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remove enumerate from proof

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christian
2019-12-23 19:59:35 +01:00
parent a67d263b1c
commit 95752d21d5
2 changed files with 14 additions and 19 deletions
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ws2019/ana/lectures/analysis.pdf
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ws2019/ana/lectures/analysis18.tex
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@@ -170,25 +170,20 @@
\end{enumerate}
\end{satz}
\begin{proof}
\begin{enumerate}
\item \begin{align*}
\cos (x+y) + i \sin (x+y) &= e^{i(x+y)} = e^{ix} \cdot e^{iy} \\
&= (\cos x + i \sin x)(\cos y + i \sin y) \\
&= \underbrace{\cos x \cos y - \sin x \sin y}_{\text{Re}} + i \underbrace{(\sin x \cos y + \cos x \sin y)}_{\text{Im}}
.\end{align*}
\item Setze $u := \frac{x+y}{2}, v := \frac{x - y}{2}$.
$x = u + v, y = u-v$.\\
\begin{align*}
\sin x - \sin y &= \sin (u+v) - \sin (u - v) \\
&= \sin u \cdot \cos v + \cos u \cdot \sin v
- (\sin u \underbrace{\cos(-v)}_{= \cos v}
+ \cos u \cdot \underbrace{\sin(-v)}_{- \sin v}) \\
&= 2 \cos u \sin v
= 2 \cos \frac{x+y}{2} \cdot \sin \frac{x - y}{2}
.\end{align*}
\end{enumerate}
\begin{proof} 1. Mit $e^{ix} = \cos x + i \sin x$ folgt direkt
\begin{align*}
\cos (x+y) + i \sin (x+y) &= e^{i(x+y)} = e^{ix} \cdot e^{iy} \\
&= (\cos x + i \sin x)(\cos y + i \sin y) \\
&= \underbrace{\cos x \cos y - \sin x \sin y}_{\text{Re}} + i \underbrace{(\sin x \cos y + \cos x \sin y)}_{\text{Im}}
\intertext{2. Setze $u := \frac{x+y}{2}, v := \frac{x - y}{2}$.
$x = u + v, y = u-v$.}
\sin x - \sin y &= \sin (u+v) - \sin (u - v) \\
&= \sin u \cdot \cos v + \cos u \cdot \sin v
- (\sin u \underbrace{\cos(-v)}_{= \cos v}
+ \cos u \cdot \underbrace{\sin(-v)}_{- \sin v}) \\
&= 2 \cos u \sin v
= 2 \cos \frac{x+y}{2} \cdot \sin \frac{x - y}{2}
.\end{align*}
\end{proof}
\end{document}