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Wolfram Alpha bestätigt das Ergebnis

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Josia Pietsch 2022-03-31 16:01:46 +02:00
parent e156619486
commit b41f724da5
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inputs/intervalle.tex
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@ -324,7 +324,7 @@ Sei $g(x) = c_0 + c_1x+ c_2x^2 + \ldots$.
Es ergibt sich für den quadratischen Term: Es ergibt sich für den quadratischen Term:
\begin{IEEEeqnarray*}{rCl} \begin{IEEEeqnarray*}{rCl}
(\sin \alpha t + \sin \beta t)^2 &=& (\sin \alpha t)^2 + 2 \sin \alpha t \sin \beta t + (\sin \beta t)^2\\ (\sin \alpha t + \sin \beta t)^2 &=& (\sin \alpha t)^2 + 2 \sin \alpha t \sin \beta t + (\sin \beta t)^2\\
&\overset{?}{=}& \frac{1}{2} - \frac{1}{2} \cos 2 \alpha t + \cos(\alpha t - \beta t) \cos (\alpha t + \beta t) &=& \frac{1}{2} - \frac{1}{2} \cos 2 \alpha t + \cos(\alpha t - \beta t) \cos (\alpha t + \beta t)
+ \frac{1}{2} - \frac{1}{2} \cos 2 \beta t\\ + \frac{1}{2} - \frac{1}{2} \cos 2 \beta t\\
\end{IEEEeqnarray*} \end{IEEEeqnarray*}