Euler ve de Moivre Formüllerinin Sonuçları
Sedat Han
FPD 1-1(2025), Sayfa [1-9]
Elektronik yayın tarihi: 28 Ekim 2025
Problemler:
Euler ve de Moivre formüllerini kullanılarak aşağıdaki denklemleri çıkarın:
\[\begin{equation}\label{1.1}\tag{1.1}
\cos x=\frac{e^{ix}+e^{-ix}}{2},
\end{equation}\]
\[\begin{equation}\label{1.2}\tag{1.2}
\sin x=\frac{e^{ix}-e^{-ix}}{2i},
\end{equation}\]
\[\begin{equation}\label{1.3}\tag{1.3}
\sin^2x+\cos^2x=1,
\end{equation}\]
\[\begin{equation}\label{1.4}\tag{1.4}
\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta,
\end{equation}\]
\[\begin{equation}\label{1.5}\tag{1.5}
\sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta,
\end{equation}\]
\[\begin{equation}\label{1.6}\tag{1.6}
\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta,
\end{equation}\]
\[\begin{equation}\label{1.7}\tag{1.7}
\cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta,
\end{equation}\]
\[\begin{equation}\label{1.8}\tag{1.8}
\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta},
\end{equation}\]
\[\begin{equation}\label{1.9}\tag{1.9}
\sin 2\alpha=2\sin\alpha\cos\alpha,
\end{equation}\]
\[\begin{equation}\label{1.10}\tag{1.10}
\cos 2\alpha=\cos^2\alpha-\sin^2\alpha,
\end{equation}\]
\[\begin{equation}\label{1.11}\tag{1.11}
\cos 2\alpha=2\cos^2\alpha-1,
\end{equation}\]
\[\begin{equation}\label{1.12}\tag{1.12}
\cos^2\alpha=\frac{1+\cos 2\alpha}{2},
\end{equation}\]
\[\begin{equation}\label{1.13}\tag{1.13}
\sin^2\alpha=\frac{1-\cos 2\alpha}{2},
\end{equation}\]
\[\begin{equation}\label{1.14}\tag{1.14}
\cosh^2\alpha-\sinh^2\alpha=1,
\end{equation}\]
\[\begin{equation}\label{1.15}\tag{1.15}
\cosh\alpha=\cosh(-\alpha)=\sqrt{\sinh^2\alpha+1},
\end{equation}\]
\[\begin{equation}\label{1.16}\tag{1.16}
\sinh\alpha=-\sinh(-\alpha)=\sqrt{\cosh^2\alpha-1},
\end{equation}\]
\[\begin{equation}\label{1.17}\tag{1.17}
\tanh\alpha=\frac{e^{\alpha}-e^{-\alpha}}{e^{\alpha}+e^{-\alpha}}=1-\frac{2}{e^{2\alpha}+1}=-\tanh(-\alpha),
\end{equation}\]
\[\begin{equation}\label{1.18}\tag{1.18}
\sin 4\alpha=4\cos^3\alpha\sin\alpha-4\cos\alpha\sin^3\alpha,
\end{equation}\]
\[\begin{equation}\label{1.19}\tag{1.19}
\cos 4\alpha=\cos^4\alpha+\sin^4\alpha-6\sin^2\alpha\cos^2\alpha,
\end{equation}\]
\[\begin{equation}\label{1.20}\tag{1.20}
\sin\alpha\sin\beta=\frac{1}{2}[\cos(\alpha-\beta)-\cos(\alpha+\beta)].
\end{equation}\]