Benutzer:Antonsusi/Tangente

(Intro)

Additionstheorem

${\displaystyle \tan(x+y)={\frac {\tan x+\tan y}{1-\tan x\;\tan y}}}$

Halbwinkelformel

${\displaystyle \tan \left({\frac {x}{2}}\right)={\frac {sin(x)}{1+cos(x)}}}$

Austauschformeln

Ausgangsformeln

Additionstheorem:	${\displaystyle \tan(x+y)={\frac {\tan x+\tan y}{1-\tan x\;\tan y}}}$
Halbwinkelformel:	${\displaystyle \tan \left({\frac {x}{2}}\right)={\frac {sin(x)}{1+cos(x)}}}$
Austauschformel Sinus	${\displaystyle \sin(x)={\frac {\tan x}{\sqrt {1+\tan ^{2}x}}}}$
Austauschformel Cosinus	${\displaystyle \cos(x)={\frac {1}{\sqrt {1+\tan ^{2}x}}}}$
Steigungswinkel	für ${\displaystyle y=mx+n}$ gilt ${\displaystyle m=\tan \phi }$

Für zwei Winkel a und b ergibt sich

Nr	a	b
2	${\displaystyle \tan \left({\frac {a}{2}}\right)={\frac {\sin(a)}{1+\cos(a)}}}$	${\displaystyle \tan \left({\frac {b}{2}}\right)={\frac {\sin(b)}{1+\cos(b)}}}$
3.1	${\displaystyle \sin(a)={\frac {\tan a}{\sqrt {1+\tan ^{2}a}}}}$	${\displaystyle \sin(b)={\frac {\tan b}{\sqrt {1+\tan ^{2}b}}}}$
3.2	${\displaystyle \cos(a)={\frac {1}{\sqrt {1+\tan ^{2}a}}}}$	${\displaystyle \cos(b)={\frac {1}{\sqrt {1+\tan ^{2}b}}}}$
4.	${\displaystyle \tan(a)=p}$	${\displaystyle \tan(b)=q}$

4 in 3.1 und 3.2 einsetzen

Nr	a	b
5.1	${\displaystyle \sin(a)={\frac {p}{\sqrt {1+p^{2}}}}}$	${\displaystyle \sin(b)={\frac {q}{\sqrt {1+q^{2}}}}}$
5.2	${\displaystyle \cos(a)={\frac {1}{\sqrt {1+p^{2}}}}}$	${\displaystyle \cos(b)={\frac {1}{\sqrt {1+q^{2}}}}}$

5.1 und 5.2 in 2 einsetzen

Nr	a	b
6	${\displaystyle \tan \left({\frac {a}{2}}\right)={\frac {\frac {p}{\sqrt {1+p^{2}}}}{1+{\frac {1}{\sqrt {1+p^{2}}}}}}}$	${\displaystyle \tan \left({\frac {b}{2}}\right)={\frac {\frac {q}{\sqrt {1+q^{2}}}}{1+{\frac {1}{\sqrt {1+q^{2}}}}}}}$
7	${\displaystyle \tan \left({\frac {a}{2}}\right)={\frac {p}{{\sqrt {1+p^{2}}}+1}}}$	${\displaystyle \tan \left({\frac {b}{2}}\right)={\frac {q}{{\sqrt {1+q^{2}}}+1}}}$

7 in das Additionstheorem einsetzen.

8

${\displaystyle \tan \left(\left({\frac {a}{2}}\right)+\left({\frac {b}{2}}\right)\right)={\frac {{\frac {p}{{\sqrt {1+p^{2}}}+1}}+{\frac {q}{{\sqrt {1+q^{2}}}+1}}}{1-{\frac {p}{{\sqrt {1+p^{2}}}+1}}\;{\frac {q}{{\sqrt {1+q^{2}}}+1}}}}}$

Die 1 mit ${\displaystyle ({\sqrt {1+p^{2}}}+1)\;({\sqrt {1+q^{2}}}+1)}$ erweitern und gleichnamig machen

9

${\displaystyle \tan \left(\left({\frac {a}{2}}\right)+\left({\frac {b}{2}}\right)\right)={\frac {{\frac {p\;({\sqrt {1+q^{2}}}+1)}{({\sqrt {1+p^{2}}}+1)\;({\sqrt {1+q^{2}}}+1)}}+{\frac {q\;({\sqrt {1+p^{2}}}+1)}{({\sqrt {1+p^{2}}}+1)\;({\sqrt {1+q^{2}}}+1)}}}{{\frac {({\sqrt {1+p^{2}}}+1)\;({\sqrt {1+q^{2}}}+1)}{({\sqrt {1+p^{2}}}+1)\;({\sqrt {1+q^{2}}}+1)}}-{\frac {p}{{\sqrt {1+p^{2}}}+1}}\;{\frac {q}{{\sqrt {1+q^{2}}}+1}}}}}$

Kürzen

10

${\displaystyle \tan \left(\left({\frac {a}{2}}\right)+\left({\frac {b}{2}}\right)\right)={\frac {p\;({\sqrt {1+q^{2}}}+1)+q\;({\sqrt {1+p^{2}}}+1)}{({\sqrt {1+p^{2}}}+1)\;({\sqrt {1+q^{2}}}+1)-p\;q}}}$

Die mittlere Tangentensteigung ${\displaystyle c}$ ist also

11

${\displaystyle c=\tan \left(\left({\frac {a}{2}}\right)+\left({\frac {b}{2}}\right)\right)={\frac {p\;({\sqrt {1+q^{2}}}+1)+q\;({\sqrt {1+p^{2}}}+1)}{({\sqrt {1+p^{2}}}+1)\;({\sqrt {1+q^{2}}}+1)-p\;q}}}$