqrt {5-2{\sqrt {5}}}}}
1
5
25
+
10
5
{\displaystyle {\tfrac {1}{5}}{\sqrt {25+10{\sqrt {5}}}}}
45
∘
{\displaystyle 45^{\circ }}
π
4
{\displaystyle {\tfrac {\pi }{4}}}
1
2
2
{\displaystyle {\tfrac {1}{2}}{\sqrt {2}}}
1
2
2
{\displaystyle {\tfrac {1}{2}}{\sqrt {2}}}
1
{\displaystyle 1\,}
1
{\displaystyle 1\,}
54
∘
{\displaystyle 54^{\circ }}
3
π
10
{\displaystyle {\tfrac {3\pi }{10}}}
1
4
(
1
+
5
)
{\displaystyle {\tfrac {1}{4}}\left(1+{\sqrt {5}}\right)}
1
4
10
−
2
5
{\displaystyle {\tfrac {1}{4}}{\sqrt {10-2{\sqrt {5}}}}}
1
5
25
+
10
5
{\displaystyle {\tfrac {1}{5}}{\sqrt {25+10{\sqrt {5}}}}}
5
−
2
5
{\displaystyle {\sqrt {5-2{\sqrt {5}}}}}
60
∘
{\displaystyle 60^{\circ }}
π
3
{\displaystyle {\tfrac {\pi }{3}}}
1
2
3
{\displaystyle {\tfrac {1}{2}}{\sqrt {3}}}
1
2
{\displaystyle {\tfrac {1}{2}}}
3
{\displaystyle {\sqrt {3}}}
1
3
3
{\displaystyle {\tfrac {1}{3}}{\sqrt {3}}}
72
∘
{\displaystyle 72^{\circ }}
2
π
5
{\displaystyle {\tfrac {2\pi }{5}}}
1
4
10
+
2
5
{\displaystyle {\tfrac {1}{4}}{\sqrt {10+2{\sqrt {5}}}}}
1
4
(
5
−
1
)
{\displaystyle {\tfrac {1}{4}}\left({\sqrt {5}}-1\right)}
5
+
2
5
{\displaystyle {\sqrt {5+2{\sqrt {5}}}}}
1
5
25
−
10
5
{\displaystyle {\tfrac {1}{5}}{\sqrt {25-10{\sqrt {5}}}}}
75
∘
{\displaystyle 75^{\circ }}
5
π
12
{\displaystyle {\tfrac {5\pi }{12}}}
1
4
(
6
+
2
)
{\displaystyle {\tfrac {1}{4}}({\sqrt {6}}+{\sqrt {2}})}
1
4
(
6
−
2
)
{\displaystyle {\tfrac {1}{4}}({\sqrt {6}}-{\sqrt {2}})}
2
+
3
{\displaystyle 2+{\sqrt {3}}}
2
−
3
{\displaystyle 2-{\sqrt {3}}}
90
∘
{\displaystyle 90^{\circ }}
π
2
{\displaystyle {\tfrac {\pi }{2}}}
1
{\displaystyle \,1}
0
{\displaystyle \,0}
±
∞
{\displaystyle \pm \infty }
0
{\displaystyle \,0}
108
∘
{\displaystyle 108^{\circ }}
3
π
5
{\displaystyle {\tfrac {3\pi }{5}}}
1
4
10
+
2
5
{\displaystyle {\tfrac {1}{4}}{\sqrt {10+2{\sqrt {5}}}}}
1
4
(
1
−
5
)
{\displaystyle {\tfrac {1}{4}}\left(1-{\sqrt {5}}\right)}
−
5
+
2
5
{\displaystyle -{\sqrt {5+2{\sqrt {5}}}}}
−
1
5
25
−
10
5
{\displaystyle -{\tfrac {1}{5}}{\sqrt {25-10{\sqrt {5}}}}}
120
∘
{\displaystyle 120^{\circ }}
2
π
3
{\displaystyle {\tfrac {2\pi }{3}}}
1
2
3
{\displaystyle {\tfrac {1}{2}}{\sqrt {3}}}
−
1
2
{\displaystyle -{\tfrac {1}{2}}}
−
3
{\displaystyle -{\sqrt {3}}}
−
1
3
3
{\displaystyle -{\tfrac {1}{3}}{\sqrt {3}}}
135
∘
{\displaystyle 135^{\circ }}
3
π
4
{\displaystyle {\tfrac {3\pi }{4}}}
1
2
2
{\displaystyle {\tfrac {1}{2}}{\sqrt {2}}}
−
1
2
2
{\displaystyle -{\tfrac {1}{2}}{\sqrt {2}}}
−
1
{\displaystyle -1\,}
−
1
{\displaystyle -1\,}
180
∘
{\displaystyle 180^{\circ }}
π
{\displaystyle \pi \,}
0
{\displaystyle \,0}
−
1
{\displaystyle \,-1}
0
{\displaystyle \,0}
±
∞
{\displaystyle \pm \infty }
270
∘
{\displaystyle 270^{\circ }}
3
π
2
{\displaystyle {\tfrac {3\pi }{2}}}
−
1
{\displaystyle \,-1}
0
{\displaystyle \,0}
±
∞
{\displaystyle \pm \infty }
0
{\displaystyle \,0}
360
∘
{\displaystyle 360^{\circ }}
2
π
{\displaystyle 2\pi }
0
{\displaystyle \,0}
1
{\displaystyle \,1}
0
{\displaystyle \,0}
±
∞
{\displaystyle \pm \infty }
Mit Hilfe der Additionstheoreme sind noch viele weitere Werte durch algebraische Ausdrücke (ggfs. mit verschachtelten Quadratwurzeln) darstellbar, insbesondere alle ganzzahligen Vielfachen von
3
∘
{\displaystyle 3^{\circ }}
.[ 2]
Die trigonometrischen Funktionen haben einfache Symmetrien:
sin
(
−
x
)
=
−
sin
(
x
)
cos
(
−
x
)
=
+
cos
(
x
)
tan
(
−
x
)
=
−
tan
(
x
)
cot
(
−
x
)
=
−
cot
(
x
)
sec
(
−
x
)
=
+
sec
(
x
)
csc
(
−
x
)
=
−
csc
(
x
)
{\displaystyle {\begin{aligned}\sin(-x)&=-\sin(x)\\\cos(-x)&=+\cos(x)\\\tan(-x)&=-\tan(x)\\\cot(-x)&=-\cot(x)\\\sec(-x)&=+\sec(x)\\\csc(-x)&=-\csc(x)\\\end{aligned}}}
sin
(
x
+
π
2
)
=
cos
x
bzw.
sin
(
x
+
90
∘
)
=
cos
x
{\displaystyle \sin \left(x+{\frac {\pi }{2}}\right)=\cos x\;\quad {\text{bzw.}}\quad \sin \left(x+90^{\circ }\right)=\cos x\;}
cos
(
x
+
π
2
)
=
−
sin
x
bzw.
cos
(
x
+
90
∘
)
=
−
sin
x
{\displaystyle \cos \left(x+{\frac {\pi }{2}}\right)=-\sin x\;\quad {\text{bzw.}}\quad \cos \left(x+90^{\circ }\right)=-\sin x\;}
tan
(
x
+
π
2
)
=
−
cot
x
bzw.
tan
(
x
+
90
∘
)
=
−
cot
x
{\displaystyle \tan \left(x+{\frac {\pi }{2}}\right)=-\cot x\;\quad {\text{bzw.}}\quad \tan \left(x+90^{\circ }\right)=-\cot x\;}
cot
(
x
+
π
2
)
=
−
tan
x
bzw.
cot
(
x
+
90
∘
)
=
−
tan
x
{\displaystyle \cot \left(x+{\frac {\pi }{2}}\right)=-\tan x\;\quad {\text{bzw.}}\quad \cot \left(x+90^{\circ }\right)=-\tan x\;}
sin
x
=
sin
(
π
−
x
)
bzw.
sin
x
=
sin
(
180
∘
−
x
)
{\displaystyle \sin x\ \;=\;\;\;\sin \left(\pi -x\right)\,\quad {\text{bzw.}}\quad \sin x\ =\;\;\;\sin \left(180^{\circ }-x\right)}
cos
x
=
−
cos
(
π
−
x
)
bzw.
cos
x
=
−
cos
(
180
∘
−
x
)
{\displaystyle \cos x\ \,=-\cos \left(\pi -x\right)\quad {\text{bzw.}}\quad \cos x\ =-\cos \left(180^{\circ }-x\right)}
tan
x
=
−
tan
(
π
−
x
)
bzw.
tan
x
=
−
tan
(
180
∘
−
x
)
{\displaystyle \tan x\ =-\tan \left(\pi -x\right)\quad {\text{bzw.}}\quad \tan x\ =-\tan \left(180^{\circ }-x\right)}
Darstellung durch den Tangens des halben Winkels
Bearbeiten
Mit der Bezeichnung
t
=
tan
x
2
{\displaystyle t=\tan {\tfrac {x}{2}}}
gelten die folgenden Beziehungen für beliebiges
x
{\displaystyle x}
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}}},}
tan
x
=
2
t
1
−
t
2
,
{\displaystyle \tan x={\frac {2t}{1-t^{2}}},}
cot
x
=
1
−
t
2
2
t
,
{\displaystyle \cot x={\frac {1-t^{2}}{2t}},}
sec
x
=
1
+
t
2
1
−
t
2
,
{\displaystyle \sec x={\frac {1+t^{2}}{1-t^{2}}},}
csc
x
=
1
+
t
2
2
t
.
{\displaystyle \csc x={\frac {1+t^{2}}{2t}}.}
Figur 1
Figur 2
Für Sinus und Kosinus lassen sich die Additionstheoreme aus der Verkettung zweier Drehungen um den Winkel
x
{\displaystyle x}
bzw.
y
{\displaystyle y}
herleiten. Das ist elementargeometrisch möglich; sehr viel einfacher ist das koordinatenweise Ablesen der Formeln aus dem Produkt zweier Drehmatrizen der Ebene
R
2
{\displaystyle \mathbb {R} ^{2}}
. Alternativ folgen die Additionstheoreme aus der Anwendung der Eulerschen Formel auf die Beziehung
e
i
(
x
+
y
)
=
e
i
x
⋅
e
i
y
{\displaystyle \textstyle e^{i(x+y)}=e^{ix}\cdot e^{iy}}
. Die Ergebnisse für das Doppelvorzeichen ergeben sich durch Anwendung der Symmetrien .[ 3]
sin
(
x
±
y
)
=
sin
x
⋅
cos
y
±
cos
x
⋅
sin
y
{\displaystyle \sin(x\pm y)=\sin x\cdot \cos y\pm \cos x\cdot \sin y}
[ 4]
cos
(
x
±
y
)
=
cos
x
⋅
cos
y
∓
sin
x
⋅
sin
y
{\displaystyle \cos(x\pm y)=\cos x\cdot \cos y\mp \sin x\cdot \sin y}
[ 4]
Geometrische Herleitungen sind in Figur 1 und Figur 2 für Winkel
α
{\displaystyle \alpha }
und
β
{\displaystyle \beta }
zwischen 0° und 90° veranschaulicht.[ 5]
Zu Figur 1:
sin
(
α
+
β
)
=
sin
α
⋅
cos
β
+
cos
α
⋅
sin
β
{\displaystyle \sin(\alpha +\beta )=\sin \alpha \cdot \cos \beta +\cos \alpha \cdot \sin \beta }
cos
(
α
+
β
)
=
cos
α
⋅
cos
β
−
sin
α
⋅
sin
β
{\displaystyle \cos(\alpha +\beta )=\cos \alpha \cdot \cos \beta -\sin \alpha \cdot \sin \beta }
Zu Figur 2:
sin
(
α
−
β
)
=
sin
α
⋅
cos
β
−
cos
α
⋅
sin
β
{\displaystyle \sin(\alpha -\beta )=\sin \alpha \cdot \cos \beta -\cos \alpha \cdot \sin \beta }
cos
(
α
−
β
)
=
cos
α
⋅
cos
β
+
sin
α
⋅
sin
β
{\displaystyle \cos(\alpha -\beta )=\cos \alpha \cdot \cos \beta +\sin \alpha \cdot \sin \beta }
Durch Erweiterung mit
1
cos
x
cos
y
{\displaystyle \textstyle {1 \over \cos x\cos y}}
bzw.
1
sin
x
sin
y
{\displaystyle \textstyle {1 \over \sin x\sin y}}
und Vereinfachung des Doppelbruchs:
tan
(
x
±
y
)
=
sin
(
x
±
y
)
cos
(
x
±
y
)
=
tan
x
±
tan
y
1
∓
tan
x
tan
y
{\displaystyle \tan(x\pm y)={\frac {\sin(x\pm y)}{\cos(x\pm y)}}={\frac {\tan x\pm \tan y}{1\mp \tan x\;\tan y}}}
cot
(
x
±
y
)
=
cos
(
x
±
y
)
sin
(
x
±
y
)
=
cot
x
cot
y
∓
1
cot
y
±
cot
x
{\displaystyle \cot(x\pm y)={\frac {\cos(x\pm y)}{\sin(x\pm y)}}={\frac {\cot x\cot y\mp 1}{\cot y\pm \cot x}}}
Für
x
=
y
{\displaystyle x=y}
folgen hieraus die Doppelwinkelfunktionen , für
y
=
π
/
2
{\displaystyle y=\pi /2}
die Phasenverschiebungen .
sin
(
x
+
y
)
⋅
sin
(
x
−
y
)
=
cos
2
y
−
cos
2
x
=
sin
2
x
−
sin
2
y
{\displaystyle \sin(x+y)\cdot \sin(x-y)=\cos ^{2}y-\cos ^{2}x=\sin ^{2}x-\sin ^{2}y}
cos
(
x
+
y
)
⋅
cos
(
x
−
y
)
=
cos
2
y
−
sin
2
x
=
cos
2
x
−
sin
2
y
{\displaystyle \cos(x+y)\cdot \cos(x-y)=\cos ^{2}y-\sin ^{2}x=\cos ^{2}x-\sin ^{2}y}
Für die Arkusfunktionen gelten folgende Additionstheoreme[ 6]
Summanden
Summenformel
Gültigkeitsbereich
arcsin
x
+
arcsin
y
=
{\displaystyle \arcsin x+\arcsin y=}
arcsin
(
x
1
−
y
2
+
y
1
−
x
2
)
{\displaystyle \arcsin \left(x{\sqrt {1-y^{2}}}+y{\sqrt {1-x^{2}}}\right)}
x
y
≤
0
{\displaystyle xy\leq 0}
oder
x
2
+
y
2
≤
1
{\displaystyle x^{2}+y^{2}\leq 1}
π
−
arcsin
(
x
1
−
y
2
+
y
1
−
x
2
)
{\displaystyle \pi -\arcsin \left(x{\sqrt {1-y^{2}}}+y{\sqrt {1-x^{2}}}\right)}
x
>
0
{\displaystyle x>0}
und
y
>
0
{\displaystyle y>0}
und
x
2
+
y
2
>
1
{\displaystyle x^{2}+y^{2}>1}
−
π
−
arcsin
(
x
1
−
y
2
+
y
1
−
x
2
)
{\displaystyle -\pi -\arcsin \left(x{\sqrt {1-y^{2}}}+y{\sqrt {1-x^{2}}}\right)}
x
<
0
{\displaystyle x<0}
und
y
<
0
{\displaystyle y<0}
und
x
2
+
y
2
>
1
{\displaystyle x^{2}+y^{2}>1}
arcsin
x
−
arcsin
y
=
{\displaystyle \arcsin x-\arcsin y=}
arcsin
(
x
1
−
y
2
−
y
1
−
x
2
)
{\displaystyle \arcsin \left(x{\sqrt {1-y^{2}}}-y{\sqrt {1-x^{2}}}\right)}
x
y
≥
0
{\displaystyle xy\geq 0}
oder
x
2
+
y
2
≤
1
{\displaystyle x^{2}+y^{2}\leq 1}
π
−
arcsin
(
x
1
−
y
2
−
y
1
−
x
2
)
{\displaystyle \pi -\arcsin \left(x{\sqrt {1-y^{2}}}-y{\sqrt {1-x^{2}}}\right)}
x
>
0
{\displaystyle x>0}
und
y
<
0
{\displaystyle y<0}
und
x
2
+
y
2
>
1
{\displaystyle x^{2}+y^{2}>1}
−
π
−
arcsin
(
x
1
−
y
2
−
y
1
−
x
2
)
{\displaystyle -\pi -\arcsin \left(x{\sqrt {1-y^{2}}}-y{\sqrt {1-x^{2}}}\right)}
x
<
0
{\displaystyle x<0}
und
y
>
0
{\displaystyle y>0}
und
x
2
+
y
2
>
1
{\displaystyle x^{2}+y^{2}>1}
arccos
x
+
arccos
y
=
{\displaystyle \arccos x+\arccos y=}
arccos
(
x
y
−
1
−
x
2
1
−
y
2
)
{\displaystyle \arccos \left(xy-{\sqrt {1-x^{2}}}{\sqrt {1-y^{2}}}\right)}
x
+
y
≥
0
{\displaystyle x+y\geq 0}
2
π
−
arccos
(
x
y
−
1
−
x
2
1
−
y
2
)
{\displaystyle 2\pi -\arccos \left(xy-{\sqrt {1-x^{2}}}{\sqrt {1-y^{2}}}\right)}
x
+
y
<
0
{\displaystyle x+y<0}
arccos
x
−
arccos
y
=
{\displaystyle \arccos x-\arccos y=}
−
arccos
(
x
y
+
1
−
x
2
1
−
y
2
)
{\displaystyle -\arccos \left(xy+{\sqrt {1-x^{2}}}{\sqrt {1-y^{2}}}\right)}
x
≥
y
{\displaystyle x\geq y}
arccos
(
x
y
+
1
−
x
2
1
−
y
2
)
{\displaystyle \arccos \left(xy+{\sqrt {1-x^{2}}}{\sqrt {1-y^{2}}}\right)}
x
<
y
{\displaystyle x<y}
arctan
x
+
arctan
y
=
{\displaystyle \arctan x+\arctan y=}
arctan
(
x
+
y
1
−
x
y
)
{\displaystyle \arctan \left({\frac {x+y}{1-xy}}\right)}
x
y
<
1
{\displaystyle xy<1}
π
+
arctan
(
x
+
y
1
−
x
y
)
{\displaystyle \pi +\arctan \left({\frac {x+y}{1-xy}}\right)}
x
>
0
{\displaystyle x>0}
und
x
y
>
1
{\displaystyle xy>1}
−
π
+
arctan
(
x
+
y
1
−
x
y
)
{\displaystyle -\pi +\arctan \left({\frac {x+y}{1-xy}}\right)}
x
<
0
{\displaystyle x<0}
und
x
y
>
1
{\displaystyle xy>1}
arctan
x
−
arctan
y
=
{\displaystyle \arctan x-\arctan y=}
arctan
(
x
−
y
1
+
x
y
)
{\displaystyle \arctan \left({\frac {x-y}{1+xy}}\right)}
x
y
>
−
1
{\displaystyle xy>-1}
π
+
arctan
(
x
−
y
1
+
x
y
)
{\displaystyle \pi +\arctan \left({\frac {x-y}{1+xy}}\right)}
x
>
0
{\displaystyle x>0}
und
x
y
<
−
1
{\displaystyle xy<-1}
−
π
+
arctan
(
x
−
y
1
+
x
y
)
{\displaystyle -\pi +\arctan \left({\frac {x-y}{1+xy}}\right)}
x
<
0
{\displaystyle x<0}
und
x
y
<
−
1
{\displaystyle xy<-1}
Figur 3
sin
(
2
x
)
=
2
sin
x
⋅
cos
x
=
2
tan
x
1
+
tan
2
x
{\displaystyle \sin(2x)=2\sin x\cdot \;\cos x={\frac {2\tan x}{1+\tan ^{2}x}}}
Eine geometrische Herleitung ist in Figur 3 für Winkel
α
{\displaystyle \alpha }
und
β
{\displaystyle \beta }
zwischen 0° und 90° veranschaulicht.[ 7]
Zu Figur 3:
Aus der Berechnung der Flächeninhalte der beiden grauen Dreiecke ergibt sich
1
2
⋅
2
sin
α
⋅
2
cos
α
=
1
2
⋅
2
⋅
sin
(
2
α
)
{\displaystyle {\frac {1}{2}}\cdot 2\sin \alpha \cdot 2\cos \alpha ={\frac {1}{2}}\cdot 2\cdot \sin(2\alpha )}
. Hieraus folgt
2
sin
α
⋅
cos
α
=
sin
(
2
α
)
{\displaystyle 2\sin \alpha \cdot \cos \alpha =\sin(2\alpha )}
.
Weitere Beziehungen:
cos
(
2
x
)
=
cos
2
x
−
sin
2
x
=
1
−
2
sin
2
x
=
2
cos
2
x
−
1
=
1
−
tan
2
x
1
+
tan
2
x
{\displaystyle \cos(2x)=\cos ^{2}x-\sin ^{2}x=1-2\sin ^{2}x=2\cos ^{2}x-1={\frac {1-\tan ^{2}x}{1+\tan ^{2}x}}}
tan
(
2
x
)
=
2
tan
x
1
−
tan
2
x
=
2
cot
x
−
tan
x
{\displaystyle \tan(2x)={\frac {2\tan x}{1-\tan ^{2}x}}={\frac {2}{\cot x-\tan x}}}
cot
(
2
x
)
=
cot
2
x
−
1
2
cot
x
=
cot
x
−
tan
x
2
{\displaystyle \cot(2x)={\frac {\cot ^{2}x-1}{2\cot x}}={\frac {\cot x-\tan x}{2}}}
Die Formeln für Vielfache berechnen sich normalerweise über die komplexen Zahlen aus der Euler-Formel
z
=
r
(
cos
ϕ
+
i
sin
ϕ
)
⟺
z
n
=
r
n
(
cos
ϕ
+
i
sin
ϕ
)
n
{\displaystyle z=r\left(\cos \phi +i\sin \phi \right)\iff z^{n}=r^{n}\left(\cos \phi +i\sin \phi \right)^{n}}
und der DeMoivre-Formel
z
n
=
r
n
(
cos
(
n
ϕ
)
+
i
sin
(
n
ϕ
)
)
{\displaystyle z^{n}=r^{n}\left(\cos \left(n\phi \right)+i\sin \left(n\phi \right)\right)}
.
Damit ergibt sich
cos
(
n
ϕ
)
+
i
sin
(
n
ϕ
)
=
(
cos
ϕ
+
i
sin
ϕ
)
n
{\displaystyle \cos \left(n\phi \right)+i\sin \left(n\phi \right)=\left(\cos \phi +i\sin \phi \right)^{n}}
.
Zerlegung in Real- und Imaginärteil liefert dann die Formeln für
cos
{\displaystyle \cos }
und
sin
{\displaystyle \sin }
bzw. die allgemeine Reihendarstellung.
Die Formel für
cos
(
n
x
)
{\displaystyle \cos(nx)}
steht über
T
n
(
cos
x
)
=
cos
(
n
x
)
{\displaystyle T_{n}(\cos x)=\cos(nx)}
[ 8] mit den Tschebyschow-Polynomen in Beziehung.
sin
(
3
x
)
=
3
sin
x
−
4
sin
3
x
{\displaystyle \sin(3x)=3\sin x-4\sin ^{3}x\,}
[ 9]
=
sin
x
(
4
cos
2
x
−
1
)
{\displaystyle =\;\sin x\left(4\cos ^{2}x-1\right)}
sin
(
4
x
)
=
8
sin
x
cos
3
x
−
4
sin
x
cos
x
{\displaystyle \sin(4x)=8\sin x\;\cos ^{3}x-4\sin x\;\cos x}
[ 10]
=
sin
x
(
8
cos
3
x
−
4
cos
x
)
{\displaystyle =\;\sin x\left(8\cos ^{3}x-4\cos x\right)}
sin
(
5
x
)
=
5
sin
x
−
20
sin
3
x
+
16
sin
5
x
{\displaystyle \sin(5x)=5\sin x-20\sin ^{3}x+16\sin ^{5}x\;}
[ 11]
=
sin
x
(
16
cos
4
x
−
12
cos
2
x
+
1
)
{\displaystyle =\;\sin x\left(16\cos ^{4}x-12\cos ^{2}x+1\right)}
sin
(
n
x
)
=
n
sin
x
cos
n
−
1
x
−
(
n
3
)
sin
3
x
cos
n
−
3
x
+
(
n
5
)
sin
5
x
cos
n
−
5
x
−
…
+
…
{\displaystyle \sin(nx)=n\;\sin x\;\cos ^{n-1}x-{n \choose 3}\sin ^{3}x\;\cos ^{n-3}x+{n \choose 5}\sin ^{5}x\;\cos ^{n-5}x\;-\ldots +\ldots }
[ 12] [ 13]
=
∑
j
=
0
⌊
n
−
1
2
⌋
(
−
1
)
j
(
n
2
j
+
1
)
sin
2
j
+
1
x
cos
n
−
2
j
−
1
x
{\displaystyle =\;\sum _{j=0}^{\left\lfloor {\frac {n-1}{2}}\right\rfloor }(-1)^{j}{n \choose 2j+1}\sin ^{2j+1}x\;\cos ^{n-2j-1}x}
=
sin
x
∑
k
=
0
⌊
n
−
1
2
⌋
(
−
1
)
k
(
n
−
k
−
1
k
)
2
n
−
2
k
−
1
cos
n
−
2
k
−
1
x
{\displaystyle =\;\sin x\sum _{k=0}^{\left\lfloor {\frac {n-1}{2}}\right\rfloor }(-1)^{k}{n-k-1 \choose k}2^{n-2k-1}\cos ^{n-2k-1}x}
sin
(
n
x
)
sin
(
x
)
=
cos
(
(
n
−
1
)
x
)
−
cos
(
(
n
+
1
)
x
)
2
{\displaystyle \sin(nx)\;\sin(x)={\frac {\cos((n-1)\,x)-\cos((n+1)\,x)}{2}}}
cos
(
3
x
)
=
4
cos
3
x
−
3
cos
x
{\displaystyle \cos(3x)=4\cos ^{3}x-3\cos x\,}
[ 14]
cos
(
4
x
)
=
8
cos
4
x
−
8
cos
2
x
+
1
{\displaystyle \cos(4x)=8\cos ^{4}x-8\cos ^{2}x+1\,}
[ 15]
cos
(
5
x
)
=
16
cos
5
x
−
20
cos
3
x
+
5
cos
x
{\displaystyle \cos(5x)=16\cos ^{5}x-20\cos ^{3}x+5\cos x\,}
[ 16]
cos
(
6
x
)
=
32
cos
6
x
−
48
cos
4
x
+
18
cos
2
x
−
1
{\displaystyle \cos(6x)=32\cos ^{6}x-48\cos ^{4}x+18\cos ^{2}x-1\,}
[ 17]
cos
(
n
x
)
=
cos
n
x
−
(
n
2
)
sin
2
x
cos
n
−
2
x
+
(
n
4
)
sin
4
x
cos
n
−
4
x
−
…
+
…
{\displaystyle \cos(nx)=\cos ^{n}x-{n \choose 2}\sin ^{2}x\;\cos ^{n-2}x+{n \choose 4}\sin ^{4}x\;\cos ^{n-4}x\;-\ldots +\ldots }
[ 13] [ 18]
=
∑
j
=
0
⌊
n
2
⌋
(
−
1
)
j
(
n
2
j
)
sin
2
j
x
cos
n
−
2
j
x
{\displaystyle =\;\sum _{j=0}^{\left\lfloor {\frac {n}{2}}\right\rfloor }(-1)^{j}{n \choose 2j}\sin ^{2j}x\;\cos ^{n-2j}x}
cos
(
n
x
)
cos
(
x
)
=
cos
(
(
n
−
1
)
x
)
+
cos
(
(
n
+
1
)
x
)
2
{\displaystyle \cos(nx)\;\cos(x)={\frac {\cos((n-1)\,x)+\cos((n+1)\,x)}{2}}}
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}}}
[ 13]
tan
(
4
x
)
=
4
tan
x
−
4
tan
3
x
1
−
6
tan
2
x
+
tan
4
x
{\displaystyle \tan(4x)={\frac {4\tan x-4\tan ^{3}x}{1-6\tan ^{2}x+\tan ^{4}x}}}
[ 13]
cot
(
3
x
)
=
cot
3
x
−
3
cot
x
3
cot
2
x
−
1
{\displaystyle \cot(3x)={\frac {\cot ^{3}x-3\cot x}{3\cot ^{2}x-1}}}
[ 13]
cot
(
4
x
)
=
cot
4
x
−
6
cot
2
x
+
1
4
cot
3
x
−
4
cot
x
{\displaystyle \cot(4x)={\frac {\cot ^{4}x-6\cot ^{2}x+1}{4\cot ^{3}x-4\cot x}}}
[ 13]
Figur 4
Zur Berechnung des Funktionswertes des halben Arguments dienen die Halbwinkelformeln [ 13] , welche sich mittels Substitution aus den Doppelwinkelformeln herleiten lassen:
sin
x
2
=
1
−
cos
x
2
für
x
∈
[
0
,
2
π
]
{\displaystyle \sin {\frac {x}{2}}={\sqrt {\frac {1-\cos x}{2}}}\quad {\text{für}}\quad x\in \left[0,2\pi \right]}
cos
x
2
=
1
+
cos
x
2
für
x
∈
[
−
π
,
π
]
{\displaystyle \cos {\frac {x}{2}}={\sqrt {\frac {1+\cos x}{2}}}\quad {\text{für}}\quad x\in \left[-\pi ,\pi \right]}
tan
x
2
=
1
−
cos
x
sin
x
=
sin
x
1
+
cos
x
für
x
∈
R
∖
π
(
2
Z
+
1
)
{\displaystyle \tan {\frac {x}{2}}={\frac {1-\cos x}{\sin x}}={\frac {\sin x}{1+\cos x}}\quad {\text{für}}\quad x\in \mathbb {R} \setminus \pi (2\mathbb {Z} +1)}
cot
x
2
=
1
+
cos
x
sin
x
=
sin
x
1
−
cos
x
für
x
∈
R
∖
2
π
Z
{\displaystyle \cot {\frac {x}{2}}={\frac {1+\cos x}{\sin x}}={\frac {\sin x}{1-\cos x}}\quad {\text{für}}\quad x\in \mathbb {R} \setminus 2\pi \mathbb {Z} }
Eine geometrische Herleitung der dritten Formel ist in Figur 4 für Winkel
α
{\displaystyle \alpha }
und
β
{\displaystyle \beta }
zwischen 0° und 90° veranschaulicht.[ 19] Aus der Berechnung der Flächeninhalte der beiden grauen Dreiecke ergibt sich unmittelbar
tan
(
α
2
)
=
sin
α
1
+
cos
α
=
1
−
cos
α
sin
α
{\displaystyle \tan \left({\frac {\alpha }{2}}\right)={\frac {\sin \alpha }{1+\cos \alpha }}={\frac {1-\cos \alpha }{\sin \alpha }}}
.
Außerdem gilt:
tan
x
2
=
tan
x
1
+
1
+
tan
2
x
für
x
∈
]
−
π
2
,
π
2
[
{\displaystyle \tan {\frac {x}{2}}={\frac {\tan x}{1+{\sqrt {1+\tan ^{2}x}}}}\quad {\text{für}}\quad x\in \left]-{\tfrac {\pi }{2}},{\tfrac {\pi }{2}}\right[}
cot
x
2
=
cot
x
+
1
+
cot
2
x
für
x
∈
]
0
,
π
[
{\displaystyle \cot {\frac {x}{2}}=\cot x+{\sqrt {1+\cot ^{2}x}}\quad {\text{für}}\quad x\in \left]0,\pi \right[}
Siehe auch: Halbwinkelsatz
Summen zweier trigonometrischer Funktionen (Identitäten)
Bearbeiten
Aus den Additionstheoremen lassen sich Identitäten ableiten, mit deren Hilfe die Summe zweier trigonometrischer Funktionen als Produkt dargestellt werden kann:[ 13]
sin
x
+
sin
y
=
2
sin
x
+
y
2
cos
x
−
y
2
sin
x
−
sin
y
=
2
cos
x
+
y
2
sin
x
−
y
2
cos
x
+
cos
y
=
2
cos
x
+
y
2
cos
x
−
y
2
cos
x
−
cos
y
=
−
2
sin
x
+
y
2
sin
x
−
y
2
{\displaystyle {\begin{aligned}\sin x+\sin y&=2\sin {\frac {x+y}{2}}\cos {\frac {x-y}{2}}\\\sin x-\sin y&=2\cos {\frac {x+y}{2}}\sin {\frac {x-y}{2}}\\\cos x+\cos y&=2\cos {\frac {x+y}{2}}\cos {\frac {x-y}{2}}\\\cos x-\cos y&=-2\sin {\frac {x+y}{2}}\sin {\frac {x-y}{2}}\end{aligned}}}
tan
x
+
tan
y
=
sin
(
x
+
y
)
cos
x
cos
y
tan
x
−
tan
y
=
sin
(
x
−
y
)
cos
x
cos
y
}
⇒
tan
x
±
tan
y
=
sin
(
x
±
y
)
cos
x
cos
y
{\displaystyle \left.{\begin{matrix}\tan x+\tan y={\dfrac {\sin(x+y)}{\cos x\cos y}}\\[1em]\tan x-\tan y={\dfrac {\sin(x-y)}{\cos x\cos y}}\end{matrix}}\right\}\Rightarrow \tan x\pm \tan y={\frac {\sin(x\pm y)}{\cos x\cos y}}}
cot
x
+
cot
y
=
sin
(
y
+
x
)
sin
x
sin
y
cot
x
−
cot
y
=
sin
(
y
−
x
)
sin
x
sin
y
}
⇒
cot
x
±
cot
y
=
sin
(
y
±
x
)
sin
x
sin
y
{\displaystyle \left.{\begin{matrix}\cot x+\cot y={\dfrac {\sin(y+x)}{\sin x\sin y}}\\[1em]\cot x-\cot y={\dfrac {\sin(y-x)}{\sin x\sin y}}\end{matrix}}\right\}\Rightarrow \cot x\pm \cot y={\frac {\sin(y\pm x)}{\sin x\sin y}}}
Daraus ergeben sich noch Spezialfälle:
cos
x
+
sin
x
=
2
⋅
sin
(
x
+
π
4
)
=
2
⋅
cos
(
x
−
π
4
)
cos
x
−
sin
x
=
2
⋅
cos
(
x
+
π
4
)
=
−
2
⋅
sin
(
x
−
π
4
)
{\displaystyle {\begin{aligned}\cos x+\sin x&={\sqrt {2}}\cdot \sin \left(x+{\frac {\pi }{4}}\right)={\sqrt {2}}\cdot \cos \left(x-{\frac {\pi }{4}}\right)\\\cos x-\sin x&={\sqrt {2}}\cdot \cos \left(x+{\frac {\pi }{4}}\right)=-{\sqrt {2}}\cdot \sin \left(x-{\frac {\pi }{4}}\right)\end{aligned}}}
Produkte der trigonometrischen Funktionen lassen sich mit folgenden Formeln berechnen:[ 13]
sin
x
sin
y
=
1
2
(
cos
(
x
−
y
)
−
cos
(
x
+
y
)
)
{\displaystyle \sin x\;\sin y={\frac {1}{2}}{\Big (}\cos(x-y)-\cos(x+y){\Big )}}
cos
x
cos
y
=
1
2
(
cos
(
x
−
y
)
+
cos
(
x
+
y
)
)
{\displaystyle \cos x\;\cos y={\frac {1}{2}}{\Big (}\cos(x-y)+\cos(x+y){\Big )}}
sin
x
cos
y
=
1
2
(
sin
(
x
−
y
)
+
sin
(
x
+
y
)
)
{\displaystyle \sin x\;\cos y={\frac {1}{2}}{\Big (}\sin(x-y)+\sin(x+y){\Big )}}
tan
x
tan
y
=
tan
x
+
tan
y
cot
x
+
cot
y
=
−
tan
x
−
tan
y
cot
x
−
cot
y
{\displaystyle \tan x\;\tan y={\frac {\tan x+\tan y}{\cot x+\cot y}}=-{\frac {\tan x-\tan y}{\cot x-\cot y}}}
cot
x
cot
y
=
cot
x
+
cot
y
tan
x
+
tan
y
=
−
cot
x
−
cot
y
tan
x
−
tan
y
{\displaystyle \cot x\;\cot y={\frac {\cot x+\cot y}{\tan x+\tan y}}=-{\frac {\cot x-\cot y}{\tan x-\tan y}}}
tan
x
cot
y
=
tan
x
+
cot
y
cot
x
+
tan
y
=
−
tan
x
−
cot
y
cot
x
−
tan
y
{\displaystyle \tan x\;\cot y={\frac {\tan x+\cot y}{\cot x+\tan y}}=-{\frac {\tan x-\cot y}{\cot x-\tan y}}}
sin
x
sin
y
sin
z
=
1
4
(
sin
(
x
+
y
−
z
)
+
sin
(
y
+
z
−
x
)
+
sin
(
z
+
x
−
y
)
−
sin
(
x
+
y
+
z
)
)
{\displaystyle \sin x\;\sin y\;\sin z={\frac {1}{4}}{\Big (}\sin(x+y-z)+\sin(y+z-x)+\sin(z+x-y)-\sin(x+y+z){\Big )}}
cos
x
cos
y
cos
z
=
1
4
(
cos
(
x
+
y
−
z
)
+
cos
(
y
+
z
−
x
)
+
cos
(
z
+
x
−
y
)
+
cos
(
x
+
y
+
z
)
)
{\displaystyle \cos x\;\cos y\;\cos z={\frac {1}{4}}{\Big (}\cos(x+y-z)+\cos(y+z-x)+\cos(z+x-y)+\cos(x+y+z){\Big )}}
sin
x
sin
y
cos
z
=
1
4
(
−
cos
(
x
+
y
−
z
)
+
cos
(
y
+
z
−
x
)
+
cos
(
z
+
x
−
y
)
−
cos
(
x
+
y
+
z
)
)
{\displaystyle \sin x\;\sin y\;\cos z={\frac {1}{4}}{\Big (}-\cos(x+y-z)+\cos(y+z-x)+\cos(z+x-y)-\cos(x+y+z){\Big )}}
sin
x
cos
y
cos
z
=
1
4
(
sin
(
x
+
y
−
z
)
−
sin
(
y
+
z
−
x
)
+
sin
(
z
+
x
−
y
)
+
sin
(
x
+
y
+
z
)
)
{\displaystyle \sin x\;\cos y\;\cos z={\frac {1}{4}}{\Big (}\sin(x+y-z)-\sin(y+z-x)+\sin(z+x-y)+\sin(x+y+z){\Big )}}
∏
m
=
1
n
cos
(
x
m
)
=
1
2
n
∑
k
1
=
1
2
⋯
∑
k
n
=
1
2
[
exp
(
i
∑
ν
=
1
n
(
−
1
)
k
ν
x
ν
)
]
=
1
2
n
−
1
∑
k
2
=
1
2
⋯
∑
k
n
=
1
2
[
cos
(
x
1
+
∑
ν
=
2
n
(
−
1
)
k
ν
x
ν
)
]
{\displaystyle \prod _{m=1}^{n}\cos(x_{m})={\frac {1}{2^{n}}}\sum _{k_{1}=1}^{2}\cdots \sum _{k_{n}=1}^{2}\left[\exp \left({\text{i}}\sum _{\nu =1}^{n}(-1)^{k_{\nu }}x_{\nu }\right)\right]={\frac {1}{2^{n-1}}}\sum _{k_{2}=1}^{2}\cdots \sum _{k_{n}=1}^{2}\left[\cos \left(x_{1}+\sum _{\nu =2}^{n}(-1)^{k_{\nu }}x_{\nu }\right)\right]}
∏
m
=
1
n
sin
(
x
m
)
=
1
(
2
i
)
n
∑
k
1
=
1
2
⋯
∑
k
n
=
1
2
[
∏
μ
=
1
n
(
−
1
)
k
μ
⋅
exp
(
i
∑
ν
=
1
n
(
−
1
)
k
ν
x
ν
)
]
=
1
2
n
−
1
∑
k
2
=
1
2
⋯
∑
k
n
=
1
2
[
∏
μ
=
2
n
(
−
1
)
k
μ
⋅
{
(
−
1
)
n
/
2
⋅
cos
(
x
1
+
∑
ν
=
2
n
(
−
1
)
k
ν
x
ν
)
gerade
n
(
−
1
)
(
n
−
1
)
/
2
⋅
sin
(
x
1
+
∑
ν
=
2
n
(
−
1
)
k
ν
x
ν
)
ungerade
n
]
{\displaystyle \prod _{m=1}^{n}\sin(x_{m})={\frac {1}{(2{\text{i}})^{n}}}\sum _{k_{1}=1}^{2}\cdots \sum _{k_{n}=1}^{2}\left[\prod _{\mu =1}^{n}(-1)^{k_{\mu }}\cdot \exp \left({\text{i}}\sum _{\nu =1}^{n}(-1)^{k_{\nu }}x_{\nu }\right)\right]={\frac {1}{2^{n-1}}}\sum _{k_{2}=1}^{2}\cdots \sum _{k_{n}=1}^{2}\left[\prod _{\mu =2}^{n}(-1)^{k_{\mu }}\cdot {\begin{cases}\displaystyle (-1)^{n/2}\cdot \cos \left(x_{1}+\sum _{\nu =2}^{n}(-1)^{k_{\nu }}x_{\nu }\right)&{\text{gerade}}\;n\\\displaystyle (-1)^{(n-1)/2}\cdot \sin \left(x_{1}+\sum _{\nu =2}^{n}(-1)^{k_{\nu }}x_{\nu }\right)&{\text{ungerade}}\;n\end{cases}}\right]}
Aus der Doppelwinkelfunktion für
sin
(
2
x
)
{\displaystyle \sin(2x)}
folgt außerdem:
sin
x
cos
x
=
1
2
sin
(
2
x
)
{\displaystyle \sin x\;\cos x={\frac {1}{2}}\sin(2x)}
sin
2
x
=
1
2
(
1
−
cos
(
2
x
)
)
{\displaystyle \sin ^{2}x={\frac {1}{2}}\ {\Big (}1-\cos(2x){\Big )}}
[ 13] [ 20]
sin
3
x
=
1
4
(
3
sin
x
−
sin
(
3
x
)
)
{\displaystyle \sin ^{3}x={\frac {1}{4}}\ {\Big (}3\,\sin x-\sin(3x){\Big )}}
[ 13] [ 21]
sin
4
x
=
1
8
(
3
−
4
cos
(
2
x
)
+
cos
(
4
x
)
)
{\displaystyle \sin ^{4}x={\frac {1}{8}}\ {\Big (}3-4\,\cos(2x)+\cos(4x){\Big )}}
[ 13] [ 22]
sin
5
x
=
1
16
(
10
sin
x
−
5
sin
(
3
x
)
+
sin
(
5
x
)
)
{\displaystyle \sin ^{5}x={\frac {1}{16}}\ {\Big (}10\,\sin x-5\,\sin(3x)+\sin(5x){\Big )}}
[ 23]
sin
6
x
=
1
32
(
10
−
15
cos
(
2
x
)
+
6
cos
(
4
x
)
−
cos
(
6
x
)
)
{\displaystyle \sin ^{6}x={\frac {1}{32}}\ {\Big (}10-15\,\cos(2x)+6\,\cos(4x)-\cos(6x){\Big )}}
[ 24]
sin
n
x
=
1
2
n
∑
k
=
0
n
(
n
k
)
cos
(
(
n
−
2
k
)
(
x
−
π
2
)
)
;
n
∈
N
{\displaystyle \sin ^{n}x={\frac {1}{2^{n}}}\,\sum _{k=0}^{n}{n \choose k}\,\cos \left((n-2k)\left(x-{\frac {\pi }{2}}\right)\right)\ ;\quad n\in \mathbb {N} }
sin
n
x
=
1
2
n
(
n
n
2
)
+
1
2
n
−
1
∑
k
=
0
n
2
−
1
(
−
1
)
n
2
−
k
(
n
k
)
cos
(
(
n
−
2
k
)
x
)
;
n
∈
N
und
n
gerade
{\displaystyle \sin ^{n}x={\frac {1}{2^{n}}}\,{n \choose {\frac {n}{2}}}+{\frac {1}{2^{n-1}}}\sum _{k=0}^{{\frac {n}{2}}-1}(-1)^{{\frac {n}{2}}-k}\,{n \choose k}\,\cos {((n-2k)x)};\quad n\in \mathbb {N} {\text{ und }}n{\text{ gerade }}}
sin
n
x
=
1
2
n
−
1
∑
k
=
0
n
−
1
2
(
−
1
)
n
−
1
2
−
k
(
n
k
)
sin
(
(
n
−
2
k
)
x
)
;
n
∈
N
und
n
ungerade
{\displaystyle \sin ^{n}x={\frac {1}{2^{n-1}}}\,\sum _{k=0}^{\frac {n-1}{2}}(-1)^{{\frac {n-1}{2}}-k}\,{n \choose k}\,\sin {((n-2k)x)};\quad n\in \mathbb {N} {\text{ und }}n{\text{ ungerade}}}
cos
2
x
=
1
2
(
1
+
cos
(
2
x
)
)
{\displaystyle \cos ^{2}x={\frac {1}{2}}\ {\Big (}1+\cos(2x){\Big )}}
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