Exercise:
Show that, $\left| \sin \theta +\sin \beta +\sin \gamma \right|\le \frac{3\sqrt{3}}{2}$ whenever $\theta +\beta +\gamma =0$
Solution: we know that $\sqrt{{{x}^{2}}}=\left| x \right|$ and ${{\sin }^{2}}x=1-{{\cos }^{2}}x$
So ${{\left( \sin \theta +\sin \beta +\sin \gamma \right)}^{2}}={{\sin }^{2}}\theta +{{\sin }^{2}}\beta +{{\sin }^{2}}\gamma +2\left( \sin \theta \sin \beta +\sin \beta \sin \gamma +\sin \theta \sin \gamma \right)$
$=1-{{\cos }^{2}}\theta +1-{{\cos }^{2}}\beta +1-{{\cos }^{2}}\gamma +2\left( \sin \theta \sin \beta +\sin \theta \sin \gamma +\sin \beta \sin \gamma \right)$
$=3-\left( {{\cos }^{2}}\theta +{{\cos }^{2}}\beta +{{\cos }^{2}}\gamma \right)+2\left( \sin \theta \sin \beta +\sin \theta \sin \gamma +\sin \beta \sin \gamma \right)$
$=3-\sum\limits_{x=\theta ,\beta ,\gamma }^{{}}{{{\cos }^{2}}x}+2\left( \sin \theta \sin \beta +\sin \theta \sin \gamma +\sin \beta \sin \gamma \right)$
Observe that $2\sin x\sin y=\cos \left( x-y \right)-\cos \left( x+y \right)$
So ${{\left( \sin \theta +\sin \beta +\sin \gamma \right)}^{2}}=3-\sum\limits_{x=\theta ,\beta ,\gamma }{{{\cos }^{2}}x}+\sum\limits_{\begin{smallmatrix}
\left( x,y \right)=\theta ,\beta ,\gamma \\
\,\,\,\,\,\,\,\,\,\,\,\,x\ne y
\end{smallmatrix}}^{{}}{\left( \cos \left( x-y \right)-\cos \left( x+y \right) \right)}$
\[=3-\sum\limits_{x=\theta ,\beta ,\gamma }{{{\cos }^{2}}x}+\sum\limits_{\begin{smallmatrix}
\left( x,y \right)=\theta ,\beta ,\gamma \\
\,\,\,\,\,\,\,\,\,x\ne y
\end{smallmatrix}}{\cos \left( x-y \right)-\sum\limits_{x=\theta ,\beta ,\gamma }{\cos x}}\]
$=3-\left( \sum\limits_{x=\theta ,\beta ,\gamma }{\left( {{\cos }^{2}}x+\cos x \right)} \right)+\sum\limits_{\begin{smallmatrix}
\left( x,y \right)=\alpha ,\beta ,\gamma \\
\,\,\,\,\,\,\,\,\,x\ne y
\end{smallmatrix}}{\cos \left( x-y \right)}$
But ${{\cos }^{2}}x+\cos x={{\cos }^{2}}x+2\frac{1}{2}\cos x+\frac{1}{4}-\frac{1}{4}={{\left( \cos x+\frac{1}{2} \right)}^{2}}-\frac{1}{4}$
So $\sum\limits_{x=\theta ,\beta ,\gamma }{{{\cos }^{2}}x+\cos x}=\sum\limits_{x=\theta ,\beta ,\gamma }{{{\left( \cos x+\frac{1}{2} \right)}^{2}}-\sum\limits_{x=\theta ,\beta ,\gamma }{\frac{1}{4}}}$
So ${{\left( \sin \theta +\sin \beta +\sin \gamma \right)}^{2}}=3+\frac{3}{4}-\sum\limits_{x=\theta ,\beta ,\gamma }^{{}}{{{\left( \cos x+\frac{1}{2} \right)}^{2}}+\sum\limits_{\begin{smallmatrix}
\left( x,y \right)=\theta ,\beta ,\gamma \\
\,\,\,\,\,\,\,\,\,\,x\ne y
\end{smallmatrix}}^{{}}{\cos \left( x-y \right)}}$
But $2{{\sin }^{2}}\left( \frac{u}{2} \right)=1-\cos u\Leftrightarrow \cos u=1-2{{\sin }^{2}}\left( \frac{u}{2} \right)$
So $\sum\limits_{\begin{smallmatrix}
\left( x,y \right)=\theta ,\beta ,\gamma \\
\,\,\,\,\,\,\,\,x\ne y
\end{smallmatrix}}{\cos }\left( x-y \right)=\sum\limits_{\begin{smallmatrix}
\left( x,y \right)=\theta ,\beta ,\gamma \\
\,\,\,\,\,\,\,\,x\ne y
\end{smallmatrix}}{1}-\sum\limits_{\begin{smallmatrix}
\left( x,y \right)=\theta ,\beta ,\gamma \\
\,\,\,\,\,\,\,\,\,x\ne y
\end{smallmatrix}}{2{{\sin }^{2}}\left( \frac{x-y}{2} \right)}$
Thus ${{\left( {{\sin }^{2}}\theta +{{\sin }^{2}}\beta +{{\sin }^{2}}\gamma \right)}^{2}}=\frac{15}{4}+3- \sum\limits_{\begin{smallmatrix}
\left( x,y \right)=\theta ,\beta ,\gamma \\
\,\,\,\,\,\,\,\,\,\,x\ne y
\end{smallmatrix}}\left[{{{\left( \cos x+\frac{1}{2} \right)}^{2}}+2{{\sin }^{2}}}\left( \frac{x-y}{2} \right) \right]$
Let $A=\sum\limits_{\begin{smallmatrix}
\left( x,y \right)=\theta ,\beta ,\gamma \\
\,\,\,\,\,\,\,\,\,x\ne y
\end{smallmatrix}}{{{\left( \cos x+\frac{1}{2} \right)}^{2}}+2{{\sin }^{2}}\left( \frac{x-y}{2} \right)>0}$ hence $-A<0\Leftrightarrow \frac{27}{4}-A<\frac{27}{4}$
thus ${{\left( \sin \theta +\sin \beta +\sin \gamma \right)}^{2}}<\frac{27}{4}\Leftrightarrow \left| \sin \theta +\sin \beta +\sin \gamma \right|<\frac{3\sqrt{3}}{2}$
Remark: Let $U=\left\{ {{a}_{1}},{{a}_{2}},....,{{a}_{n}} \right\}$ be the permutation we define the cyclic sum of $f\left( {{a}_{1}},{{a}_{2}},...{{a}_{n}} \right)$ to be
$\sum\limits_{U}{f\left( {{a}_{1}},{{a}_{2}},....,{{a}_{n}} \right)=f\left( {{a}_{1}},{{a}_{2}},...,{{a}_{n}} \right)+f\left( {{a}_{2}},{{a}_{3}},...,,{{a}_{n}},{{a}_{1}} \right)+....+f\left( {{a}_{n}},{{a}_{1}},...,{{a}_{n-1}} \right)}$
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