Bölüm 1 - İçindekiler (17 Temmuz 2025)
Euler ve de Moivre Formüllerinin Sonuçları
Problem 1: Euler ve de Moivre formüllerini kullanılarak aşağıdaki denklemleri çıkarın:
\[\cos x=\frac{e^{ix}+e^{-ix}}{2}, \tag{1}\]
\[\sin x=\frac{e^{ix}-e^{-ix}}{2i}, \tag{2}\]
\[\sin^2x+\cos^2x=1, \tag{3}\]
\[\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta, \tag{4}\]
\[\sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta, \tag{5}\]
\[\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta, \tag{6}\]
\[\cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta, \tag{7}\]
\[\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}, \tag{8}\]
\[\sin 2\alpha=2\sin\alpha\cos\alpha, \tag{9}\]
\[\cos 2\alpha=\cos^2\alpha-\sin^2\alpha, \tag{10}\]
\[\cos 2\alpha=2\cos^2\alpha-1, \tag{11}\]
\[\cos^2\alpha=\frac{1+\cos 2\alpha}{2}, \tag{12}\]
\[\sin^2\alpha=\frac{1-\cos 2\alpha}{2}, \tag{13}\]
\[\cosh^2\alpha-\sinh^2\alpha=1, \tag{14}\]
\[\cosh\alpha=\cosh(-\alpha)=\sqrt{\sinh^2\alpha+1}, \tag{15}\]
\[\sinh\alpha=-\sinh(-\alpha)=\sqrt{\cosh^2\alpha-1}, \tag{16}\]
\[ \tanh\alpha=\frac{e^{\alpha}-e^{-\alpha}}{e^{\alpha}+e^{-\alpha}}=1-\frac{2}{e^{2\alpha}+1}=-\tanh(-\alpha), \tag{17}\]
\[\sin 4\alpha=4\cos^3\alpha\sin\alpha-4\cos\alpha\sin^3\alpha, \tag{18}\]
\[\cos 4\alpha=\cos^4\alpha+\sin^4\alpha-6\sin^2\alpha\cos^2\alpha, \tag{19}\]
\[\sin\alpha\sin\beta=\frac{1}{2}[\cos(\alpha-\beta)-\cos(\alpha+\beta)]. \tag{20}\]