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\date{\today{}}
\author{Thomas Kuster}
\title{UWIS, Mathematik 1, Lösung Serie 12}

\begin{document}
\markright{UWIS, Mathematik 1, Lösung Serie 12}
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\maketitle

\section{DGL mit Taylorpolynomen lösen}
\subsection{}
\subsection{}

\section{}
\subsection{exakte Werte}
\begin{equation*}\begin{split}
\cos{75\grad{}} + i\sin{75\grad{}} =& e^{i75\grad{}} = e^{i\left(\frac{\pi}{4}\frac{\pi}{6}\right)} = e^{i\left(\frac{\pi}{4}\right)} e^{i\left(\frac{\pi}{6}\right)}\\
=& \left(\cos{\frac{\pi}{4}}+i\sin{\frac{\pi}{4}}\right)\left(\cos{\frac{\pi}{6}}+i\sin{\frac{\pi}{6}}\right)\\
=& \left(\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}\right)\left(\frac{\sqrt{3}}{2}+i\frac{1}{2}\right)\\
=& \frac{\sqrt{3}}{2\sqrt{2}}+i\frac{1}{2\sqrt{2}}+i\frac{\sqrt{3}}{2\sqrt{2}}-\frac{1}{2\sqrt{2}}
= \frac{\sqrt{3}-1}{2\sqrt{2}} + i\frac{\sqrt{3}+1}{2\sqrt{2}}
\end{split}\end{equation*}
Somit ist $\sin{75\grad{}} = \frac{\sqrt{3}+1}{2\sqrt{2}}$ und $\cos{75\grad{}} = \frac{\sqrt{3}-1}{2\sqrt{2}}$.

\begin{equation*}\begin{split}
\cos{105\grad{}} + i\sin{105\grad{}} =& e^{i(75\grad{}+30\grad{})} = \left(\frac{\sqrt{3}-1}{2\sqrt{2}} + i\frac{\sqrt{3}+1}{2\sqrt{2}}\right) e^{i\frac{\pi}{6}}\\
=& \left(\frac{\sqrt{3}-1}{2\sqrt{2}} + i\frac{\sqrt{3}+1}{2\sqrt{2}}\right)\left(\frac{\sqrt{3}}{2}+i\frac{1}{2}\right)\\
=& \frac{\left(\sqrt{3}-1\right)\sqrt{3}}{4\sqrt{2}}+i\frac{\sqrt{3}-1}{4\sqrt{2}}+i\frac{\left(\sqrt{3}+1\right)\sqrt{3}}{4\sqrt{2}}-\frac{\sqrt{3}+1}{4\sqrt{2}}\\
=&\frac{3-\sqrt{3}-1-\sqrt{3}}{4\sqrt{2}} + i\frac{\sqrt{3}-1+\sqrt{3}+3}{4\sqrt{2}}\\
=&\frac{2-2\sqrt{3}}{4\sqrt{2}} + i\frac{2\sqrt{3}-2}{4\sqrt{2}} = \frac{1-\sqrt{3}}{2\sqrt{2}} + i\frac{\sqrt{3}+1}{2\sqrt{2}}\\
\end{split}\end{equation*}
Somit ist $\sin{105\grad{}} = \frac{\sqrt{3}+1}{2\sqrt{2}}$ und $\cos{105\grad{}} = \frac{1-\sqrt{3}}{2\sqrt{2}}$.

\subsection{Funktionen von $3\alpha$}
\begin{equation*}\begin{split}
e^{i3\alpha} =& e^{i\alpha}e^{i\alpha}e^{i\alpha} = (\cos{\alpha}+i\sin{\alpha})(\cos{\alpha}+i\sin{\alpha})(\cos{\alpha}+i\sin{\alpha})\\
=& (\cos^2{\alpha}+2i\sin{\alpha}\cos{\alpha}+\sin^2{\alpha})(\cos{\alpha}+i\sin{\alpha})\\
=& \cos^3{\alpha}+2i\sin{\alpha}\cos^2{\alpha}-\sin^2{\alpha}\cos{\alpha}+i\sin{\alpha}\cos^2{\alpha}\\
&\qquad+2i^2\sin^2{\alpha}\cos{\alpha}-i\sin^3{\alpha}\\
=& \cos^3{\alpha}-3\sin^2{\alpha}\cos{\alpha} +i(3\sin{\alpha}\cos^2{\alpha}-\sin^3{\alpha})\\
=& \cos^3{\alpha}-3(1-\cos^2{\alpha})\cos{\alpha} +i(3\sin{\alpha}(1-\sin^2{\alpha})-\sin^3{\alpha})\\
=& 4\cos^3{\alpha}-3\cos{\alpha} +i(3\sin{\alpha}-4\sin^3{\alpha})
\end{split}\end{equation*}
\begin{eqnarray*}
\sin{3\alpha} & = & 3\sin{\alpha}-4\sin^3{\alpha}\\
\cos{3\alpha} & = & 4\cos^3{\alpha}-3\cos{\alpha}
\end{eqnarray*}

\subsection{Linearfaktorzerlegung von $x^4-2x^2-8$}
Substitution mit $y = x^2 \Rightarrow x^4-2x^2-8 = y^2-2y-8 = (y-2)(y+4)$. Durch Rücksubstitution ergibt sich:
\begin{eqnarray*}
\sqrt{y_1} & = & \pm \sqrt{2}\\
\sqrt{y_2} & = & \pm i2\\
\end{eqnarray*}
Somit ist die Linearfaktorzerlegung $(x+\sqrt{2})(x-\sqrt{2})(x+i2)(x-i2)$.

\subsection{Polynom mit gegebenen Nullstellen}
\begin{equation*}\begin{split}
(x-1)(x-(1+i))(x-(1-i)) = (x-1)(x-1-i)(x-1+i)\\
= (x-1)(x^2-x+ix-x+1-i-ix+i+1)\\
= (x-1)(x^2-2x+2) = x^3-2x^2+2x-x^2+2x-2\\
= x^3-3x^2+4x-2
\end{split}\end{equation*}

\section{Differentialgleichung}


\end{document}
