Question Number 72021 by mathmax by abdo last updated on 23/Oct/19
$${calculate}\:{interms}\:{of}\:{n}\:\:{U}_{{n}} =\sum_{\mathrm{1}\leqslant{i}<{j}\leqslant{n}} \:{sin}\left(\frac{{i}\pi}{{n}}\right){sin}\left(\frac{{j}\pi}{{n}}\right) \\ $$
Commented by mathmax by abdo last updated on 26/Oct/19
$${we}\:{have}\:\left(\sum_{{i}=\mathrm{1}} ^{{n}} {sin}\left({x}_{{i}} \right)\right)^{\mathrm{2}} =\sum_{{i}=\mathrm{1}} ^{{n}} \:{sin}^{\mathrm{2}} \left({x}_{{i}} \right)\:+\mathrm{2}\sum_{\mathrm{1}\leqslant{i}<{j}\leqslant{n}} \:{sin}\left(\:\:{x}_{{i}} \right){sin}\left({x}_{{j}} \right) \\ $$$${let}\:{x}_{{i}} =\frac{{i}\pi}{{n}}\:\Rightarrow\left(\sum_{{i}=\mathrm{1}} ^{{n}} {sin}\left(\frac{{i}\pi}{{n}}\right)^{\mathrm{2}} \right)=\sum_{{i}=\mathrm{1}} ^{{n}} {sin}^{\mathrm{2}} \left(\frac{{i}\pi}{{n}}\right)+\mathrm{2}\sum_{\mathrm{1}\leqslant{i}<{j}\leqslant{n}} {sin}\left(\frac{{i}\pi}{{n}}\right){sin}\left(\frac{{j}\pi}{{n}}\right) \\ $$$$\Rightarrow\sum_{\mathrm{1}\leqslant{i}<{j}\leqslant{j}} {sin}\left(\frac{{i}\pi}{{n}}\right){sin}\left(\frac{{j}\pi}{{n}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{\:\left(\sum_{{i}=\mathrm{1}} ^{{n}} \:{sin}\left(\frac{{i}\pi}{{n}}\right)\right)^{\mathrm{2}} −\sum_{{i}=\mathrm{1}} ^{{n}} \:{sin}^{\mathrm{2}} \left(\frac{{i}\pi}{{n}}\right)\right\} \\ $$$${let}\:{A}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} {sin}\left(\frac{{k}\pi}{{n}}\right)\:\Rightarrow{A}_{{n}} ={Im}\left(\sum_{{k}=\mathrm{0}} ^{{n}} \:{e}^{\frac{{ik}\pi}{{n}}} \right)\:{and} \\ $$$$\sum_{{i}=\mathrm{0}} ^{{n}} \:{e}^{\frac{{ik}\pi}{{n}}} \:=\sum_{{i}.\mathrm{0}} ^{{n}} \:\left({e}^{{i}\frac{\pi}{{n}}} \right)^{{k}} \:=\frac{\mathrm{1}−\left({e}^{{i}\frac{\pi}{{n}}} \right)^{{n}+\mathrm{1}} }{\mathrm{1}−{e}^{\frac{{i}\pi}{{n}}} }\:=\frac{\mathrm{1}−{e}^{{i}\frac{\left({n}+\mathrm{1}\right)\pi}{{n}}} }{\mathrm{1}−{e}^{\frac{{i}\pi}{{n}}} } \\ $$$$=\frac{\mathrm{1}−{cos}\left(\frac{\left({n}+\mathrm{1}\right)\pi}{{n}}\right)−{isin}\left(\frac{\left({n}+\mathrm{1}\right)\pi}{{n}}\right)}{\mathrm{1}−{cos}\left(\frac{\pi}{{n}}\right)−{isin}\left(\frac{\pi}{{n}}\right)} \\ $$$$=\frac{\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{\left({n}+\mathrm{1}\right)\pi}{\mathrm{2}{n}}\right)−\mathrm{2}{isin}\left(\frac{\left({n}+\mathrm{1}\right)\pi}{\mathrm{2}{n}}\right){cos}\left(\frac{\left({n}+\mathrm{1}\right)\pi}{\mathrm{2}{n}}\right)}{\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}{n}}\right)−\mathrm{2}{isin}\left(\frac{\pi}{\mathrm{2}{n}}\right){cos}\left(\frac{\pi}{\mathrm{2}{n}}\right)}\: \\ $$$$=\frac{−{isin}\left(\frac{\left({n}+\mathrm{1}\right)\pi}{\mathrm{2}{n}}\right){e}^{{i}\frac{\left({n}+\mathrm{1}\right)\pi}{\mathrm{2}{n}}} }{−{isin}\left(\frac{\pi}{\mathrm{2}{n}}\right){e}^{\frac{{i}\pi}{\mathrm{2}{n}}} }\:=\frac{{sin}\left(\frac{\left({n}+\mathrm{1}\right)\pi}{\mathrm{2}{n}}\right)}{{sin}\left(\frac{\pi}{\mathrm{2}{n}}\right)}\left({i}\right)\:\Rightarrow \\ $$$${A}_{{n}} =\frac{{sin}\left(\frac{\left({n}+\mathrm{1}\right)\pi}{\mathrm{2}{n}}\right)}{{sin}\left(\frac{\pi}{\mathrm{2}{n}}\right)}\:\:{let}\:\:{calculate}\:\sum_{{k}=\mathrm{1}} ^{{n}} \:{sin}^{\mathrm{2}} \left(\frac{{k}\pi}{{n}}\right)…{be}\:{continued}… \\ $$$$ \\ $$
Commented by mathmax by abdo last updated on 26/Oct/19
$${we}\:{have}\:\sum_{{k}=\mathrm{1}} ^{{n}} \:{sin}^{\mathrm{2}} \left(\frac{{k}\pi}{{n}}\right)=\sum_{{k}=\mathrm{1}} ^{{n}} \frac{\mathrm{1}−{cos}\left(\frac{\mathrm{2}{k}\pi}{{n}}\right)}{\mathrm{2}} \\ $$$$=\frac{{n}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\sum_{{k}=\mathrm{1}} ^{{n}} \:{cos}\left(\frac{\mathrm{2}{k}\pi}{{n}}\right)\:{and} \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \:{cos}\left(\frac{\mathrm{2}{k}\pi}{{n}}\right)={Re}\left(\sum_{{k}=\mathrm{0}} ^{{n}} \:{e}^{\frac{{i}\mathrm{2}{k}\pi}{{n}}} −\mathrm{1}\right) \\ $$$$\sum_{{k}=\mathrm{0}} ^{{n}} \:{e}^{\frac{{i}\mathrm{2}{k}\pi}{{n}}} \:=\frac{\mathrm{1}−\left({e}^{\frac{{i}\mathrm{2}\pi}{{n}}} \right)^{{n}+\mathrm{1}} }{\mathrm{1}−{e}^{\frac{{i}\mathrm{2}\pi}{{n}}} }\:=\frac{\mathrm{1}−{e}^{\frac{{i}\left(\mathrm{2}\pi\right)\left({n}+\mathrm{1}\right)}{{n}}} }{\mathrm{1}−{e}^{\frac{{i}\mathrm{2}\pi}{{n}}} } \\ $$$$=\frac{\mathrm{1}−{cos}\left(\frac{\mathrm{2}\left({n}+\mathrm{1}\right)\pi}{{n}}\right)−{isin}\left(\frac{\mathrm{2}\left({n}+\mathrm{1}\right)\pi}{{n}}\right)}{\mathrm{1}−{cos}\left(\frac{\mathrm{2}\pi}{{n}}\right)−{isin}\left(\frac{\mathrm{2}\pi}{{n}}\right)} \\ $$$$=\frac{\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{\left({n}+\mathrm{1}\right)\pi}{{n}}\right)−\mathrm{2}{isin}\left(\frac{\left({n}+\mathrm{1}\right)\pi}{{n}}\right){cos}\left(\frac{\left({n}+\mathrm{1}\right)\pi}{{n}}\right)}{\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{\pi}{{n}}\right)−\mathrm{2}{isin}\left(\frac{\pi}{{n}}\right){cos}\left(\frac{\pi}{{n}}\right)} \\ $$$$=\frac{−{isin}\left(\frac{\left({n}+\mathrm{1}\right)\pi}{{n}}\right){e}^{{i}\frac{\left({n}+\mathrm{1}\right)\pi}{{n}}} }{−{isin}\left(\frac{\pi}{{n}}\right){e}^{\frac{{i}\pi}{{n}}} }\:=\frac{{sin}\left(\frac{\left({n}+\mathrm{1}\right)\pi}{{n}}\right)}{{sin}\left(\frac{\pi}{{n}}\right)}×\left(−\mathrm{1}\right)\:\Rightarrow \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \:{cos}\left(\frac{\mathrm{2}{k}\pi}{{n}}\right)=−\mathrm{1}−\frac{{sin}\left(\frac{\left({n}+\mathrm{1}\right)\pi}{{n}}\right)}{{sin}\left(\frac{\pi}{{n}}\right)}\:=−\mathrm{1}−\frac{{sin}\left(\pi+\frac{\pi}{{n}}\right)}{{sin}\left(\frac{\pi}{{n}}\right)} \\ $$$$=−\mathrm{1}+\mathrm{1}\:=\mathrm{0}\:\Rightarrow\sum_{{k}=\mathrm{1}} ^{{n}} \:{sin}^{\mathrm{2}} \left(\frac{{k}\pi}{{n}}\right)=\frac{{n}}{\mathrm{2}}\:\Rightarrow \\ $$$$\sum_{\mathrm{1}\leqslant{i}<{j}\leqslant{n}} \:\:\:{sin}\left(\frac{{i}\pi}{{n}}\right){sin}\left(\frac{{j}\pi}{{n}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{\:\:\frac{{sin}^{\mathrm{2}} \left(\frac{\left({n}+\mathrm{1}\right)\pi}{\mathrm{2}{n}}\right)}{{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}{n}}\right)}\:−\frac{{n}^{\mathrm{2}} }{\mathrm{4}}\right\} \\ $$