Benutzer:Antonsusi/Tangente

(Intro)

Additionstheorem

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

Halbwinkelformel

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

Austauschformeln

Ausgangsformeln

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

Für zwei Winkel a und b ergibt sich

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

4 in 3.1 und 3.2 einsetzen

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

5.1 und 5.2 in 2 einsetzen

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

7 in das Additionstheorem einsetzen.

8

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

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

9

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

Kürzen

10

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

Die mittlere Tangentensteigung $c$ ist also

11

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