Question

If \[
x+ y+ z = \pi \text { then} \\
\sin(2x) + \sin(2y) + \sin(2 z) = 4 \sin(x) \sin(y) \sin(z)
\]
Try to mimic the proof of
http://openstudy.com/study#/updates/4fbcc61ae4b0556534319ae4

Answer 1

\[ 4 \frac{1}{-8i}\left ( \frac{e^{i2x}-1}{e^{ix}}\times \frac{e^{i2y}-1}{e^{iy}}\times\frac{e^{i2z}-1}{e^{iz}} \right ) \]
\[ \frac{-1}{2i} \left ( \frac{e^{i(2x+2y+2z)} - e^{i2(x+y)}- e^{i2(z+y)}- e^{i2(x+z)} + e^{i2x} + e^{i2z}+ e^{i2y}- 1}{e^{i(x+y+z)}} \right ) \]
\[ \frac{-1}{2i} \left ( \frac{e^{i2\pi} - e^{i2(\pi - z)}- e^{i2(\pi - y)}- e^{i2(\pi - x)} + e^{i2x} + e^{i2z}+ e^{i2y}- 1}{-1} \right ) \]
\[ \frac{1}{2i} \left ( e^{i2\pi} - e^{i2(\pi - z)}- e^{i2(\pi - y)}- e^{i2(\pi - x)} + e^{i2x} + e^{i2z}+ e^{i2y} - 1 \right ) \]
\[ \frac{1}{2i} \left ( 1 - e^{i2(\pi - z)}- e^{i2(\pi - y)}- e^{i2(\pi - x)} + e^{i2x} + e^{i2z}+ e^{i2y} - 1 \right ) \]
\[ \frac{1}{2i} \left ( (e^{i2x} - e^{-i2x}) + (e^{i2y} - e^{-i2y}) +(e^{i2z} - e^{-i2z}) \right ) \]
\[ \sin 2x + \sin 2y + \sin 2z \]