\[ \begin{array}{ll} \sin z_1\cos z_2+\sin z_2\cos z_1 \\[1.6ex] \quad\displaystyle =\,\frac{1}{4i}\,(e^{iz_1}-e^{-iz_1})(e^{iz_2}+e^{-iz_2}) +\frac{1}{4i}\,(e^{iz_2}-e^{-iz_2})(e^{iz_1}+e^{-iz_1}) \\[1.6ex] \quad\displaystyle =\,\frac{1}{4i}\,(e^{i(z_1+z_2)}+e^{i(z_1-z_2)}-e^{-i(z_1-z_2)}-e^{-i(z_1+z_2)}) \\[1.6ex] \qquad\quad\displaystyle +\,\frac{1}{4i}\,(e^{i(z_2+z_1)}+e^{i(z_2-z_1)}-e^{-i(z_2-z_1)}-e^{-i(z_2+z_1)}) \\[1.6ex] \quad\displaystyle =\,\frac{1}{2i}\,(e^{i(z_1+z_2)}-e^{-i(z_1+z_2)}) \\[1.6ex] \quad\displaystyle =\,\sin(z_1+z_2). \end{array} \]

\[ \begin{array}{ll} \cos z_1\cos z_2-\sin z_1\sin z_2 \\[1.6ex] \quad\displaystyle =\,\frac{1}{4}\,(e^{iz_1}+e^{-iz_1})(e^{iz_2}+e^{-iz_2})+\frac{1}{4}\,(e^{iz_1}-e^{-iz_1})(e^{iz_2}-e^{-iz_2}) \\[1.6ex] \quad\displaystyle =\,\frac{1}{4}\,(e^{i(z_1+z_2)}+e^{i(z_1-z_2)}+e^{-i(z_1-z_2)}+e^{-i(z_1+z_2)}) \\[1.6ex] \qquad\quad\displaystyle +\,\frac{1}{4}\,(e^{i(z_1+z_2)}-e^{i(z_1-z_2)}-e^{-i(z_1-z_2)}+e^{-i(z_1+z_2)}) \\[1.6ex] \quad\displaystyle =\,\frac{1}{2}\,(e^{i(z_1+z_2)}+e^{-i(z_1+z_2)}) \\[1.6ex] \quad\displaystyle =\,\cos(z_1+z_2). \end{array} \] Damit ist alles gezeigt.\( \qquad\Box \)

Zunächst ist für alle \( z\in\mathbb C \) \[ \begin{array}{lll} \sin^2z+\cos^2z & = & \displaystyle -\,\frac{1}{4}\,(e^{iz}-e^{-iz})^2+\frac{1}{4}\,(e^{iz}+e^{-iz})^2 \\[1.6ex] & = & \displaystyle -\,\frac{1}{4}\,(e^{2iz}+e^{-2iz}-2)+\frac{1}{4}\,(e^{2iz}+e^{-2iz}+2) \\[1.6ex] & = & \frac{1}{4}\cdot 2+\frac{1}{4}\cdot 2\,=\,1. \end{array} \]

\[ \sin 2z=\sin(z+z)=\sin z\cos z+\sin z\cos z=2\sin z\cos z. \]

\[ \cos 2z=\cos(z+z)=\cos z\cos z-\sin z\sin z=\cos^2z-\sin^2z. \]

\[ 1-\cos 2z=(\sin^2z+\cos^2z)-(\cos^2z-\sin^2z)=2\sin^2z. \]

\[ 1+\cos 2z=(\sin^2z+\cos^2z)+(\cos^2z-\sin^2z)=2\cos^2z. \] Damit ist alles gezeigt.\( \qquad\Box \)

\begin{align} \cos x &= \sum_{k=0}^\infty\frac{(-1)^k}{(2k)!}\,x^{2k}=1-\frac{x^2}{2!}+\frac{x^4}{4!}+\ldots\,, \\[1.6ex] \sin x &= \sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)!}\,x^{2k+1}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\ldots \end{align}

\[ \cos 0=1,\quad \sin 0=0. \]

\[ \frac{d}{dx}\,\cos x=-\sin x\lt 0,\quad \frac{d}{dx}\,\sin x=\cos x\gt 0 \quad\mbox{für alle}\ 0\le x\le\frac{\pi}{2}\,, \]

\[ \cos\frac{\pi}{2}=0,\quad \sin\frac{\pi}{2}=\sqrt{1-\cos^2\frac{\pi}{2}}=1. \]

\begin{align} &\cos\left(\frac{\pi}{2}-x\right)=\cos\frac{\pi}{2}\cos x-\sin\frac{\pi}{2}\sin x=-\sin(-x)=\sin x, \\[1.6ex] &\sin\left(\frac{\pi}{2}-x\right)=\cos\frac{\pi}{2}\sin x+\sin\frac{\pi}{2}\cos x=\cos x. \end{align}

\[ \frac{d}{dx}\,\sin x=\cos x\gt 0\quad\mbox{für alle}\ 0\lt x\lt\frac{\pi}{2} \]

\[ \frac{d}{dx}\,\cos x=-\sin x\lt 0\quad\mbox{für alle}\ 0\lt x\lt\frac{\pi}{2} \]

\[ \sin\left(x-\frac{\pi}{2}\right)=-\cos x,\quad \cos x=-\sin\left(x-\frac{\pi}{2}\right) \]

Damit ist alles gezeigt.\( \qquad\Box \)

\[ \cos^2\frac{z}{2}=\frac{1+\cos z}{2}\,. \]

\[ \sin^2\frac{z}{2}=\frac{1-\cos z}{2}\,. \]

\[ \cos\frac{z}{2}=\sqrt{\frac{1+\cos x}{2}}\,,\quad x\in[-\pi,\pi]. \]

\[ \sin\frac{x}{2}=\sqrt{\frac{1-\cos x}{2}}\,,\quad x\in[0,2\pi]. \] Damit ist alles gezeigt.\( \qquad\Box \)

\[ \cos\frac{\pi}{4}=\sqrt{\frac{1+\cos\frac{\pi}{2}}{2}}=\sqrt{\frac{1}{2}}=\frac{\sqrt{2}}{2}\,. \]

\[ \sin\frac{\pi}{4}=\sqrt{\frac{1-\cos\frac{\pi}{2}}{2}}=\sqrt{\frac{1}{2}}=\frac{\sqrt{2}}{2}\,. \]

\[ \cos\frac{\pi}{8} =\sqrt{\frac{1+\cos\frac{\pi}{2}}{2}} =\sqrt{\frac{1+\frac{\sqrt{2}}{2}}{2}} =\sqrt{\frac{2}{4}+\frac{\sqrt{2}}{4}} =\frac{\sqrt{2+\sqrt{2}}}{2} \]

\[ \sin\frac{\pi}{8} =\sqrt{\frac{1-\cos\frac{\pi}{4}}{2}} =\sqrt{\frac{1-\frac{\sqrt{2}}{2}}{2}} =\sqrt{\frac{2}{4}-\frac{\sqrt{2}}{4}} =\frac{\sqrt{2-\sqrt{2}}}{2}\,. \]

\begin{align} \sin(3x) &= \sin(x+2x) =\sin x\cos 2x+\sin 2x\cos x \\[0.6ex] &= \sin x(\cos^2x-\sin^2x)+2\sin x\cos^2x =3\sin x\cos^2x-\sin^3 x \\[0.6ex] &= 3\sin x(1-\sin^2x)-\sin^3x =3\sin x-4\sin^3x \end{align}

\[ 1=\sin\frac{\pi}{2}=3\sin\frac{\pi}{6}-4\sin^3\frac{\pi}{6}\,. \]

\[ f(x):=4x^3-3x+1,\quad x\in[0,1], \]

\[ f'(x)=12x^2-3,\quad f''(x)=24x. \]

\[ \sin\frac{\pi}{6}=\frac{1}{2}\,,\quad \cos\frac{\pi}{6}=\sqrt{1-\frac{1}{4}}=\frac{\sqrt{3}}{2}\,. \]

\[ \cos\frac{\pi}{12}=\sqrt{\frac{1+\cos\frac{\pi}{6}}{2}}=\sqrt{\frac{1}{2}+\frac{\sqrt{3}}{4}}=\frac{\sqrt{2+\sqrt{3}}}{2} \]

\[ \sin\frac{\pi}{12}=\sqrt{1-\frac{2+\sqrt{3}}{4}}=\sqrt{\frac{2-\sqrt{3}}{4}}=\frac{\sqrt{2-\sqrt{3}}}{2}\,. \] Das war gesucht.\( \qquad\Box \)

Es sei \( z=x+iy. \) Wir ermitteln zunächst \[ z^2=(x+iy)^2=x^2-y^2+2ixy. \] Es bedeute \( \psi \) das Argument von \( z^2, \) also \[ z=|z|(\cos\psi+i\sin\psi)=x^2-y^2+2ixy \] bzw. nach Vergleich \[ \cos\psi=\frac{x^2-y^2}{x^2+y^2}\,,\quad \sin\psi=\frac{2xy}{x^2+y^2}\,. \] Es ist also insbesondere \[ \cos\psi =\frac{x^2-y^2}{x^2+y^2} =\left(\frac{x}{\sqrt{x^2+y^2}}\right)^2-\left(\frac{y}{\sqrt{x^2+y^2}}\right)^2 =\left(\frac{\mbox{Re}\,z}{|z|}\right)^2-\left(\frac{\mbox{Im}\,z}{|z|}\right)^2 \] und damit \[ \cos\psi=\cos^2\varphi-\sin^2\varphi. \] Das vergleichen wir mit der Winkelverdopplungsformel \( \cos 2\varphi=\cos^2\varphi-\sin^2\varphi \) und erhalten \[ \cos\psi=\cos 2\varphi. \] Analog berechnen wir \[ \sin\psi =\frac{2xy}{x^2+y^2} =2\,\frac{x}{\sqrt{x^2+y^2}}\,\frac{y}{\sqrt{x^2+y^2}} =2\cos\varphi\sin\varphi, \] und auch das vergleichen wir mit der Winkelverdopplungsformel \( \sin 2\varphi=2\sin\varphi\cos\varphi, \) also \[ \sin\psi=\sin 2\varphi. \] Das war gefordert.\( \qquad\Box \)

Die Behauptung ist für \( k=1 \) richtig und für \( k=2 \) nach PA 64. Sei also nun \[ z^k=r^k(\cos k\varphi+i\sin k\varphi) \] für alle \( k=1,\ldots,n \) für ein \( n\in\mathbb N \) richtig. Unter Benutzung der Additionstheoreme berechnen wir \begin{align} z^{n+1} &= z\cdot z^n=r(\cos\varphi+i\sin\varphi)\cdot r^n(\cos n\varphi+i\sin n\varphi) \\[0.6ex] &= r^{n+1}\,\big\{(\cos\varphi\cos n\varphi-\sin\varphi\sin n\varphi)+i(\cos\varphi\sin n\varphi+\sin\varphi\cos n\varphi)\big\} \\[0.6ex] &= r^{n+1}\,\big\{\cos(\varphi+n\varphi)+i\sin(\varphi+n\varphi)\big\} \\[0.6ex] &= r^{n+1}\,\big\{\cos(n+1)\varphi+i\sin(n+1)\varphi\}\,. \end{align} Das Prinzip der vollständigen Induktion beweist die Behauptung.\( \qquad\Box \)

\[ z^{100}=\sqrt{2}^{100}\left(\cos\frac{100\pi}{4}+i\sin\frac{100\pi}{4}\right)=\sqrt{2}^{100}(\cos\pi+i\sin\pi)=-\sqrt{2}^{100}\,. \]

\[ z^{90}=1^{90}\left(\cos\frac{90\pi}{4}+i\sin\frac{90\pi}{4}\right)=\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}=i. \]

\[ z^{84}=1^{84}\left(\cos\frac{84\pi}{6}+i\sin\frac{84\pi}{6}\right)=\cos 0+i\sin 0=1. \] Damit sind alle gesuchten Größen ermittelt.\( \qquad\Box \)

\[ \cos z=\frac{1}{2}\,(e^{iz}+e^{-iz}),\quad \sin z=\frac{1}{2i}\,(e^{iz}-e^{-iz}) \]

\begin{align} \cos(iz) &= \frac{1}{2}\,(e^{i\cdot iz}+e^{-i\cdot iz})=\frac{1}{2}\,(e^{-z}+e^{z})=\frac{1}{2}\,(e^z+e^{-z})=\cosh z, \\[1.6ex] \sin(iz) &= \frac{1}{2i}\,(e^{i\cdot iz}-e^{-i\cdot iz})=\frac{1}{2i}\,(e^{-z}-e^{z})=-\,\frac{1}{2i}\,(e^z-e^{-z}) \\[1.6ex] &= \frac{i}{2}\,(e^z-e^{-z})=i\sinh z. \end{align}

\[ \begin{array}{l} \sin x\cos+i\cos x\sinh y \\[1.6ex] \quad\displaystyle =\,\frac{1}{4i}\,(e^{ix}-e^{-ix})(e^y+e^{-y})+\frac{i}{4}\,(e^{ix}+e^{-ix})(e^y-e^{-y}) \\[1.6ex] \quad\displaystyle =\,\frac{1}{4i}\,(e^{ix+y}+e^{ix-y}-e^{-ix+y}-e^{-ix-y}) \\[1.6ex] \qquad\quad\displaystyle -\,\frac{1}{4i}\,(e^{ix+y}-e^{ix-y}+e^{-ix+y}-e^{-ix-y}) \\[1.6ex] \quad\displaystyle =\,\frac{1}{4i}\,(2e^{ix-y}-2e^{-ix+y}) \,=\,\frac{1}{2i}\,(e^{ix-y}-e^{-ix+y}) \\[1.6ex] \quad\displaystyle =\,\frac{1}{2i}\,(e^{ix+i\cdot iy}-e^{-ix-i\cdot iy}) \,=\,\frac{1}{2i}\,(e^{i(x+iy)}-e^{-i(x+iy)}) \\[1.6ex] \quad\displaystyle =\,\frac{1}{2i}\,(e^{iz}-e^{-iz}) \,=\,\sin z \end{array} \]

\[ \begin{array}{l} \cos x\cosh y-i\sin x\sinh y \\[1.6ex] \quad\displaystyle =\,\frac{1}{4}\,(e^{ix}+e^{-ix})(e^y+e^{-y}-\frac{i}{4i}\,(e^{ix}-e^{-ix})(e^y-e^{-y}) \\[1.6ex] \quad\displaystyle =\,\frac{1}{4}\,(e^{ix+y}+e^{ix-y}+e^{-ix+y}+e^{-ix-y}) \\[1.6ex] \qquad\quad\displaystyle -\,\frac{1}{4}\,(e^{ix+y}-e^{ix-y}-e^{-ix+y}+e^{-ix-y}) \\[1.6ex] \quad\displaystyle =\,\frac{1}{4}\,(2e^{ix-y}+2e^{-ix+y}) \,=\,\frac{1}{2}\,(e^{ix-y}+e^{-ix+y}) \\[1.6ex] \quad\displaystyle =\,\frac{1}{2}\,(e^{ix+i\cdot iy}+e^{-ix-i\cdot iy}) \,=\,\frac{1}{2}\,(e^{i(x+iy)}+e^{-i(x+iy)}) \\[1.6ex] \quad\displaystyle =\,\frac{1}{2}\,(e^{iz}+e^{-iz}) \,=\,\cos z. \end{array} \]

\[ \sin(iy)=i\sinh y\quad\mbox{und damit}\quad|\sin(iy)|=|\sinh y|\longrightarrow\infty\ \mbox{für}\ y\to\infty \]

\[ \cos(iy)=\cosh y\quad\mbox{und damit}\quad|\cos(iy)|=\cosh y\longrightarrow\infty\ \mbox{für}\ y\to\infty\,. \]

\[ \tan\colon\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\longrightarrow\mathbb R,\quad \arctan\colon\mathbb R\longrightarrow\left(-\frac{\pi}{2},\frac{\pi}{2}\right). \]

\[ \frac{d}{dx}\,\tan x =\frac{d}{dx}\,\frac{\sin x}{\cos x} =\frac{cos^2x+\sin^2x}{\cos^2x} =\frac{1}{\cos^2x}\,. \]

\[ \frac{1}{\cos^2x} =\frac{\sin^2x+\cos^2x}{\cos^2x} =\frac{\sin^2}{\cos^2x}+1 =\tan^2x+1 \]

\[ \frac{d}{dy}\,\arctan y =\frac{1}{\frac{1}{\cos^2(\arctan y)}} =\frac{1}{\tan^2(\arctan y)+1} =\frac{1}{1+y^2}\,. \]

Damit ist alles gezeigt.\( \qquad\Box \)