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Beweis für:
sin
(
α
±
β
)
=
sin
α
⋅
cos
β
±
sin
β
⋅
cos
α
{\displaystyle \sin(\alpha \pm \beta )=\sin \alpha \cdot \cos \beta \pm \sin \beta \cdot \cos \alpha }
△
A
O
B
{\displaystyle \triangle {AOB}}
(1)
sin
(
α
+
β
)
=
A
B
¯
1
=
A
B
¯
{\displaystyle \sin(\alpha +\beta )={\frac {\overline {AB}}{1}}={\overline {AB}}}
Im rechtwinkligen Dreieck
△
B
O
D
{\displaystyle \triangle {BOD}}
(2)
sin
β
=
B
D
¯
1
=
B
D
¯
{\displaystyle \sin \beta ={\frac {\overline {BD}}{1}}={\overline {BD}}}
und
(3)
cos
β
=
O
D
¯
1
=
O
D
¯
{\displaystyle \cos \beta ={\frac {\overline {OD}}{1}}={\overline {OD}}}
Im rechtwinkligen Dreieck
△
O
C
D
{\displaystyle \triangle {OCD}}
(4.1)
sin
α
=
C
D
¯
O
D
¯
{\displaystyle \sin \alpha ={\frac {\overline {CD}}{\overline {OD}}}}
(3) eingesetzt
(4.2)
sin
α
=
C
D
¯
cos
β
{\displaystyle \sin \alpha ={\frac {\overline {CD}}{\cos \beta }}}
(4.3)
sin
α
⋅
cos
β
=
C
D
¯
{\displaystyle \sin \alpha \cdot \cos \beta ={\overline {CD}}}
Die Dreiecke
△
B
S
D
{\displaystyle \triangle {BSD}}
△
O
S
A
{\displaystyle \triangle {OSA}}
∠
B
S
D
=
∠
O
S
A
{\displaystyle \angle {BSD}=\angle {OSA}}
(5.0)
∠
S
O
A
=
∠
S
B
D
=
α
{\displaystyle \angle {SOA}=\angle {SBD}=\alpha }
△
E
D
B
{\displaystyle \triangle {EDB}}
(5.1)
cos
α
=
B
E
¯
B
D
¯
{\displaystyle \cos \alpha ={\frac {\overline {BE}}{\overline {BD}}}}
(2) eingesetzt
(5.2)
cos
α
=
B
E
¯
sin
β
{\displaystyle \cos \alpha ={\frac {\overline {BE}}{\sin \beta }}}
(5.3)
sin
β
⋅
cos
α
=
B
E
¯
{\displaystyle \sin \beta \cdot \cos \alpha ={\overline {BE}}}
(6.1)
A
B
¯
=
C
D
¯
+
B
E
¯
{\displaystyle {\overline {AB}}={\overline {CD}}+{\overline {BE}}}
(4.3) und (5.3) eingesetzt
(6.2)
A
B
¯
=
sin
α
⋅
cos
β
+
sin
β
⋅
cos
α
{\displaystyle {\overline {AB}}=\sin \alpha \cdot \cos \beta +\sin \beta \cdot \cos \alpha }
in (1) eingesetzt
(7)
sin
(
α
+
β
)
=
sin
α
⋅
cos
β
+
sin
β
⋅
cos
α
{\displaystyle \sin(\alpha +\beta )=\sin \alpha \cdot \cos \beta +\sin \beta \cdot \cos \alpha }
β
{\displaystyle \beta }
(8)
sin
(
α
−
β
)
=
sin
α
⋅
cos
(
−
β
)
+
sin
(
−
β
)
⋅
cos
α
{\displaystyle \sin(\alpha -\beta )=\sin \alpha \cdot \cos(-\beta )+\sin(-\beta )\cdot \cos \alpha }
(9a)
cos
(
−
β
)
=
+
cos
β
{\displaystyle \cos(-\beta )=+\cos \beta \,}
und
(9b)
sin
(
−
β
)
=
−
sin
β
{\displaystyle \sin(-\beta )=-\sin \beta \,}
eingesetzt in (8)
(10)
sin
(
α
−
β
)
=
sin
α
⋅
cos
β
−
sin
β
⋅
cos
α
{\displaystyle \sin(\alpha -\beta )=\sin \alpha \cdot \cos \beta -\sin \beta \cdot \cos \alpha }
(7) und (10) zusammengefasst
(11)
sin
(
α
±
β
)
=
sin
α
⋅
cos
β
±
sin
β
⋅
cos
α
{\displaystyle \sin(\alpha \pm \beta )=\sin \alpha \cdot \cos \beta \pm \sin \beta \cdot \cos \alpha }
bei
α
=
β
{\displaystyle \alpha =\beta }
(12)
sin
2
α
=
sin
α
⋅
cos
α
+
sin
α
⋅
cos
α
{\displaystyle \sin 2\alpha =\sin \alpha \cdot \cos \alpha +\sin \alpha \cdot \cos \alpha }
(13)
sin
2
α
=
2
sin
α
⋅
cos
α
{\displaystyle \sin 2\alpha =2\sin \alpha \cdot \cos \alpha }
Beweis für:
sin
(
α
+
β
)
⋅
sin
(
α
−
β
)
{\displaystyle \sin(\alpha +\beta )\cdot \sin(\alpha -\beta )}
(14.1)
sin
(
α
+
β
)
⋅
sin
(
α
−
β
)
=
(
sin
α
⋅
cos
β
+
cos
α
⋅
sin
β
)
(
sin
α
⋅
cos
β
−
cos
α
⋅
sin
β
)
{\displaystyle \sin(\alpha +\beta )\cdot \sin(\alpha -\beta )=(\sin \alpha \cdot \cos \beta +\cos \alpha \cdot \sin \beta )(\sin \alpha \cdot \cos \beta -\cos \alpha \cdot \sin \beta )}
nach dem dritten binomischem Lehrsatz
(14.2)
sin
(
α
+
β
)
⋅
sin
(
α
−
β
)
=
sin
2
α
⋅
cos
2
β
−
cos
2
α
⋅
sin
2
β
{\displaystyle \sin(\alpha +\beta )\cdot \sin(\alpha -\beta )=\sin ^{2}\alpha \cdot \cos ^{2}\beta -\cos ^{2}\alpha \cdot \sin ^{2}\beta }
weil
(15.1)
sin
2
α
=
1
−
cos
2
α
{\displaystyle \sin ^{2}\alpha =1-\cos ^{2}\alpha }
(15.2)
cos
2
α
=
1
−
sin
2
α
{\displaystyle \cos ^{2}\alpha =1-\sin ^{2}\alpha }
in (14.2) eingesetzt
Beweis für:
tan
(
α
±
β
)
=
tan
(
α
)
±
tan
(
β
)
1
∓
tan
(
α
)
⋅
tan
(
β
)
{\displaystyle \tan \left(\alpha \pm \beta \right)={\frac {\tan \left(\alpha \right)\pm \tan \left(\beta \right)}{1\mp \tan \left(\alpha \right)\cdot \tan \left(\beta \right)}}}
Es gilt:
(1)
tan
(
α
+
β
)
=
sin
(
α
+
β
)
cos
(
α
+
β
)
{\displaystyle \tan(\alpha +\beta )={\frac {\sin(\alpha +\beta )}{\cos(\alpha +\beta )}}}
Nach den Additonstheoremen (Sinus) und (Kosinus)
(2.1)
sin
(
α
+
β
)
=
sin
α
⋅
cos
β
+
sin
β
⋅
cos
α
{\displaystyle \sin(\alpha +\beta )=\sin \alpha \cdot \cos \beta +\sin \beta \cdot \cos \alpha }
(2.2)
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 }
in (1) eingestzt
(3)
tan
(
α
+
β
)
=
sin
α
⋅
cos
β
+
sin
β
⋅
cos
α
cos
α
⋅
cos
β
−
sin
α
⋅
sin
β
{\displaystyle \tan(\alpha +\beta )={\frac {\sin \alpha \cdot \cos \beta +\sin \beta \cdot \cos \alpha }{\cos \alpha \cdot \cos \beta -\sin \alpha \cdot \sin \beta }}}
Zähler und Nenner durch
cos
α
⋅
cos
β
{\displaystyle \cos \alpha \cdot \cos \beta }
(4.1)
tan
(
α
+
β
)
=
sin
α
⋅
cos
β
+
sin
β
⋅
cos
α
cos
α
⋅
cos
β
cos
α
⋅
cos
β
−
sin
α
⋅
sin
β
cos
α
⋅
cos
β
{\displaystyle \tan(\alpha +\beta )={\frac {\frac {\sin \alpha \cdot \cos \beta +\sin \beta \cdot \cos \alpha }{\cos \alpha \cdot \cos \beta }}{\frac {\cos \alpha \cdot \cos \beta -\sin \alpha \cdot \sin \beta }{\cos \alpha \cdot \cos \beta }}}}
(4.2)
tan
(
α
+
β
)
=
sin
α
cos
α
⋅
1
+
1
⋅
sin
β
cos
β
1
−
sin
α
⋅
sin
β
cos
α
⋅
cos
β
{\displaystyle \tan(\alpha +\beta )={\frac {{\frac {\sin \alpha }{\cos \alpha }}\cdot 1+1\cdot {\frac {\sin \beta }{\cos \beta }}}{1-{\frac {\sin \alpha \cdot \sin \beta }{\cos \alpha \cdot \cos \beta }}}}}
mit
tan
=
sin
cos
{\displaystyle \tan ={\frac {\sin }{\cos }}}
(5)
tan
(
α
+
β
)
=
tan
α
+
tan
β
1
−
tan
α
⋅
tan
β
{\displaystyle \tan(\alpha +\beta )={\frac {\tan \alpha +\tan \beta }{1-\tan \alpha \cdot \tan \beta }}}
Wenn Winkel
β
{\displaystyle \beta }
(6)
tan
(
α
−
β
)
=
tan
α
+
tan
(
−
β
)
1
−
tan
α
⋅
tan
(
−
β
)
{\displaystyle \tan(\alpha -\beta )={\frac {\tan \alpha +\tan(-\beta )}{1-\tan \alpha \cdot \tan(-\beta )}}}
weil
(7)
tan
(
−
β
)
=
−
tan
β
{\displaystyle \tan(-\beta )=-\tan \beta }
(8)
tan
(
α
−
β
)
=
tan
α
−
tan
β
1
+
tan
α
⋅
tan
β
{\displaystyle \tan(\alpha -\beta )={\frac {\tan \alpha -\tan \beta }{1+\tan \alpha \cdot \tan \beta }}}
tan
(
α
±
β
)
=
tan
(
α
)
±
tan
(
β
)
1
∓
tan
(
α
)
⋅
tan
(
β
)
{\displaystyle \tan \left(\alpha \pm \beta \right)={\frac {\tan \left(\alpha \right)\pm \tan \left(\beta \right)}{1\mp \tan \left(\alpha \right)\cdot \tan \left(\beta \right)}}}
Es gilt:
(1)
cot
(
α
+
β
)
=
cos
(
α
+
β
)
sin
(
α
+
β
)
{\displaystyle \cot(\alpha +\beta )={\frac {\cos(\alpha +\beta )}{\sin(\alpha +\beta )}}}
Nach den Additonstheoremen (Sinus) und (Kosinus)
(2.1)
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 }
(2.2)
sin
(
α
+
β
)
=
sin
α
⋅
cos
β
+
sin
β
⋅
cos
α
{\displaystyle \sin(\alpha +\beta )=\sin \alpha \cdot \cos \beta +\sin \beta \cdot \cos \alpha }
in (1) eingestzt
(3)
cot
(
α
+
β
)
=
cos
α
⋅
cos
β
−
sin
α
⋅
sin
β
sin
α
⋅
cos
β
+
sin
β
⋅
cos
α
{\displaystyle \cot(\alpha +\beta )={\frac {\cos \alpha \cdot \cos \beta -\sin \alpha \cdot \sin \beta }{\sin \alpha \cdot \cos \beta +\sin \beta \cdot \cos \alpha }}}
Zähler und Nenner durch
sin
α
⋅
sin
β
{\displaystyle \sin \alpha \cdot \sin \beta }
(4.1)
cot
(
α
+
β
)
=
cos
α
⋅
cos
β
−
sin
α
⋅
sin
β
sin
α
⋅
sin
β
sin
α
⋅
cos
β
+
sin
β
⋅
cos
α
sin
α
⋅
sin
β
{\displaystyle \cot(\alpha +\beta )={\frac {\frac {\cos \alpha \cdot \cos \beta -\sin \alpha \cdot \sin \beta }{\sin \alpha \cdot \sin \beta }}{\frac {\sin \alpha \cdot \cos \beta +\sin \beta \cdot \cos \alpha }{\sin \alpha \cdot \sin \beta }}}}
(4.2)
cot
(
α
+
β
)
=
cos
α
⋅
cos
β
sin
α
⋅
sin
β
−
1
1
⋅
cos
β
sin
β
+
1
⋅
cos
α
sin
α
{\displaystyle \cot(\alpha +\beta )={\frac {{\frac {\cos \alpha \cdot \cos \beta }{\sin \alpha \cdot \sin \beta }}-1}{1\cdot {\frac {\cos \beta }{\sin \beta }}+1\cdot {\frac {\cos \alpha }{\sin \alpha }}}}}
mit
cot
=
cos
sin
{\displaystyle \cot ={\frac {\cos }{\sin }}}
(5)
cot
(
α
+
β
)
=
cot
α
⋅
cot
β
−
1
cot
β
+
cot
α
{\displaystyle \cot(\alpha +\beta )={\frac {\cot \alpha \cdot \cot \beta -1}{\cot \beta +\cot \alpha }}}
β
{\displaystyle \beta }
(6)
cot
(
α
−
β
)
=
cot
α
⋅
cot
(
−
β
)
−
1
cot
(
−
β
)
+
cot
α
{\displaystyle \cot(\alpha -\beta )={\frac {\cot \alpha \cdot \cot(-\beta )-1}{\cot(-\beta )+\cot \alpha }}}
weil
(7)
cot
(
−
β
)
=
−
cot
β
{\displaystyle \cot(-\beta )=-\cot \beta }
(8.1)
cot
(
α
−
β
)
=
−
cot
α
⋅
cot
β
−
1
−
cot
β
+
cot
α
{\displaystyle \cot(\alpha -\beta )={\frac {-\cot \alpha \cdot \cot \beta -1}{-\cot \beta +\cot \alpha }}}
Zähler und Nenner mal -1
(8.2)
cot
(
α
−
β
)
=
−
cot
α
⋅
cot
β
−
1
−
cot
β
+
cot
α
{\displaystyle \cot(\alpha -\beta )={\frac {-\cot \alpha \cdot \cot \beta -1}{-\cot \beta +\cot \alpha }}}
(9)
cot
(
α
±
β
)
=
cot
α
⋅
cot
β
∓
1
cot
β
±
cot
α
{\displaystyle \cot(\alpha \pm \beta )={\frac {\cot \alpha \cdot \cot \beta \mp 1}{\cot \beta \pm \cot \alpha }}}
