Formelsammlung Trigonometrie

Dreieckberechnung

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} $ABC$ habe die Seiten a = B C {\displaystyle a=BC} $a=BC$ , b = C A {\displaystyle b=CA} $b=CA$ und c = A B {\displaystyle c=AB} $c=AB$ , die Winkel α {\displaystyle \alpha } $\alpha$ , β {\displaystyle \beta } $\beta$ und γ {\displaystyle \gamma } $\gamma$ bei den Ecken A {\displaystyle A} $A$ , B {\displaystyle B} $B$ und C {\displaystyle C} $C$ . Ferner seien r {\displaystyle r} $r$ der Umkreisradius, ρ {\displaystyle \rho } $\rho$ der Inkreisradius und ρ a {\displaystyle \rho _{a}} $\rho _{a}$ , ρ b {\displaystyle \rho _{b}} $\rho _{b}$ und ρ c {\displaystyle \rho _{c}} $\rho _{c}$ die Ankreisradien (und zwar die Radien der Ankreise, die den Ecken A {\displaystyle A} $A$ , B {\displaystyle B} $B$ bzw. C {\displaystyle C} $C$ gegenüberliegen) des Dreiecks A B C {\displaystyle ABC} $ABC$ . Die Variable s {\displaystyle s} $s$ steht für den halben Umfang des Dreiecks A B C {\displaystyle ABC} $ABC$ :

s = a + b + c 2 {\displaystyle s={\frac {a+b+c}{2}}}

s={\frac {a+b+c}{2}}

Schließlich wird die Fläche des Dreiecks A B C {\displaystyle ABC} $ABC$ mit F {\displaystyle F} $F$ 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} $r$ , den Inkreisradius ρ {\displaystyle \rho } $\rho$ und die drei Ankreisradien ρ a {\displaystyle \rho _{a}} $\rho _{a}$ , ρ b {\displaystyle \rho _{b}} $\rho _{b}$ , ρ c {\displaystyle \rho _{c}} $\rho _{c}$ benutzt werden. Oft werden davon abweichend für dieselben Größen auch die Bezeichnungen R {\displaystyle R} $R$ , r {\displaystyle r} $r$ , r a {\displaystyle r_{a}} $r_{a}$ , r b {\displaystyle r_{b}} $r_{b}$ , r c {\displaystyle r_{c}} $r_{c}$ verwendet.

Winkelsumme

α + β + γ = 180 ∘ {\displaystyle \alpha +\beta +\gamma =180^{\circ }}

\alpha +\beta +\gamma =180^{\circ }

Sinussatz

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}}}

{\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 }} $\alpha =90^{\circ }$

sin ⁡ β = b a {\displaystyle \sin \beta ={\frac {b}{a}}}

\sin \beta ={\frac {b}{a}}

sin ⁡ γ = c a {\displaystyle \sin \gamma ={\frac {c}{a}}}

\sin \gamma ={\frac {c}{a}}

wenn β = 90 ∘ {\displaystyle \beta =90^{\circ }} $\beta =90^{\circ }$

sin ⁡ α = a b {\displaystyle \sin \alpha ={\frac {a}{b}}}

\sin \alpha ={\frac {a}{b}}

sin ⁡ γ = c b {\displaystyle \sin \gamma ={\frac {c}{b}}}

\sin \gamma ={\frac {c}{b}}

wenn γ = 90 ∘ {\displaystyle \gamma =90^{\circ }} $\gamma =90^{\circ }$

sin ⁡ α = a c {\displaystyle \sin \alpha ={\frac {a}{c}}}

\sin \alpha ={\frac {a}{c}}

sin ⁡ β = b c {\displaystyle \sin \beta ={\frac {b}{c}}}

\sin \beta ={\frac {b}{c}}

Kosinussatz

Formel 1:

a 2 = b 2 + c 2 − 2 b c cos ⁡ α {\displaystyle a^{2}=b^{2}+c^{2}-2bc\ \cos \alpha }

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 }

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 }

c^{2}=a^{2}+b^{2}-2ab\ \cos \gamma

Formel 2:

wenn α = 90 ∘ {\displaystyle \alpha =90^{\circ }} $\alpha =90^{\circ }$

cos ⁡ β = c a {\displaystyle \cos \beta ={\frac {c}{a}}}

\cos \beta ={\frac {c}{a}}

cos ⁡ γ = b a {\displaystyle \cos \gamma ={\frac {b}{a}}}

\cos \gamma ={\frac {b}{a}}

wenn β = 90 ∘ {\displaystyle \beta =90^{\circ }} $\beta =90^{\circ }$

cos ⁡ α = c b {\displaystyle \cos \alpha ={\frac {c}{b}}}

\cos \alpha ={\frac {c}{b}}

cos ⁡ γ = a b {\displaystyle \cos \gamma ={\frac {a}{b}}}

\cos \gamma ={\frac {a}{b}}

wenn γ = 90 ∘ {\displaystyle \gamma =90^{\circ }} $\gamma =90^{\circ }$

a 2 + b 2 = c 2 {\displaystyle a^{2}+b^{2}=c^{2}}

a^{2}+b^{2}=c^{2}

(Satz des Pythagoras) cos ⁡ α = b c {\displaystyle \cos \alpha ={\frac {b}{c}}}

\cos \alpha ={\frac {b}{c}}

cos ⁡ β = a c {\displaystyle \cos \beta ={\frac {a}{c}}}

\cos \beta ={\frac {a}{c}}

Projektionssatz

a = b cos ⁡ γ + c cos ⁡ β {\displaystyle a=b\,\cos \gamma +c\,\cos \beta }

a=b\,\cos \gamma +c\,\cos \beta

b = c cos ⁡ α + a cos ⁡ γ {\displaystyle b=c\,\cos \alpha +a\,\cos \gamma }

b=c\,\cos \alpha +a\,\cos \gamma

c = a cos ⁡ β + b cos ⁡ α {\displaystyle c=a\,\cos \beta +b\,\cos \alpha }

c=a\,\cos \beta +b\,\cos \alpha

Die Mollweideschen Formeln

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}}}}}

{\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}}}}}

{\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}}}}

Tangenssatz

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}}}}}

{\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}}} ${\frac {a+b}{a-b}}$ und c + a c − a {\displaystyle {\frac {c+a}{c-a}}} ${\frac {c+a}{c-a}}$ :

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}}}}}

{\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}}}}}

{\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)} $\tan(-x)=-\tan(x)$ 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}}}}}

{\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 }} $\alpha =90^{\circ }$

tan ⁡ β = b c {\displaystyle \tan \beta ={\frac {b}{c}}}

\tan \beta ={\frac {b}{c}}

tan ⁡ γ = c b {\displaystyle \tan \gamma ={\frac {c}{b}}}

\tan \gamma ={\frac {c}{b}}

wenn β = 90 ∘ {\displaystyle \beta =90^{\circ }} $\beta =90^{\circ }$

tan ⁡ α = a c {\displaystyle \tan \alpha ={\frac {a}{c}}}

\tan \alpha ={\frac {a}{c}}

tan ⁡ γ = c a {\displaystyle \tan \gamma ={\frac {c}{a}}}

\tan \gamma ={\frac {c}{a}}

wenn γ = 90 ∘ {\displaystyle \gamma =90^{\circ }} $\gamma =90^{\circ }$

tan ⁡ α = a b {\displaystyle \tan \alpha ={\frac {a}{b}}}

\tan \alpha ={\frac {a}{b}}

tan ⁡ β = b a {\displaystyle \tan \beta ={\frac {b}{a}}}

\tan \beta ={\frac {b}{a}}

Formeln mit dem halben Umfang

Im Folgenden bedeutet s {\displaystyle s} $s$ immer die Hälfte des Umfangs des Dreiecks A B C {\displaystyle ABC} $ABC$ , also s = a + b + c 2 {\displaystyle s={\frac {a+b+c}{2}}} $s={\frac {a+b+c}{2}}$ .

s − a = b + c − a 2 {\displaystyle s-a={\frac {b+c-a}{2}}}

s-a={\frac {b+c-a}{2}}

s − b = c + a − b 2 {\displaystyle s-b={\frac {c+a-b}{2}}}

s-b={\frac {c+a-b}{2}}

s − c = a + b − c 2 {\displaystyle s-c={\frac {a+b-c}{2}}}

s-c={\frac {a+b-c}{2}}

( s − b ) + ( s − c ) = a {\displaystyle \left(s-b\right)+\left(s-c\right)=a}

\left(s-b\right)+\left(s-c\right)=a

( s − c ) + ( s − a ) = b {\displaystyle \left(s-c\right)+\left(s-a\right)=b}

\left(s-c\right)+\left(s-a\right)=b

( s − a ) + ( s − b ) = c {\displaystyle \left(s-a\right)+\left(s-b\right)=c}

\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}

\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 {\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 {\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}}}}

\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 {\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 {\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}}}}

\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 {\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 {\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)}}}}

\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=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}}}

s-a=4r\cos {\frac {\alpha }{2}}\sin {\frac {\beta }{2}}\sin {\frac {\gamma }{2}}

Flächeninhalt und Umkreisradius

Der Flächeninhalt des Dreiecks wird hier mit F {\displaystyle F} $F$ bezeichnet (nicht, wie heute üblich, mit A {\displaystyle A} $A$ , um eine Verwechselung mit der Dreiecksecke A {\displaystyle A} $A$ 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={\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)}}}

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={\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}}

F={\frac {1}{2}}ah_{a}={\frac {1}{2}}bh_{b}={\frac {1}{2}}ch_{c}

, wobei h a {\displaystyle h_{a}}

h_{a}

, h b {\displaystyle h_{b}}

h_{b}

und h c {\displaystyle h_{c}}

h_{c}

die Längen der von A {\displaystyle A}

A

, B {\displaystyle B}

B

bzw. C {\displaystyle C}

C

ausgehenden Höhen des Dreiecks A B C {\displaystyle ABC}

ABC

sind. F = 2 r 2 sin α sin β sin γ {\displaystyle F=2r^{2}\sin \,\alpha \,\sin \,\beta \,\sin \,\gamma }

F=2r^{2}\sin \,\alpha \,\sin \,\beta \,\sin \,\gamma

F = a b c 4 r {\displaystyle F={\frac {abc}{4r}}}

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=\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={\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=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=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 )}}}}

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}}}}

{\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={\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 )}}}}

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}}} ${\frac {a}{\sin \alpha }}={\frac {b}{\sin \beta }}={\frac {c}{\sin \gamma }}=2r={\frac {abc}{2F}}$

a = 2 r sin ⁡ α {\displaystyle a=2r\,\sin \alpha }

a=2r\,\sin \alpha

b = 2 r sin ⁡ β {\displaystyle b=2r\,\sin \beta }

b=2r\,\sin \beta

c = 2 r sin ⁡ γ {\displaystyle c=2r\,\sin \gamma }

c=2r\,\sin \gamma

r = a b c 4 F {\displaystyle r={\frac {abc}{4F}}}

r={\frac {abc}{4F}}

In- und Ankreisradien

In diesem Abschnitt werden Formeln aufgelistet, in denen der Inkreisradius ρ {\displaystyle \rho } $\rho$ und die Ankreisradien ρ a {\displaystyle \rho _{a}} $\rho _{a}$ , ρ b {\displaystyle \rho _{b}} $\rho _{b}$ und ρ c {\displaystyle \rho _{c}} $\rho _{c}$ des Dreiecks A B C {\displaystyle ABC} $ABC$ 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}}}

\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}}}

\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)}

\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}}}

\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}}}}

\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}}}}}

\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}

a\cdot b+b\cdot c+c\cdot a=s^{2}+\rho ^{2}+4\cdot \rho \cdot r

Wichtige Ungleichung: 2 ρ ≤ r {\displaystyle 2\rho \leq r} $2\rho \leq r$ ; Gleichheit tritt nur dann ein, wenn Dreieck A B C {\displaystyle ABC} $ABC$ 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}}}

\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}}}

\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)}

\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)}}}

\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}}}}

\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}} $\rho _{a}$ gilt in analoger Form für ρ b {\displaystyle \rho _{b}} $\rho _{b}$ und ρ c {\displaystyle \rho _{c}} $\rho _{c}$ .

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}}}}

{\frac {1}{\rho }}={\frac {1}{\rho _{a}}}+{\frac {1}{\rho _{b}}}+{\frac {1}{\rho _{c}}}

Höhen

Die Längen der von A {\displaystyle A} $A$ , B {\displaystyle B} $B$ bzw. C {\displaystyle C} $C$ ausgehenden Höhen des Dreiecks A B C {\displaystyle ABC} $ABC$ werden mit h a {\displaystyle h_{a}} $h_{a}$ , h b {\displaystyle h_{b}} $h_{b}$ und h c {\displaystyle h_{c}} $h_{c}$ 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_{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_{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_{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 }}}

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}}

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}}}}

{\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} $ABC$ einen rechten Winkel bei C {\displaystyle C} $C$ (ist also γ = 90 ∘ {\displaystyle \gamma =90^{\circ }} $\gamma =90^{\circ }$ ), dann gilt

h c = a b c {\displaystyle h_{c}={\frac {ab}{c}}}

h_{c}={\frac {ab}{c}}

h a = b {\displaystyle h_{a}=b}

h_{a}=b

h b = a {\displaystyle h_{b}=a}

h_{b}=a

Seitenhalbierende

Die Längen der von A {\displaystyle A} $A$ , B {\displaystyle B} $B$ bzw. C {\displaystyle C} $C$ ausgehenden Seitenhalbierenden des Dreiecks A B C {\displaystyle ABC} $ABC$ werden s a {\displaystyle s_{a}} $s_{a}$ , s b {\displaystyle s_{b}} $s_{b}$ und s c {\displaystyle s_{c}} $s_{c}$ 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_{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_{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_{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)}

s_{a}^{2}+s_{b}^{2}+s_{c}^{2}={\frac {3}{4}}\left(a^{2}+b^{2}+c^{2}\right)

Winkelhalbierende

Wir bezeichnen mit w α {\displaystyle w_{\alpha }} $w_{\alpha }$ , w β {\displaystyle w_{\beta }} $w_{\beta }$ und w γ {\displaystyle w_{\gamma }} $w_{\gamma }$ die Längen der von A {\displaystyle A} $A$ , B {\displaystyle B} $B$ bzw. C {\displaystyle C} $C$ ausgehenden Winkelhalbierenden im Dreieck A B C {\displaystyle ABC} $ABC$ .

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_{\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_{\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}}}

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}}

Allgemeine Trigonometrie in der Ebene

Die trigonometrischen Funktionen am Einheitskreis:

C P ¯ = sin ⁡ b {\displaystyle {\overline {CP}}=\sin b} ${\overline {CP}}=\sin b$	S P ¯ = cos ⁡ b {\displaystyle {\overline {SP}}=\cos b} ${\overline {SP}}=\cos b$
D T ¯ = tan ⁡ b {\displaystyle {\overline {DT}}=\tan b} ${\overline {DT}}=\tan b$	E K ¯ = cot ⁡ b {\displaystyle {\overline {EK}}=\cot b} ${\overline {EK}}=\cot b$
O T ¯ = sec b {\displaystyle {\overline {OT}}=\operatorname {sec} \,b} ${\overline {OT}}=\operatorname {sec} \,b$	O K ¯ = csc b {\displaystyle {\overline {OK}}=\operatorname {csc} \,b} ${\overline {OK}}=\operatorname {csc} \,b$

Periodizität

sin ⁡ x = sin ⁡ ( x + 2 n π ) ; n ∈ Z {\displaystyle \sin x\quad =\quad \sin(x+2n\pi );\quad n\in \mathbb {Z} }

\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} }

\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} }

\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} }

\cot x\quad =\quad \cot(x+n\pi );\quad n\in \mathbb {Z}

Gegenseitige Darstellung

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}}}

\tan x={\frac {\sin x}{\cos x}}

sin 2 ⁡ x + cos 2 ⁡ x = 1 {\displaystyle \sin ^{2}x+\cos ^{2}x=1}

\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+\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}

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}}}

\sin x\;=\;{\sqrt {1-\cos ^{2}x}}

für

x ∈ 3 π 2 , 2 π 270 ∘ , 360 ∘ {\frac {3\pi }{2}},2\pi \right270^{\circ },360^{\circ }{\frac {3\pi }{2}},2\pi \right270^{\circ },360^{\circ } π 2 , 3 π 2 90 ∘ , 270 ∘ {\frac {\pi }{2}},{\frac {3\pi }{2}}\right90^{\circ },270^{\circ }{\frac {\pi }{2}},{\frac {3\pi }{2}}\right90^{\circ },270^{\circ } 3 π 2 , 2 π 270 ∘ , 360 ∘ {\frac {3\pi }{2}},2\pi \right270^{\circ },360^{\circ }{\frac {3\pi }{2}},2\pi \right270^{\circ },360^{\circ } π 2 , 3 π 2 90 ∘ , 270 ∘ {\frac {\pi }{2}},{\frac {3\pi }{2}}\right90^{\circ },270^{\circ }{\frac {\pi }{2}},{\frac {3\pi }{2}}\right90^{\circ },270^{\circ } π 2 , π 90 ∘ , 180 ∘ {\frac {\pi }{2}},\pi \right90^{\circ },180^{\circ }{\frac {\pi }{2}},\pi \right90^{\circ },180^{\circ } 3 π 2 , 2 π 270 ∘ , 360 ∘ {\frac {3\pi }{2}},2\pi \right270^{\circ },360^{\circ }{\frac {3\pi }{2}},2\pi \right270^{\circ },360^{\circ } 3 π 2 , 2 π 270 ∘ , 360 ∘ {\frac {3\pi }{2}},2\pi \right270^{\circ },360^{\circ }{\frac {3\pi }{2}},2\pi \right270^{\circ },360^{\circ } π 2 , 3 π 2 90 ∘ , 270 ∘ {\frac {\pi }{2}},{\frac {3\pi }{2}}\right90^{\circ },270^{\circ }{\frac {\pi }{2}},{\frac {3\pi }{2}}\right90^{\circ },270^{\circ } 0 ∘ , 180 ∘ 0^{\circ },180^{\circ }\right0^{\circ },180^{\circ }\right 180 ∘ , 360 ∘ 180^{\circ },360^{\circ }\right180^{\circ },360^{\circ }\right 270 ∘ , 360 ∘ ] {\displaystyle \cos x>0\quad {\text{für}}\quad x\in \left270^{\circ },360^{\circ }\right]}

\cos x>0\quad {\text{für}}\quad x\in \left270^{\circ },360^{\circ }\right]

cos ⁡ x < 0 für x ∈ ] 90 ∘ , 270 ∘ 90^{\circ },270^{\circ }\right90^{\circ },270^{\circ }\right 0 ∘ , 90 ∘ 180 ∘ , 270 ∘ 0^{\circ },90^{\circ }\right180^{\circ },270^{\circ }\right0^{\circ },90^{\circ }\right180^{\circ },270^{\circ }\right 90 ∘ , 180 ∘ 270 ∘ , 360 ∘ 90^{\circ },180^{\circ }\right270^{\circ },360^{\circ }\right90^{\circ },180^{\circ }\right270^{\circ },360^{\circ }\right

Symmetrien

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}}}

{\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}}

Phasenverschiebungen

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\;}

\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\;}

\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\;}

\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\;}

\cot \left(x+{\frac {\pi }{2}}\right)=-\tan x\;\quad {\text{bzw.}}\quad \cot \left(x+90^{\circ }\right)=-\tan x\;

Rückführung auf spitze Winkel

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)}

\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)}

\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)}

\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

Mit der Bezeichnung t = tan ⁡ x 2 {\displaystyle t=\tan {\tfrac {x}{2}}} $t=\tan {\tfrac {x}{2}}$ gelten die folgenden Beziehungen für beliebiges x {\displaystyle x} $x$

sin ⁡ x = 2 t 1 + t 2 , {\displaystyle \sin x={\frac {2t}{1+t^{2}}},} $\sin x={\frac {2t}{1+t^{2}}},$		cos ⁡ x = 1 − t 2 1 + t 2 , {\displaystyle \cos x={\frac {1-t^{2}}{1+t^{2}}},} $\cos x={\frac {1-t^{2}}{1+t^{2}}},$
tan ⁡ x = 2 t 1 − t 2 , {\displaystyle \tan x={\frac {2t}{1-t^{2}}},} $\tan x={\frac {2t}{1-t^{2}}},$		cot ⁡ x = 1 − t 2 2 t , {\displaystyle \cot x={\frac {1-t^{2}}{2t}},} $\cot x={\frac {1-t^{2}}{2t}},$
sec ⁡ x = 1 + t 2 1 − t 2 , {\displaystyle \sec x={\frac {1+t^{2}}{1-t^{2}}},} $\sec x={\frac {1+t^{2}}{1-t^{2}}},$		csc ⁡ x = 1 + t 2 2 t . {\displaystyle \csc x={\frac {1+t^{2}}{2t}}.} $\csc x={\frac {1+t^{2}}{2t}}.$

Additionstheoreme

Figur 1

Figur 2

Für Sinus und Kosinus lassen sich die Additionstheoreme aus der Verkettung zweier Drehungen um den Winkel x {\displaystyle x} $x$ bzw. y {\displaystyle y} $y$ 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}} $\mathbb {R} ^{2}$ . 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}} $\textstyle e^{i(x+y)}=e^{ix}\cdot e^{iy}$ . Die Ergebnisse für das Doppelvorzeichen ergeben sich durch Anwendung der Symmetrien.

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}

\sin(x\pm y)=\sin x\cdot \cos y\pm \cos x\cdot \sin y

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}

\cos(x\pm y)=\cos x\cdot \cos y\mp \sin x\cdot \sin y

Geometrische Herleitungen sind in Figur 1 und Figur 2 für Winkel α {\displaystyle \alpha } $\alpha$ und β {\displaystyle \beta } $\beta$ zwischen 0° und 90° veranschaulicht.

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 }

\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 }

\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 }

\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 }

\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}} $\textstyle {1 \over \cos x\cos y}$ bzw. 1 sin ⁡ x sin ⁡ y {\displaystyle \textstyle {1 \over \sin x\sin y}} $\textstyle {1 \over \sin x\sin y}$ 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}}}

\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}}}

\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} $x=y$ folgen hieraus die Doppelwinkelfunktionen, für y = π / 2 {\displaystyle y=\pi /2} $y=\pi /2$ 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}

\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}

\cos(x+y)\cdot \cos(x-y)=\cos ^{2}y-\sin ^{2}x=\cos ^{2}x-\sin ^{2}y

Additionstheoreme für Arkusfunktionen

Für die Arkusfunktionen gelten folgende Additionstheoreme

Summanden	Summenformel	Gültigkeitsbereich
arcsin ⁡ x + arcsin ⁡ y = {\displaystyle \arcsin x+\arcsin y=} $\arcsin x+\arcsin y=$	arcsin ⁡ ( x 1 − y 2 + y 1 − x 2 ) {\displaystyle \arcsin \left(x{\sqrt {1-y^{2}}}+y{\sqrt {1-x^{2}}}\right)} $\arcsin \left(x{\sqrt {1-y^{2}}}+y{\sqrt {1-x^{2}}}\right)$	x y ≤ 0 {\displaystyle xy\leq 0} $xy\leq 0$ oder x 2 + y 2 ≤ 1 {\displaystyle x^{2}+y^{2}\leq 1} $x^{2}+y^{2}\leq 1$
	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)} $\pi -\arcsin \left(x{\sqrt {1-y^{2}}}+y{\sqrt {1-x^{2}}}\right)$	x > 0 {\displaystyle x>0} $x>0$ und y > 0 {\displaystyle y>0} $y>0$ und x 2 + y 2 > 1 {\displaystyle x^{2}+y^{2}>1} $x^{2}+y^{2}>1$
	− 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)} $-\pi -\arcsin \left(x{\sqrt {1-y^{2}}}+y{\sqrt {1-x^{2}}}\right)$	x < 0 {\displaystyle x<0} $x<0$ und y < 0 {\displaystyle y<0} $y<0$ und x 2 + y 2 > 1 {\displaystyle x^{2}+y^{2}>1} $x^{2}+y^{2}>1$
arcsin ⁡ x − arcsin ⁡ y = {\displaystyle \arcsin x-\arcsin y=} $\arcsin x-\arcsin y=$	arcsin ⁡ ( x 1 − y 2 − y 1 − x 2 ) {\displaystyle \arcsin \left(x{\sqrt {1-y^{2}}}-y{\sqrt {1-x^{2}}}\right)} $\arcsin \left(x{\sqrt {1-y^{2}}}-y{\sqrt {1-x^{2}}}\right)$	x y ≥ 0 {\displaystyle xy\geq 0} $xy\geq 0$ oder x 2 + y 2 ≤ 1 {\displaystyle x^{2}+y^{2}\leq 1} $x^{2}+y^{2}\leq 1$
	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)} $\pi -\arcsin \left(x{\sqrt {1-y^{2}}}-y{\sqrt {1-x^{2}}}\right)$	x > 0 {\displaystyle x>0} $x>0$ und y < 0 {\displaystyle y<0} $y<0$ und x 2 + y 2 > 1 {\displaystyle x^{2}+y^{2}>1} $x^{2}+y^{2}>1$
	− 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)} $-\pi -\arcsin \left(x{\sqrt {1-y^{2}}}-y{\sqrt {1-x^{2}}}\right)$	x < 0 {\displaystyle x<0} $x<0$ und y > 0 {\displaystyle y>0} $y>0$ und x 2 + y 2 > 1 {\displaystyle x^{2}+y^{2}>1} $x^{2}+y^{2}>1$
arccos ⁡ x + arccos ⁡ y = {\displaystyle \arccos x+\arccos y=} $\arccos x+\arccos y=$	arccos ⁡ ( x y − 1 − x 2 1 − y 2 ) {\displaystyle \arccos \left(xy-{\sqrt {1-x^{2}}}{\sqrt {1-y^{2}}}\right)} $\arccos \left(xy-{\sqrt {1-x^{2}}}{\sqrt {1-y^{2}}}\right)$	x + y ≥ 0 {\displaystyle x+y\geq 0} $x+y\geq 0$
	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)} $2\pi -\arccos \left(xy-{\sqrt {1-x^{2}}}{\sqrt {1-y^{2}}}\right)$	x + y < 0 {\displaystyle x+y<0} $x+y<0$
arccos ⁡ x − arccos ⁡ y = {\displaystyle \arccos x-\arccos y=} $\arccos x-\arccos y=$	− arccos ⁡ ( x y + 1 − x 2 1 − y 2 ) {\displaystyle -\arccos \left(xy+{\sqrt {1-x^{2}}}{\sqrt {1-y^{2}}}\right)} $-\arccos \left(xy+{\sqrt {1-x^{2}}}{\sqrt {1-y^{2}}}\right)$	x ≥ y {\displaystyle x\geq y} $x\geq y$
	arccos ⁡ ( x y + 1 − x 2 1 − y 2 ) {\displaystyle \arccos \left(xy+{\sqrt {1-x^{2}}}{\sqrt {1-y^{2}}}\right)} $\arccos \left(xy+{\sqrt {1-x^{2}}}{\sqrt {1-y^{2}}}\right)$	x < y {\displaystyle x<y} $x<y$
arctan ⁡ x + arctan ⁡ y = {\displaystyle \arctan x+\arctan y=} $\arctan x+\arctan y=$	arctan ⁡ ( x + y 1 − x y ) {\displaystyle \arctan \left({\frac {x+y}{1-xy}}\right)} $\arctan \left({\frac {x+y}{1-xy}}\right)$	x y < 1 {\displaystyle xy<1} $xy<1$
	π + arctan ⁡ ( x + y 1 − x y ) {\displaystyle \pi +\arctan \left({\frac {x+y}{1-xy}}\right)} $\pi +\arctan \left({\frac {x+y}{1-xy}}\right)$	x > 0 {\displaystyle x>0} $x>0$ und x y > 1 {\displaystyle xy>1} $xy>1$
	− π + arctan ⁡ ( x + y 1 − x y ) {\displaystyle -\pi +\arctan \left({\frac {x+y}{1-xy}}\right)} $-\pi +\arctan \left({\frac {x+y}{1-xy}}\right)$	x < 0 {\displaystyle x<0} $x<0$ und x y > 1 {\displaystyle xy>1} $xy>1$
arctan ⁡ x − arctan ⁡ y = {\displaystyle \arctan x-\arctan y=} $\arctan x-\arctan y=$	arctan ⁡ ( x − y 1 + x y ) {\displaystyle \arctan \left({\frac {x-y}{1+xy}}\right)} $\arctan \left({\frac {x-y}{1+xy}}\right)$	x y > − 1 {\displaystyle xy>-1} $xy>-1$
	π + arctan ⁡ ( x − y 1 + x y ) {\displaystyle \pi +\arctan \left({\frac {x-y}{1+xy}}\right)} $\pi +\arctan \left({\frac {x-y}{1+xy}}\right)$	x > 0 {\displaystyle x>0} $x>0$ und x y < − 1 {\displaystyle xy<-1} $xy<-1$
	− π + arctan ⁡ ( x − y 1 + x y ) {\displaystyle -\pi +\arctan \left({\frac {x-y}{1+xy}}\right)} $-\pi +\arctan \left({\frac {x-y}{1+xy}}\right)$	x < 0 {\displaystyle x<0} $x<0$ und x y < − 1 {\displaystyle xy<-1} $xy<-1$

Doppelwinkelfunktionen

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}}}

\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 } $\alpha$ und β {\displaystyle \beta } $\beta$ zwischen 0° und 90° veranschaulicht.

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 )}

{\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 )}

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}}}

\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}}}

\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}}}

\cot(2x)={\frac {\cot ^{2}x-1}{2\cot x}}={\frac {\cot x-\tan x}{2}}

Winkelfunktionen für weitere Vielfache

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}} $z=r\left(\cos \phi +i\sin \phi \right)\iff z^{n}=r^{n}\left(\cos \phi +i\sin \phi \right)^{n}$ 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)} $z^{n}=r^{n}\left(\cos \left(n\phi \right)+i\sin \left(n\phi \right)\right)$ . 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}} $\cos \left(n\phi \right)+i\sin \left(n\phi \right)=\left(\cos \phi +i\sin \phi \right)^{n}$ . Zerlegung in Real- und Imaginärteil liefert dann die Formeln für cos {\displaystyle \cos } $\cos$ und sin {\displaystyle \sin } $\sin$ bzw. die allgemeine Reihendarstellung.

Die Formel für cos ⁡ ( n x ) {\displaystyle \cos(nx)} $\cos(nx)$ steht über T n ( cos ⁡ x ) = cos ⁡ ( n x ) {\displaystyle T_{n}(\cos x)=\cos(nx)} $T_{n}(\cos x)=\cos(nx)$ 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\,}

\sin(3x)=3\sin x-4\sin ^{3}x\,

= sin ⁡ x ( 4 cos 2 ⁡ x − 1 ) {\displaystyle =\;\sin x\left(4\cos ^{2}x-1\right)}

=\;\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}

\sin(4x)=8\sin x\;\cos ^{3}x-4\sin x\;\cos x

= sin ⁡ x ( 8 cos 3 ⁡ x − 4 cos ⁡ x ) {\displaystyle =\;\sin x\left(8\cos ^{3}x-4\cos x\right)}

=\;\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\;}

\sin(5x)=5\sin x-20\sin ^{3}x+16\sin ^{5}x\;

= 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 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 }

\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

= ∑ 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}

=\;\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 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

cos ⁡ ( 3 x ) = 4 cos 3 ⁡ x − 3 cos ⁡ x {\displaystyle \cos(3x)=4\cos ^{3}x-3\cos x\,}

\cos(3x)=4\cos ^{3}x-3\cos x\,

cos ⁡ ( 4 x ) = 8 cos 4 ⁡ x − 8 cos 2 ⁡ x + 1 {\displaystyle \cos(4x)=8\cos ^{4}x-8\cos ^{2}x+1\,}

\cos(4x)=8\cos ^{4}x-8\cos ^{2}x+1\,

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\,}

\cos(5x)=16\cos ^{5}x-20\cos ^{3}x+5\cos x\,

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\,}

\cos(6x)=32\cos ^{6}x-48\cos ^{4}x+18\cos ^{2}x-1\,

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 }

\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

= ∑ 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}

=\;\sum _{j=0}^{\left\lfloor {\frac {n}{2}}\right\rfloor }(-1)^{j}{n \choose 2j}\sin ^{2j}x\;\cos ^{n-2j}x

tan ⁡ ( 3 x ) = 3 tan ⁡ x − tan 3 ⁡ x 1 − 3 tan 2 ⁡ x {\displaystyle \tan(3x)={\frac {3\tan x-\tan ^{3}x}{1-3\tan ^{2}x}}}

\tan(3x)={\frac {3\tan x-\tan ^{3}x}{1-3\tan ^{2}x}}

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}}}

\tan(4x)={\frac {4\tan x-4\tan ^{3}x}{1-6\tan ^{2}x+\tan ^{4}x}}

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}}}

\cot(3x)={\frac {\cot ^{3}x-3\cot x}{3\cot ^{2}x-1}}

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}}}

\cot(4x)={\frac {\cot ^{4}x-6\cot ^{2}x+1}{4\cot ^{3}x-4\cot x}}

Halbwinkelformeln

Figur 4

Zur Berechnung des Funktionswertes des halben Arguments dienen die Halbwinkelformeln, welche sich mittels Substitution aus den Doppelwinkelformeln herleiten lassen:

sin ⁡ x 2 = 1 − cos ⁡ x 2 für x ∈ {\displaystyle \sin {\frac {x}{2}}={\sqrt {\frac {1-\cos x}{2}}}\quad {\text{für}}\quad x\in \left}

\sin {\frac {x}{2}}={\sqrt {\frac {1-\cos x}{2}}}\quad {\text{für}}\quad x\in \left

cos ⁡ x 2 = 1 + cos ⁡ x 2 für x ∈ {\displaystyle \cos {\frac {x}{2}}={\sqrt {\frac {1+\cos x}{2}}}\quad {\text{für}}\quad x\in \left}

\cos {\frac {x}{2}}={\sqrt {\frac {1+\cos x}{2}}}\quad {\text{für}}\quad x\in \left

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)}

\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} }

\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 } $\alpha$ und β {\displaystyle \beta } $\beta$ zwischen 0° und 90° veranschaulicht. 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 }}} $\tan \left({\frac {\alpha }{2}}\right)={\frac {\sin \alpha }{1+\cos \alpha }}={\frac {1-\cos \alpha }{\sin \alpha }}$ .

Außerdem gilt:

tan ⁡ x 2 = tan ⁡ x 1 + 1 + tan 2 ⁡ x für x ∈ ] − π 2 , π 2 -{\tfrac {\pi }{2}},{\tfrac {\pi }{2}}\right-{\tfrac {\pi }{2}},{\tfrac {\pi }{2}}\right 0 , π 0,\pi \right0,\pi \right

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}}}

{\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}}\\\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}}}

\left.{\begin{matrix}\tan x+\tan y={\dfrac {\sin(x+y)}{\cos x\cos y}}\\\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}}\\\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}}}

\left.{\begin{matrix}\cot x+\cot y={\dfrac {\sin(y+x)}{\sin x\sin y}}\\\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}}}

{\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 Winkelfunktionen

Produkte der trigonometrischen Funktionen lassen sich mit folgenden Formeln berechnen:

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 )}}

\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 )}}

\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 )}}

\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}}}

\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}}}

\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}}}

\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 )}}

\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 )}}

\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\;\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 )}}

\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 = 1 2 n − 1 ∑ k 2 = 1 2 ⋯ ∑ k n = 1 2 {\displaystyle \prod _{m=1}^{n}\cos(x_{m})={\frac {1}{2^{n}}}\sum _{k_{1}=1}^{2}\cdots \sum _{k_{n}=1}^{2}\left={\frac {1}{2^{n-1}}}\sum _{k_{2}=1}^{2}\cdots \sum _{k_{n}=1}^{2}\left}

\prod _{m=1}^{n}\cos(x_{m})={\frac {1}{2^{n}}}\sum _{k_{1}=1}^{2}\cdots \sum _{k_{n}=1}^{2}\left={\frac {1}{2^{n-1}}}\sum _{k_{2}=1}^{2}\cdots \sum _{k_{n}=1}^{2}\left

∏ m = 1 n sin ⁡ ( x m ) = 1 ( 2 i ) n ∑ k 1 = 1 2 ⋯ ∑ k n = 1 2 = 1 2 n − 1 ∑ k 2 = 1 2 ⋯ ∑ k n = 1 2 {\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={\frac {1}{2^{n-1}}}\sum _{k_{2}=1}^{2}\cdots \sum _{k_{n}=1}^{2}\left}

\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={\frac {1}{2^{n-1}}}\sum _{k_{2}=1}^{2}\cdots \sum _{k_{n}=1}^{2}\left

Aus der Doppelwinkelfunktion für sin ⁡ ( 2 x ) {\displaystyle \sin(2x)} $\sin(2x)$ folgt außerdem:

sin ⁡ x cos ⁡ x = 1 2 sin ⁡ ( 2 x ) {\displaystyle \sin x\;\cos x={\frac {1}{2}}\sin(2x)}

\sin x\;\cos x={\frac {1}{2}}\sin(2x)

Potenzen der Winkelfunktionen

Sinus sin 2 ⁡ x = 1 2 ( 1 − cos ⁡ ( 2 x ) ) {\displaystyle \sin ^{2}x={\frac {1}{2}}\ {\Big (}1-\cos(2x){\Big )}}

\sin ^{2}x={\frac {1}{2}}\ {\Big (}1-\cos(2x){\Big )}

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 )}}

\sin ^{3}x={\frac {1}{4}}\ {\Big (}3\,\sin x-\sin(3x){\Big )}

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 )}}

\sin ^{4}x={\frac {1}{8}}\ {\Big (}3-4\,\cos(2x)+\cos(4x){\Big )}

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 )}}

\sin ^{5}x={\frac {1}{16}}\ {\Big (}10\,\sin x-5\,\sin(3x)+\sin(5x){\Big )}

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 )}}

\sin ^{6}x={\frac {1}{32}}\ {\Big (}10-15\,\cos(2x)+6\,\cos(4x)-\cos(6x){\Big )}

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={\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={\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}}}

\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}}

Kosinus cos 2 ⁡ x = 1 2 ( 1 + cos ⁡ ( 2 x ) ) {\displaystyle \cos ^{2}x={\frac {1}{2}}\ {\Big (}1+\cos(2x){\Big )}}

\cos ^{2}x={\frac {1}{2}}\ {\Big (}1+\cos(2x){\Big )}

cos 3 ⁡ x = 1 4 ( 3 cos ⁡ x + cos ⁡ ( 3 x ) ) {\displaystyle \cos ^{3}x={\frac {1}{4}}\ {\Big (}3\,\cos x+\cos(3x){\Big )}}

\cos ^{3}x={\frac {1}{4}}\ {\Big (}3\,\cos x+\cos(3x){\Big )}

cos 4 ⁡ x = 1 8 ( 3 + 4 cos ⁡ ( 2 x ) + cos ⁡ ( 4 x ) ) {\displaystyle \cos ^{4}x={\frac {1}{8}}\ {\Big (}3+4\,\cos(2x)+\cos(4x){\Big )}}

\cos ^{4}x={\frac {1}{8}}\ {\Big (}3+4\,\cos(2x)+\cos(4x){\Big )}

cos 5 ⁡ x = 1 16 ( 10 cos ⁡ x + 5 cos ⁡ ( 3 x ) + cos ⁡ ( 5 x ) ) {\displaystyle \cos ^{5}x={\frac {1}{16}}\ {\Big (}10\,\cos x+5\,\cos(3x)+\cos(5x){\Big )}}

\cos ^{5}x={\frac {1}{16}}\ {\Big (}10\,\cos x+5\,\cos(3x)+\cos(5x){\Big )}

cos 6 ⁡ x = 1 32 ( 10 + 15 cos ⁡ ( 2 x ) + 6 cos ⁡ ( 4 x ) + cos ⁡ ( 6 x ) ) {\displaystyle \cos ^{6}x={\frac {1}{32}}\ {\Big (}10+15\,\cos(2x)+6\,\cos(4x)+\cos(6x){\Big )}}

\cos ^{6}x={\frac {1}{32}}\ {\Big (}10+15\,\cos(2x)+6\,\cos(4x)+\cos(6x){\Big )}

cos n ⁡ x = 1 2 n ∑ k = 0 n ( n k ) cos ⁡ ( ( n − 2 k ) x ) ; n ∈ N {\displaystyle \cos ^{n}x={\frac {1}{2^{n}}}\,\sum _{k=0}^{n}{n \choose k}\,\cos((n-2k)x);\quad n\in \mathbb {N} }

\cos ^{n}x={\frac {1}{2^{n}}}\,\sum _{k=0}^{n}{n \choose k}\,\cos((n-2k)x);\quad n\in \mathbb {N}

cos n ⁡ x = 1 2 n ( n n 2 ) + 1 2 n − 1 ∑ k = 0 n 2 − 1 ( n k ) cos ⁡ ( ( n − 2 k ) x ) ; n ∈ N und n gerade {\displaystyle \cos ^{n}x={\frac {1}{2^{n}}}\,{n \choose {\frac {n}{2}}}+{\frac {1}{2^{n-1}}}\sum _{k=0}^{{\frac {n}{2}}-1}{n \choose k}\,\cos {((n-2k)x)};\quad n\in \mathbb {N} {\text{ und }}n{\text{ gerade }}}

\cos ^{n}x={\frac {1}{2^{n}}}\,{n \choose {\frac {n}{2}}}+{\frac {1}{2^{n-1}}}\sum _{k=0}^{{\frac {n}{2}}-1}{n \choose k}\,\cos {((n-2k)x)};\quad n\in \mathbb {N} {\text{ und }}n{\text{ gerade }}

cos n ⁡ x = 1 2 n − 1 ∑ k = 0 n − 1 2 ( n k ) cos ⁡ ( ( n − 2 k ) x ) ; n ∈ N und n ungerade {\displaystyle \cos ^{n}x={\frac {1}{2^{n-1}}}\,\sum _{k=0}^{\frac {n-1}{2}}{n \choose k}\,\cos {((n-2k)x)};\quad n\in \mathbb {N} {\text{ und }}n{\text{ ungerade}}}

\cos ^{n}x={\frac {1}{2^{n-1}}}\,\sum _{k=0}^{\frac {n-1}{2}}{n \choose k}\,\cos {((n-2k)x)};\quad n\in \mathbb {N} {\text{ und }}n{\text{ ungerade}}

Tangens tan 2 ⁡ x = 1 − cos ⁡ ( 2 x ) 1 + cos ⁡ ( 2 x ) = sec 2 ⁡ ( x ) − 1 {\displaystyle \tan ^{2}x={\frac {1-\cos(2x)}{1+\cos(2x)}}=\sec ^{2}(x)-1}

\tan ^{2}x={\frac {1-\cos(2x)}{1+\cos(2x)}}=\sec ^{2}(x)-1

Umrechnung in andere trigonometrische Funktionen

sin ⁡ ( arccos ⁡ x ) = cos ⁡ ( arcsin ⁡ x ) = 1 − x 2 {\displaystyle \sin(\arccos x)=\cos(\arcsin x)={\sqrt {1-x^{2}}}}

\sin(\arccos x)=\cos(\arcsin x)={\sqrt {1-x^{2}}}

sin ⁡ ( arctan ⁡ x ) = cos ⁡ ( arccot ⁡ x ) = x 1 + x 2 {\displaystyle \sin(\arctan x)=\cos(\operatorname {arccot} x)={\frac {x}{\sqrt {1+x^{2}}}}}

\sin(\arctan x)=\cos(\operatorname {arccot} x)={\frac {x}{\sqrt {1+x^{2}}}}

sin ⁡ ( arccot ⁡ x ) = cos ⁡ ( arctan ⁡ x ) = 1 1 + x 2 {\displaystyle \sin(\operatorname {arccot} x)=\cos(\arctan x)={\frac {1}{\sqrt {1+x^{2}}}}}

\sin(\operatorname {arccot} x)=\cos(\arctan x)={\frac {1}{\sqrt {1+x^{2}}}}

tan ⁡ ( arcsin ⁡ x ) = cot ⁡ ( arccos ⁡ x ) = x 1 − x 2 {\displaystyle \tan(\arcsin x)=\cot(\arccos x)={\frac {x}{\sqrt {1-x^{2}}}}}

\tan(\arcsin x)=\cot(\arccos x)={\frac {x}{\sqrt {1-x^{2}}}}

tan ⁡ ( arccos ⁡ x ) = cot ⁡ ( arcsin ⁡ x ) = 1 − x 2 x {\displaystyle \tan(\arccos x)=\cot(\arcsin x)={\frac {\sqrt {1-x^{2}}}{x}}}

\tan(\arccos x)=\cot(\arcsin x)={\frac {\sqrt {1-x^{2}}}{x}}

tan ⁡ ( arccot ⁡ x ) = cot ⁡ ( arctan ⁡ x ) = 1 x {\displaystyle \tan(\operatorname {arccot} x)=\cot(\arctan x)={\frac {1}{x}}}

\tan(\operatorname {arccot} x)=\cot(\arctan x)={\frac {1}{x}}

Weitere Formeln für den Fall α + β + γ = 180°

Die folgenden Formeln gelten für beliebige ebene Dreiecke und folgen nach längeren Termumformungen aus α + β + γ = 180 ∘ {\displaystyle \alpha +\beta +\gamma =180^{\circ }} $\alpha +\beta +\gamma =180^{\circ }$ , solange die in den Formeln vorkommenden Funktionen wohldefiniert sind (Letzteres betrifft nur die Formeln, in denen Tangens und Kotangens vorkommen).

tan ⁡ α + tan ⁡ β + tan ⁡ γ = tan ⁡ α tan ⁡ β tan ⁡ γ cot ⁡ α 2 + cot ⁡ β 2 + cot ⁡ γ 2 = cot ⁡ α 2 cot ⁡ β 2 cot ⁡ γ 2 {\displaystyle {\begin{aligned}\tan \alpha +\tan \beta +\tan \gamma &=\tan \alpha \tan \beta \tan \gamma \\\cot {\frac {\alpha }{2}}+\cot {\frac {\beta }{2}}+\cot {\frac {\gamma }{2}}&=\cot {\frac {\alpha }{2}}\cot {\frac {\beta }{2}}\cot {\frac {\gamma }{2}}\end{aligned}}}

{\begin{aligned}\tan \alpha +\tan \beta +\tan \gamma &=\tan \alpha \tan \beta \tan \gamma \\\cot {\frac {\alpha }{2}}+\cot {\frac {\beta }{2}}+\cot {\frac {\gamma }{2}}&=\cot {\frac {\alpha }{2}}\cot {\frac {\beta }{2}}\cot {\frac {\gamma }{2}}\end{aligned}}

cot ⁡ β cot ⁡ γ + cot ⁡ γ cot ⁡ α + cot ⁡ α cot ⁡ β = 1 tan ⁡ β 2 tan ⁡ γ 2 + tan ⁡ γ 2 tan ⁡ α 2 + tan ⁡ α 2 tan ⁡ β 2 = 1 {\displaystyle {\begin{aligned}\cot \beta \cot \gamma +\cot \gamma \cot \alpha +\cot \alpha \cot \beta &=1\\\tan {\frac {\beta }{2}}\tan {\frac {\gamma }{2}}+\tan {\frac {\gamma }{2}}\tan {\frac {\alpha }{2}}+\tan {\frac {\alpha }{2}}\tan {\frac {\beta }{2}}&=1\end{aligned}}}

{\begin{aligned}\cot \beta \cot \gamma +\cot \gamma \cot \alpha +\cot \alpha \cot \beta &=1\\\tan {\frac {\beta }{2}}\tan {\frac {\gamma }{2}}+\tan {\frac {\gamma }{2}}\tan {\frac {\alpha }{2}}+\tan {\frac {\alpha }{2}}\tan {\frac {\beta }{2}}&=1\end{aligned}}

sin ⁡ α + sin ⁡ β + sin ⁡ γ = 4 cos ⁡ α 2 cos ⁡ β 2 cos ⁡ γ 2 − sin ⁡ α + sin ⁡ β + sin ⁡ γ = 4 cos ⁡ α 2 sin ⁡ β 2 sin ⁡ γ 2 {\displaystyle {\begin{aligned}\sin \alpha +\sin \beta +\sin \gamma &=4\cos {\frac {\alpha }{2}}\cos {\frac {\beta }{2}}\cos {\frac {\gamma }{2}}\\-\sin \alpha +\sin \beta +\sin \gamma &=4\cos {\frac {\alpha }{2}}\sin {\frac {\beta }{2}}\sin {\frac {\gamma }{2}}\end{aligned}}}

{\begin{aligned}\sin \alpha +\sin \beta +\sin \gamma &=4\cos {\frac {\alpha }{2}}\cos {\frac {\beta }{2}}\cos {\frac {\gamma }{2}}\\-\sin \alpha +\sin \beta +\sin \gamma &=4\cos {\frac {\alpha }{2}}\sin {\frac {\beta }{2}}\sin {\frac {\gamma }{2}}\end{aligned}}

cos ⁡ α + cos ⁡ β + cos ⁡ γ = 4 sin ⁡ α 2 sin ⁡ β 2 sin ⁡ γ 2 + 1 − cos ⁡ α + cos ⁡ β + cos ⁡ γ = 4 sin ⁡ α 2 cos ⁡ β 2 cos ⁡ γ 2 − 1 {\displaystyle {\begin{aligned}\cos \alpha +\cos \beta +\cos \gamma &=4\sin {\frac {\alpha }{2}}\sin {\frac {\beta }{2}}\sin {\frac {\gamma }{2}}+1\\-\cos \alpha +\cos \beta +\cos \gamma &=4\sin {\frac {\alpha }{2}}\cos {\frac {\beta }{2}}\cos {\frac {\gamma }{2}}-1\end{aligned}}}

{\begin{aligned}\cos \alpha +\cos \beta +\cos \gamma &=4\sin {\frac {\alpha }{2}}\sin {\frac {\beta }{2}}\sin {\frac {\gamma }{2}}+1\\-\cos \alpha +\cos \beta +\cos \gamma &=4\sin {\frac {\alpha }{2}}\cos {\frac {\beta }{2}}\cos {\frac {\gamma }{2}}-1\end{aligned}}

sin ⁡ 2 α + sin ⁡ 2 β + sin ⁡ 2 γ = 4 sin ⁡ α sin ⁡ β sin ⁡ γ − sin ⁡ 2 α + sin ⁡ 2 β + sin ⁡ 2 γ = 4 sin ⁡ α cos ⁡ β cos ⁡ γ {\displaystyle {\begin{aligned}\sin 2\alpha +\sin 2\beta +\sin 2\gamma &=4\sin \alpha \sin \beta \sin \gamma \\-\sin 2\alpha +\sin 2\beta +\sin 2\gamma &=4\sin \alpha \cos \beta \cos \gamma \end{aligned}}}

{\begin{aligned}\sin 2\alpha +\sin 2\beta +\sin 2\gamma &=4\sin \alpha \sin \beta \sin \gamma \\-\sin 2\alpha +\sin 2\beta +\sin 2\gamma &=4\sin \alpha \cos \beta \cos \gamma \end{aligned}}

cos ⁡ 2 α + cos ⁡ 2 β + cos ⁡ 2 γ = − 4 cos ⁡ α cos ⁡ β cos ⁡ γ − 1 − cos ⁡ 2 α + cos ⁡ 2 β + cos ⁡ 2 γ = − 4 cos ⁡ α sin ⁡ β sin ⁡ γ + 1 {\displaystyle {\begin{aligned}\cos 2\alpha +\cos 2\beta +\cos 2\gamma &=-4\cos \alpha \cos \beta \cos \gamma -1\\-\cos 2\alpha +\cos 2\beta +\cos 2\gamma &=-4\cos \alpha \sin \beta \sin \gamma +1\end{aligned}}}

{\begin{aligned}\cos 2\alpha +\cos 2\beta +\cos 2\gamma &=-4\cos \alpha \cos \beta \cos \gamma -1\\-\cos 2\alpha +\cos 2\beta +\cos 2\gamma &=-4\cos \alpha \sin \beta \sin \gamma +1\end{aligned}}

sin 2 ⁡ α + sin 2 ⁡ β + sin 2 ⁡ γ = 2 cos ⁡ α cos ⁡ β cos ⁡ γ + 2 − sin 2 ⁡ α + sin 2 ⁡ β + sin 2 ⁡ γ = 2 cos ⁡ α sin ⁡ β sin ⁡ γ {\displaystyle {\begin{aligned}\sin ^{2}\alpha +\sin ^{2}\beta +\sin ^{2}\gamma &=2\cos \alpha \cos \beta \cos \gamma +2\\-\sin ^{2}\alpha +\sin ^{2}\beta +\sin ^{2}\gamma &=2\cos \alpha \sin \beta \sin \gamma \end{aligned}}}

{\begin{aligned}\sin ^{2}\alpha +\sin ^{2}\beta +\sin ^{2}\gamma &=2\cos \alpha \cos \beta \cos \gamma +2\\-\sin ^{2}\alpha +\sin ^{2}\beta +\sin ^{2}\gamma &=2\cos \alpha \sin \beta \sin \gamma \end{aligned}}

cos 2 ⁡ α + cos 2 ⁡ β + cos 2 ⁡ γ = − 2 cos ⁡ α cos ⁡ β cos ⁡ γ + 1 − cos 2 ⁡ α + cos 2 ⁡ β + cos 2 ⁡ γ = − 2 cos ⁡ α sin ⁡ β sin ⁡ γ + 1 {\displaystyle {\begin{aligned}\cos ^{2}\alpha +\cos ^{2}\beta +\cos ^{2}\gamma &=-2\cos \alpha \cos \beta \cos \gamma +1\\-\cos ^{2}\alpha +\cos ^{2}\beta +\cos ^{2}\gamma &=-2\cos \alpha \sin \beta \sin \gamma +1\end{aligned}}}

{\begin{aligned}\cos ^{2}\alpha +\cos ^{2}\beta +\cos ^{2}\gamma &=-2\cos \alpha \cos \beta \cos \gamma +1\\-\cos ^{2}\alpha +\cos ^{2}\beta +\cos ^{2}\gamma &=-2\cos \alpha \sin \beta \sin \gamma +1\end{aligned}}

sin 2 ⁡ 2 α + sin 2 ⁡ 2 β + sin 2 ⁡ 2 γ = − 2 cos ⁡ 2 α cos ⁡ 2 β cos ⁡ 2 γ + 2 − sin 2 ⁡ 2 α + sin 2 ⁡ 2 β + sin 2 ⁡ 2 γ = − 2 cos ⁡ 2 α sin ⁡ 2 β sin ⁡ 2 γ {\displaystyle {\begin{aligned}\sin ^{2}2\alpha +\sin ^{2}2\beta +\sin ^{2}2\gamma &=-2\cos 2\alpha \cos 2\beta \cos 2\gamma +2\\-\sin ^{2}2\alpha +\sin ^{2}2\beta +\sin ^{2}2\gamma &=-2\cos 2\alpha \sin 2\beta \sin 2\gamma \end{aligned}}}

{\begin{aligned}\sin ^{2}2\alpha +\sin ^{2}2\beta +\sin ^{2}2\gamma &=-2\cos 2\alpha \cos 2\beta \cos 2\gamma +2\\-\sin ^{2}2\alpha +\sin ^{2}2\beta +\sin ^{2}2\gamma &=-2\cos 2\alpha \sin 2\beta \sin 2\gamma \end{aligned}}

cos 2 ⁡ 2 α + cos 2 ⁡ 2 β + cos 2 ⁡ 2 γ = 2 cos ⁡ 2 α cos ⁡ 2 β cos ⁡ 2 γ + 1 − cos 2 ⁡ 2 α + cos 2 ⁡ 2 β + cos 2 ⁡ 2 γ = 2 cos ⁡ 2 α sin ⁡ 2 β sin ⁡ 2 γ + 1 {\displaystyle {\begin{aligned}\cos ^{2}2\alpha +\cos ^{2}2\beta +\cos ^{2}2\gamma &=2\cos 2\alpha \cos 2\beta \cos 2\gamma +1\\-\cos ^{2}2\alpha +\cos ^{2}2\beta +\cos ^{2}2\gamma &=2\cos 2\alpha \sin 2\beta \sin 2\gamma +1\end{aligned}}}

{\begin{aligned}\cos ^{2}2\alpha +\cos ^{2}2\beta +\cos ^{2}2\gamma &=2\cos 2\alpha \cos 2\beta \cos 2\gamma +1\\-\cos ^{2}2\alpha +\cos ^{2}2\beta +\cos ^{2}2\gamma &=2\cos 2\alpha \sin 2\beta \sin 2\gamma +1\end{aligned}}

sin 2 ⁡ α 2 + sin 2 ⁡ β 2 + sin 2 ⁡ γ 2 = − 2 sin ⁡ α 2 sin ⁡ β 2 sin ⁡ γ 2 + 1 − sin 2 ⁡ α 2 + sin 2 ⁡ β 2 + sin 2 ⁡ γ 2 = − 2 sin ⁡ α 2 cos ⁡ β 2 cos ⁡ γ 2 + 1 {\displaystyle {\begin{aligned}\sin ^{2}{\frac {\alpha }{2}}+\sin ^{2}{\frac {\beta }{2}}+\sin ^{2}{\frac {\gamma }{2}}&=-2\sin {\frac {\alpha }{2}}\sin {\frac {\beta }{2}}\sin {\frac {\gamma }{2}}+1\\-\sin ^{2}{\frac {\alpha }{2}}+\sin ^{2}{\frac {\beta }{2}}+\sin ^{2}{\frac {\gamma }{2}}&=-2\sin {\frac {\alpha }{2}}\cos {\frac {\beta }{2}}\cos {\frac {\gamma }{2}}+1\end{aligned}}}

{\begin{aligned}\sin ^{2}{\frac {\alpha }{2}}+\sin ^{2}{\frac {\beta }{2}}+\sin ^{2}{\frac {\gamma }{2}}&=-2\sin {\frac {\alpha }{2}}\sin {\frac {\beta }{2}}\sin {\frac {\gamma }{2}}+1\\-\sin ^{2}{\frac {\alpha }{2}}+\sin ^{2}{\frac {\beta }{2}}+\sin ^{2}{\frac {\gamma }{2}}&=-2\sin {\frac {\alpha }{2}}\cos {\frac {\beta }{2}}\cos {\frac {\gamma }{2}}+1\end{aligned}}

cos 2 ⁡ α 2 + cos 2 ⁡ β 2 + cos 2 ⁡ γ 2 = 2 sin ⁡ α 2 sin ⁡ β 2 sin ⁡ γ 2 + 2 − cos 2 ⁡ α 2 + cos 2 ⁡ β 2 + cos 2 ⁡ γ 2 = 2 sin ⁡ α 2 cos ⁡ β 2 cos ⁡ γ 2 {\displaystyle {\begin{aligned}\cos ^{2}{\frac {\alpha }{2}}+\cos ^{2}{\frac {\beta }{2}}+\cos ^{2}{\frac {\gamma }{2}}&=2\sin {\frac {\alpha }{2}}\sin {\frac {\beta }{2}}\sin {\frac {\gamma }{2}}+2\\-\cos ^{2}{\frac {\alpha }{2}}+\cos ^{2}{\frac {\beta }{2}}+\cos ^{2}{\frac {\gamma }{2}}&=2\sin {\frac {\alpha }{2}}\cos {\frac {\beta }{2}}\cos {\frac {\gamma }{2}}\end{aligned}}}

{\begin{aligned}\cos ^{2}{\frac {\alpha }{2}}+\cos ^{2}{\frac {\beta }{2}}+\cos ^{2}{\frac {\gamma }{2}}&=2\sin {\frac {\alpha }{2}}\sin {\frac {\beta }{2}}\sin {\frac {\gamma }{2}}+2\\-\cos ^{2}{\frac {\alpha }{2}}+\cos ^{2}{\frac {\beta }{2}}+\cos ^{2}{\frac {\gamma }{2}}&=2\sin {\frac {\alpha }{2}}\cos {\frac {\beta }{2}}\cos {\frac {\gamma }{2}}\end{aligned}}

Sinusoid und Linearkombination mit gleicher Phase

a sin ⁡ α + b cos ⁡ α = { a 2 + b 2 sin ⁡ ( α + arctan ⁡ ( b a ) ) , für alle a > 0 a 2 + b 2 cos ⁡ ( α − arctan ⁡ ( a b ) ) , für alle b > 0 {\displaystyle {\begin{aligned}a\sin \alpha +b\cos \alpha =&{\begin{cases}{\sqrt {a^{2}+b^{2}}}\sin \left(\alpha +\arctan \left({\tfrac {b}{a}}\right)\right)&{\text{, für alle }}a>0\\{\sqrt {a^{2}+b^{2}}}\cos \left(\alpha -\arctan \left({\tfrac {a}{b}}\right)\right)&{\text{, für alle }}b>0\end{cases}}\end{aligned}}}

{\begin{aligned}a\sin \alpha +b\cos \alpha =&{\begin{cases}{\sqrt {a^{2}+b^{2}}}\sin \left(\alpha +\arctan \left({\tfrac {b}{a}}\right)\right)&{\text{, für alle }}a>0\\{\sqrt {a^{2}+b^{2}}}\cos \left(\alpha -\arctan \left({\tfrac {a}{b}}\right)\right)&{\text{, für alle }}b>0\end{cases}}\end{aligned}}

a cos ⁡ α + b sin ⁡ α = sgn ⁡ ( a ) a 2 + b 2 cos ⁡ ( α + arctan ⁡ ( − b a ) ) {\displaystyle {\begin{aligned}a\cos \alpha +b\sin \alpha =\operatorname {sgn}(a){\sqrt {a^{2}+b^{2}}}\cos \left(\alpha +\arctan \left(-{\tfrac {b}{a}}\right)\right)\end{aligned}}}

{\begin{aligned}a\cos \alpha +b\sin \alpha =\operatorname {sgn}(a){\sqrt {a^{2}+b^{2}}}\cos \left(\alpha +\arctan \left(-{\tfrac {b}{a}}\right)\right)\end{aligned}}

a sin ⁡ ( x + α ) + b sin ⁡ ( x + β ) = a 2 + b 2 + 2 a b cos ⁡ ( α − β ) ⋅ sin ⁡ ( x + δ ) , {\displaystyle a\sin(x+\alpha )+b\sin(x+\beta )={\sqrt {a^{2}+b^{2}+2ab\cos(\alpha -\beta )}}\cdot \sin(x+\delta ),}

a\sin(x+\alpha )+b\sin(x+\beta )={\sqrt {a^{2}+b^{2}+2ab\cos(\alpha -\beta )}}\cdot \sin(x+\delta ),

wobei δ = atan2 ⁡ ( a sin ⁡ α + b sin ⁡ β , a cos ⁡ α + b cos ⁡ β ) . {\displaystyle \delta =\operatorname {atan2} (a\sin \alpha +b\sin \beta ,a\cos \alpha +b\cos \beta ).} $\delta =\operatorname {atan2} (a\sin \alpha +b\sin \beta ,a\cos \alpha +b\cos \beta ).$

Allgemeiner ist

∑ i a i sin ⁡ ( x + δ i ) = a sin ⁡ ( x + δ ) , {\displaystyle \sum _{i}a_{i}\sin(x+\delta _{i})=a\sin(x+\delta ),}

\sum _{i}a_{i}\sin(x+\delta _{i})=a\sin(x+\delta ),

wobei

a 2 = ∑ i , j a i a j cos ⁡ ( δ i − δ j ) {\displaystyle a^{2}=\sum _{i,j}a_{i}a_{j}\cos(\delta _{i}-\delta _{j})}

a^{2}=\sum _{i,j}a_{i}a_{j}\cos(\delta _{i}-\delta _{j})

und

δ = atan2 ⁡ ( ∑ i a i sin ⁡ δ i , ∑ i a i cos ⁡ δ i ) . {\displaystyle \delta =\operatorname {atan2} \left(\sum _{i}a_{i}\sin \delta _{i},\sum _{i}a_{i}\cos \delta _{i}\right).}

\delta =\operatorname {atan2} \left(\sum _{i}a_{i}\sin \delta _{i},\sum _{i}a_{i}\cos \delta _{i}\right).

Ableitungen und Stammfunktionen

Siehe Formelsammlung Ableitungen und Stammfunktionen

Bestimmte Integrale

Die Lösungen der nachfolgenden bestimmten Integrale stehen im Zusammenhang mit der Euler’schen Betafunktion, welche weiterhin mit der Gammafunktion verknüpft ist. Das zweite Integral ist z. B. in der Physik bei der Berechnung von Kräften zwischen zylinderförmigen Dauermagneten unter Verwendung der sogenannten Multipol-Entwicklung hilfreich.

∫ 0 π / 2 cos ν 1 ⁡ φ sin ν 2 ⁡ φ d φ = 1 2 ⋅ B ⁡ ( ν 1 + 1 2 , ν 2 + 1 2 ) , Re ( ν j + 1 2 ) > 0 {\displaystyle \int _{0}^{\pi /2}\cos ^{\nu _{1}}\varphi \sin ^{\nu _{2}}\varphi \;{\text{d}}\varphi ={\frac {1}{2}}\cdot \operatorname {B} \left({\frac {\nu _{1}+1}{2}},{\frac {\nu _{2}+1}{2}}\right)\;,\quad {\text{Re}}\left({\frac {\nu _{j}+1}{2}}\right)>0}

\int _{0}^{\pi /2}\cos ^{\nu _{1}}\varphi \sin ^{\nu _{2}}\varphi \;{\text{d}}\varphi ={\frac {1}{2}}\cdot \operatorname {B} \left({\frac {\nu _{1}+1}{2}},{\frac {\nu _{2}+1}{2}}\right)\;,\quad {\text{Re}}\left({\frac {\nu _{j}+1}{2}}\right)>0

∫ 0 π cos ν 1 ⁡ φ sin ν 2 ⁡ φ d φ = B ⁡ ( ν 1 + 1 2 , ν 2 + 1 2 ) ⋅ 1 + ( − 1 ) ν 1 2 , ν j = 0 , 1 , 2 , 3 , … {\displaystyle \int _{0}^{\pi }\cos ^{\nu _{1}}\varphi \sin ^{\nu _{2}}\varphi \;{\text{d}}\varphi =\operatorname {B} \left({\frac {\nu _{1}+1}{2}},{\frac {\nu _{2}+1}{2}}\right)\cdot {\frac {1+(-1)^{\nu _{1}}}{2}},\quad \nu _{j}=0,1,2,3,\dots }

\int _{0}^{\pi }\cos ^{\nu _{1}}\varphi \sin ^{\nu _{2}}\varphi \;{\text{d}}\varphi =\operatorname {B} \left({\frac {\nu _{1}+1}{2}},{\frac {\nu _{2}+1}{2}}\right)\cdot {\frac {1+(-1)^{\nu _{1}}}{2}},\quad \nu _{j}=0,1,2,3,\dots

∫ 0 2 π cos ν 1 ⁡ φ sin ν 2 ⁡ φ d φ = 2 ⋅ B ⁡ ( ν 1 + 1 2 , ν 2 + 1 2 ) ⋅ 1 + ( − 1 ) ν 1 2 ⋅ 1 + ( − 1 ) ν 2 2 , ν j = 0 , 1 , 2 , 3 , … {\displaystyle \int _{0}^{2\pi }\cos ^{\nu _{1}}\varphi \sin ^{\nu _{2}}\varphi \;{\text{d}}\varphi =2\cdot \operatorname {B} \left({\frac {\nu _{1}+1}{2}},{\frac {\nu _{2}+1}{2}}\right)\cdot {\frac {1+(-1)^{\nu _{1}}}{2}}\cdot {\frac {1+(-1)^{\nu _{2}}}{2}},\quad \nu _{j}=0,1,2,3,\dots }

\int _{0}^{2\pi }\cos ^{\nu _{1}}\varphi \sin ^{\nu _{2}}\varphi \;{\text{d}}\varphi =2\cdot \operatorname {B} \left({\frac {\nu _{1}+1}{2}},{\frac {\nu _{2}+1}{2}}\right)\cdot {\frac {1+(-1)^{\nu _{1}}}{2}}\cdot {\frac {1+(-1)^{\nu _{2}}}{2}},\quad \nu _{j}=0,1,2,3,\dots

Reihenentwicklung

Der Sinus (rot) verglichen mit seinem 7. Taylorpolynom (grün)

Wie auch sonst in der Analysis werden alle Winkel im Bogenmaß angegeben.

Man kann zeigen, dass der Kosinus die Ableitung des Sinus darstellt und die Ableitung des Kosinus der negative Sinus ist. Hat man diese Ableitungen, kann man die Taylorreihe entwickeln (am einfachsten mit dem Entwicklungspunkt x = 0 {\displaystyle x=0} $x=0$ ) und zeigen, dass die folgenden Identitäten für alle x {\displaystyle x} $x$ aus den reellen Zahlen gelten. Mit diesen Reihen werden die trigonometrischen Funktionen für komplexe Argumente definiert ( B n {\displaystyle B_{n}} $B_{n}$ bzw. β n {\displaystyle \beta _{n}} $\beta _{n}$ bezeichnet dabei die Bernoulli-Zahlen):

sin ⁡ x = ∑ n = 0 ∞ ( − 1 ) n ( 2 n + 1 ) ! x 2 n + 1 = x − x 3 3 ! + x 5 5 ! − x 7 7 ! ± ⋯ , | x | < ∞ {\displaystyle {\begin{aligned}\sin x&=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n+1)!}}x^{2n+1}\\&=x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}-{\frac {x^{7}}{7!}}\pm \cdots \;,\qquad |x|<\infty \end{aligned}}}

{\begin{aligned}\sin x&=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n+1)!}}x^{2n+1}\\&=x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}-{\frac {x^{7}}{7!}}\pm \cdots \;,\qquad |x|<\infty \end{aligned}}

cos ⁡ x = ∑ n = 0 ∞ ( − 1 ) n ( 2 n ) ! x 2 n = 1 − x 2 2 ! + x 4 4 ! − x 6 6 ! ± ⋯ , | x | < ∞ {\displaystyle {\begin{aligned}\cos x&=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n)!}}x^{2n}\\&=1-{\frac {x^{2}}{2!}}+{\frac {x^{4}}{4!}}-{\frac {x^{6}}{6!}}\pm \cdots \;,\qquad |x|<\infty \end{aligned}}}

{\begin{aligned}\cos x&=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n)!}}x^{2n}\\&=1-{\frac {x^{2}}{2!}}+{\frac {x^{4}}{4!}}-{\frac {x^{6}}{6!}}\pm \cdots \;,\qquad |x|<\infty \end{aligned}}

tan ⁡ x = ∑ n = 1 ∞ ( − 1 ) n 2 2 n ( 1 − 2 2 n ) β 2 n ( 2 n ) ! x 2 n − 1 = ∑ n = 1 ∞ ( − 1 ) n − 1 2 2 n ( 2 2 n − 1 ) B 2 n ( 2 n ) ! x 2 n − 1 = x + 1 3 x 3 + 2 15 x 5 + 17 315 x 7 + 62 2835 x 9 + ⋯ | x | < π 2 {\displaystyle {\begin{aligned}\tan x&=\sum _{n=1}^{\infty }(-1)^{n}{\frac {2^{2n}(1-2^{2n})\beta _{2n}}{(2n)!}}x^{2n-1}=\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}2^{2n}(2^{2n}-1)B_{2n}}{(2n)!}}x^{2n-1}\\&=x+{\frac {1}{3}}x^{3}+{\frac {2}{15}}x^{5}+{\frac {17}{315}}x^{7}+{\frac {62}{2835}}x^{9}+\,\cdots \qquad |x|<{\tfrac {\pi }{2}}\end{aligned}}}

{\begin{aligned}\tan x&=\sum _{n=1}^{\infty }(-1)^{n}{\frac {2^{2n}(1-2^{2n})\beta _{2n}}{(2n)!}}x^{2n-1}=\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}2^{2n}(2^{2n}-1)B_{2n}}{(2n)!}}x^{2n-1}\\&=x+{\frac {1}{3}}x^{3}+{\frac {2}{15}}x^{5}+{\frac {17}{315}}x^{7}+{\frac {62}{2835}}x^{9}+\,\cdots \qquad |x|<{\tfrac {\pi }{2}}\end{aligned}}

cot ⁡ x = 1 x − ∑ n = 1 ∞ ( − 1 ) n − 1 2 2 n β 2 n ( 2 n ) ! x 2 n − 1 = 1 x − ∑ n = 1 ∞ ( − 1 ) n − 1 2 2 n B 2 n ( 2 n ) ! x 2 n − 1 = 1 x − 1 3 x − 1 45 x 3 − 2 945 x 5 − 1 4725 x 7 − ⋯ , 0 < | x | < π {\displaystyle {\begin{aligned}\cot x&={\frac {1}{x}}-\sum _{n=1}^{\infty }{\frac {\left(-1\right)^{n-1}2^{2n}\beta _{2n}}{(2n)!}}x^{2n-1}={\frac {1}{x}}-\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}2^{2n}B_{2n}}{(2n)!}}x^{2n-1}\\&={\frac {1}{x}}-{\frac {1}{3}}x-{\frac {1}{45}}x^{3}-{\frac {2}{945}}x^{5}-{\frac {1}{4725}}x^{7}-\,\cdots ,\qquad 0<|x|<\pi \end{aligned}}}

{\begin{aligned}\cot x&={\frac {1}{x}}-\sum _{n=1}^{\infty }{\frac {\left(-1\right)^{n-1}2^{2n}\beta _{2n}}{(2n)!}}x^{2n-1}={\frac {1}{x}}-\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}2^{2n}B_{2n}}{(2n)!}}x^{2n-1}\\&={\frac {1}{x}}-{\frac {1}{3}}x-{\frac {1}{45}}x^{3}-{\frac {2}{945}}x^{5}-{\frac {1}{4725}}x^{7}-\,\cdots ,\qquad 0<|x|<\pi \end{aligned}}

Produktentwicklung

sin ⁡ ( x ) = x ∏ k = 1 ∞ ( 1 − x 2 k 2 π 2 ) {\displaystyle \sin(x)=x\prod _{k=1}^{\infty }\left(1-{\frac {x^{2}}{k^{2}\pi ^{2}}}\right)}

\sin(x)=x\prod _{k=1}^{\infty }\left(1-{\frac {x^{2}}{k^{2}\pi ^{2}}}\right)

cos ⁡ ( x ) = ∏ k = 1 ∞ ( 1 − 4 x 2 ( 2 k − 1 ) 2 π 2 ) {\displaystyle \cos(x)=\prod _{k=1}^{\infty }\left(1-{\frac {4x^{2}}{(2k-1)^{2}\pi ^{2}}}\right)}

\cos(x)=\prod _{k=1}^{\infty }\left(1-{\frac {4x^{2}}{(2k-1)^{2}\pi ^{2}}}\right)

sin ⁡ ( x ) = ∏ n = − ∞ ∞ ( x + n π π 2 + n π ) {\displaystyle \sin(x)=\prod _{n=-\infty }^{\infty }\left({\frac {x+n\pi }{{\frac {\pi }{2}}+n\pi }}\right)}

\sin(x)=\prod _{n=-\infty }^{\infty }\left({\frac {x+n\pi }{{\frac {\pi }{2}}+n\pi }}\right)

cos ⁡ ( x ) = ∏ n = − ∞ ∞ ( x + n π + π 2 π 2 + n π ) {\displaystyle \cos(x)=\prod _{n=-\infty }^{\infty }\left({\frac {x+n\pi +{\frac {\pi }{2}}}{{\frac {\pi }{2}}+n\pi }}\right)}

\cos(x)=\prod _{n=-\infty }^{\infty }\left({\frac {x+n\pi +{\frac {\pi }{2}}}{{\frac {\pi }{2}}+n\pi }}\right)

tan ⁡ ( x ) = ∏ n = − ∞ ∞ ( x + n π x + n π + π 2 ) {\displaystyle \tan(x)=\prod _{n=-\infty }^{\infty }\left({\frac {x+n\pi }{x+n\pi +{\frac {\pi }{2}}}}\right)}

\tan(x)=\prod _{n=-\infty }^{\infty }\left({\frac {x+n\pi }{x+n\pi +{\frac {\pi }{2}}}}\right)

csc ⁡ ( x ) = ∏ n = − ∞ ∞ ( π 2 + n π x + n π ) {\displaystyle \csc(x)=\prod _{n=-\infty }^{\infty }\left({\frac {{\frac {\pi }{2}}+n\pi }{x+n\pi }}\right)}

\csc(x)=\prod _{n=-\infty }^{\infty }\left({\frac {{\frac {\pi }{2}}+n\pi }{x+n\pi }}\right)

sec ⁡ ( x ) = ∏ n = − ∞ ∞ ( π 2 + n π x + n π + π 2 ) {\displaystyle \sec(x)=\prod _{n=-\infty }^{\infty }\left({\frac {{\frac {\pi }{2}}+n\pi }{x+n\pi +{\frac {\pi }{2}}}}\right)}

\sec(x)=\prod _{n=-\infty }^{\infty }\left({\frac {{\frac {\pi }{2}}+n\pi }{x+n\pi +{\frac {\pi }{2}}}}\right)

cot ⁡ ( x ) = ∏ n = − ∞ ∞ ( x + n π + π 2 x + n π ) {\displaystyle \cot(x)=\prod _{n=-\infty }^{\infty }\left({\frac {x+n\pi +{\frac {\pi }{2}}}{x+n\pi }}\right)}

\cot(x)=\prod _{n=-\infty }^{\infty }\left({\frac {x+n\pi +{\frac {\pi }{2}}}{x+n\pi }}\right)

Zusammenhang mit der komplexen Exponentialfunktion

Ferner besteht zwischen den Funktionen sin ⁡ x {\displaystyle \sin x} $\sin x$ , cos ⁡ x {\displaystyle \cos x} $\cos x$ und der komplexen Exponentialfunktion exp ⁡ ( i x ) {\displaystyle \exp(\mathrm {i} x)} $\exp(\mathrm {i} x)$ folgender Zusammenhang:

exp ⁡ ( ± i x ) = cos ⁡ x ± i sin ⁡ x = e ± i x {\displaystyle \exp(\pm \mathrm {i} x)=\cos x\pm \mathrm {i} \sin x=e^{\pm \mathrm {i} x}}

\exp(\pm \mathrm {i} x)=\cos x\pm \mathrm {i} \sin x=e^{\pm \mathrm {i} x}

(Eulersche Formel)

Weiterhin wird cos ⁡ x + i sin ⁡ x =: cis ⁡ ( x ) {\displaystyle \cos {x}+\mathrm {i} \sin {x}=:\operatorname {cis} (x)} $\cos {x}+\mathrm {i} \sin {x}=:\operatorname {cis} (x)$ geschrieben.

Auf Grund der oben genannten Symmetrien gilt weiter:

cos ⁡ x = exp ⁡ ( i x ) + exp ⁡ ( − i x ) 2 {\displaystyle \cos x={\frac {\exp(\mathrm {i} x)+\exp(-\mathrm {i} x)}{2}}}

\cos x={\frac {\exp(\mathrm {i} x)+\exp(-\mathrm {i} x)}{2}}

sin ⁡ x = exp ⁡ ( i x ) − exp ⁡ ( − i x ) 2 i {\displaystyle \sin x={\frac {\exp(\mathrm {i} x)-\exp(-\mathrm {i} x)}{2\mathrm {i} }}}

\sin x={\frac {\exp(\mathrm {i} x)-\exp(-\mathrm {i} x)}{2\mathrm {i} }}

Mit diesen Beziehungen können einige Additionstheoreme besonders einfach und elegant hergeleitet werden.

Sphärische Trigonometrie

Eine Formelsammlung für das rechtwinklige und das allgemeine Dreieck auf der Kugeloberfläche findet sich in einem eigenen Kapitel.

Literatur, Weblinks

Abramowitz-Stegun: online (Formeln, Theorieteil – ohne den reinen Tabellenteil); eine HTML- oder PDF-Version kann (legal) heruntergeladen werden.

Einzelnachweise

↑ Die Wurzel 2006/04+05, 104ff., ohne Beweis
↑ Joachim Mohr: Kosinus-, Sinus und Tangenswerte, abgerufen am 1. Juni 2016
↑ Ausführliche Beweise in Wikibooks Beweisarchiv.
↑ a b Otto Forster: Analysis 1. Differential- und Integralrechnung einer Veränderlichen. vieweg 1983, Seite 87.
↑ Roger B. Nelsen: Beweise ohne Worte, Deutschsprachige Ausgabe herausgegeben von Nicola Oswald, Springer Spektrum, Springer-Verlag Berlin Heidelberg 2016, ISBN 978-3-662-50330-0, Seite 44
↑ I. N. Bronstein, K. A. Semendjajew: Taschenbuch der Mathematik. 19. Auflage, 1979. B.G. Teubner Verlagsgesellschaft, Leipzig. S. 237.
↑ Roger B. Nelsen: Beweise ohne Worte, Deutschsprachige Ausgabe herausgegeben von Nicola Oswald, Springer Spektrum, Springer-Verlag Berlin Heidelberg 2016, ISBN 978-3-662-50330-0, Seite 46
↑ Milton Abramowitz and Irene A. Stegun, 22.3.15, (s. a. oben „Weblinks“)
↑ Milton Abramowitz and Irene A. Stegun, 4.3.27, (s. a. oben „Weblinks“)
↑ Milton Abramowitz and Irene A. Stegun, 4.3.29, (s. a. oben „Weblinks“)
↑ I. S. Gradshteyn and I. M. Ryzhik, Table of Integrals, Series, and Products, Academic Press, 5th edition (1994). ISBN 0-12-294755-X 1.333.4
↑ I. S. Gradshteyn and I. M. Ryzhik, ebenda 1.331.3 (Bei dieser Formel enthält Gradshteyn/Ryzhik allerdings einen Vorzeichenfehler)
↑ a b c d e f g h i j k l m n o I. N. Bronstein, K. A. Semendjajew, Taschenbuch der Mathematik, B. G. Teubner Verlagsgesellschaft Leipzig. 19. Auflage 1979. 2.5.2.1.3
↑ Milton Abramowitz and Irene A. Stegun, 4.3.28, (s. a. oben „Weblinks“)
↑ Milton Abramowitz and Irene A. Stegun, 4.3.30, (s. a. oben „Weblinks“)
↑ I. S. Gradshteyn and I. M. Ryzhik, ebenda 1.335.4
↑ I. S. Gradshteyn and I. M. Ryzhik, ebenda 1.335.5
↑ I. S. Gradshteyn and I. M. Ryzhik, ebenda 1.331.3
↑ Roger B. Nelsen: Beweise ohne Worte, Deutschsprachige Ausgabe herausgegeben von Nicola Oswald, Springer Spektrum, Springer-Verlag Berlin Heidelberg 2016, ISBN 978-3-662-50330-0, Seite 49
↑ I. S. Gradshteyn and I. M. Ryzhik, ebenda 1.321.1
↑ I. S. Gradshteyn and I. M. Ryzhik, ebenda 1.321.2
↑ I. S. Gradshteyn and I. M. Ryzhik, ebenda 1.321.3
↑ I. S. Gradshteyn and I. M. Ryzhik, ebenda 1.321.4
↑ I. S. Gradshteyn and I. M. Ryzhik, ebenda 1.321.5
↑ I. S. Gradshteyn and I. M. Ryzhik, ebenda 1.323.1
↑ I. S. Gradshteyn and I. M. Ryzhik, ebenda 1.323.2
↑ I. S. Gradshteyn and I. M. Ryzhik, ebenda 1.323.3
↑ I. S. Gradshteyn and I. M. Ryzhik, ebenda 1.323.4
↑ I. S. Gradshteyn and I. M. Ryzhik, ebenda 1.323.5
↑ Weisstein, Eric W.: Harmonic Addition Theorem. Abgerufen am 20. Januar 2018 (englisch).
↑ Milton Abramowitz and Irene A. Stegun, 4.3.67, (s. a. oben „Weblinks“)
↑ Milton Abramowitz and Irene A. Stegun, 4.3.70, (s. a. oben „Weblinks“)
↑ Herbert Amann, Joachim Escher: Analysis I, Birkhäuser Verlag, Basel 2006, 3. Auflage, S. 292 und 298