Beweisarchiv: Geometrie
Beweis für:
tan ( α ± β ) = tan α ± tan β 1 ∓ tan α ⋅ tan β {\displaystyle \tan(\alpha \pm \beta )={\frac {\tan \alpha \ \pm \tan \beta }{1\mp \tan \alpha \cdot \tan \beta }}}
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 } geteilt
(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 }}} eingesetzt
(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 } negativ:
(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 }}}
(5) und (8) zusammengefasst
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)}}}
Additionstheoreme
geteilt
eingesetzt
negativ:
