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July 20th, 2018, 01:01 AM   #11
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$\displaystyle \begin{align*}\sin2x+\sin2y+\sin2z&=2\sin x\cos x+2\sin y\cos y+2\sin z\cos z \\
&=-2\sin x\cos(y+z)-2\sin y\cos(x+z)-2\sin z\cos(x+y) \\
&=-2\sin x(\cos y\cos z-\sin y\sin z)-2\sin y(\cos x\cos z-\sin x\sin z)-2\sin z(\cos x\cos y-\sin x\sin y) \\
&=-2\sin x\cos y\cos z-2\sin y\cos x\cos z-2\sin z\cos x\cos y+6\sin x\sin y\sin z \\
&=-\sin x(\cos(y-z)+\cos(y+z))-\sin y(\cos(x-z)+\cos(x+z))-\sin z(\cos(x-y)+\cos(x+y))+6\sin x\sin y\sin z \\
&=-\sin x\cos(y-z)-\sin x\cos(y+z)-\sin y\cos(x-z)-\sin y\cos(x+z)-\sin z\cos(x-y)-\sin z\cos(x+y)+6\sin x\sin y\sin z \\
&=-\frac{\sin(x+y-z)}{2}-\frac{\sin(x-y+z)}{2}-\frac{\sin(x+y+z)}{2}-\frac{\sin(x-y-z)}{2}-\frac{\sin(x+y-z)}{2}-\frac{\sin(y-x+z)}{2} \\
&-\frac{\sin(x+y+z)}{2}-\frac{\sin(y-x-z)}{2}-\frac{\sin(x+z-y)}{2}-\frac{\sin(z+y-x)}{2}-\frac{\sin(x+y+z)}{2}-\frac{\sin(z-x-y)}{2}+6\sin x\sin y\sin z \\
&=-\frac{\sin2x}{2}-\frac{\sin2y}{2}-\frac{\sin2z}{2}+6\sin x\sin y\sin z \\
\frac32(\sin2x+\sin2y+\sin2z)&=6\sin x\sin y\sin z \\
\sin2x+\sin2y+\sin2z&=4\sin x\sin y\sin z\end{align*}$

Last edited by greg1313; July 20th, 2018 at 07:52 AM.
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