Abakus findet, dass der Beweis zu einfach war, darum jetzt der 2. Anlauf
a)
$$sin (z)^2= \frac{e^{i2z}+e^{-i2z}-2}{-4}=\frac{1}{2} (1- \frac{e^{i2z}+e^{-i2z}}{2}=\frac{1}{2} (1-\cos (2z))$$
b)
$$2 \sin \frac{z+w}{2} \cos \frac{z-w}{2} =$$
$$ \frac{(e^{i(z+w)/2}-e^{-i(z+w)/2})(e^{i(z-w)/2}+e^{-i(z-w)/2})}{2i}=$$
$$ \frac{e^{iz}+e^{iw}-e^{-iz}-e^{-iw}}{2i}$$$$= \sin z+\sin w $$
c)
$$ \sin (z)(3-4 \sin(z)^{2}) =$$
$$ \frac{e^{iz}-e^{-iz}}{2i} (3-4\frac{e^{i2z}+e^{-i2z}-2}{-4})=$$
$$ \frac{e^{iz}-e^{-iz}}{2i} (1+e^{i2z}+e^{-i2z})=$$
$$ \frac{e^{iz}-e^{-iz} + e^{i3z}-e^{-i3z}-e^{iz}+e^{-iz}}{2i}=$$
$$ \frac{e^{i3z}-e^{-i3z}}{2i}=\sin (3z)$$
(a)
\( \sin (z)^{2}=\frac{1}{2}(1-\cos (2 z) \)
\(1-2 \sin (z)^{2}=\cos (2 z) \)
\(\cos(z)^2- \sin (z)^{2}=\cos (2 z) \)
(b)
\( \sin z+\sin w= \sin ( \frac{z+w}{2} + \frac{z-w}{2}) +\sin ( \frac{z+w}{2} - \frac{z-w}{2} ) = 2 \sin \frac{z+w}{2} \cos \frac{z-w}{2} \)
(c)
\( \sin (2z)= 2 \ sin (z) \cos ( z) \)
\( \cos(2z)=1-2\sin (z)^2 \)
\( \sin (3 z) =\sin (z)\left(3-4 \sin(z)^{2}\right) \).
