Question Number 220947 by fantastic last updated on 21/May/25
$$\underset{{k}=\mathrm{1}} {\overset{\mathrm{13}} {\sum}}\:\:\frac{\mathrm{1}}{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{4}}+\frac{\left({k}−\mathrm{1}\right)\pi}{\mathrm{6}}\right)\mathrm{sin}\:\left(\frac{\pi}{\mathrm{4}}+\frac{{k}\pi}{\mathrm{6}}\right)} \\ $$
Answered by MrGaster last updated on 21/May/25
![=Σ_(k=1) ^(13) 2[cot((π/4)+(((k−1)π)/6))−cot((π/4)+((kπ)/6))] =2[cot((π/4))−cot((π/4)+((13π)/6))] =2[1−cot((π/4)+(π/6)2π)] =2[1−cot(((5π)/(12)))] cot(((5π)/(12)))=2−(√3) =2[1−(2−(√3))] =2((√3)−1) =2(√3)−2 (2)=2Σ_(k=1) ^(13) [cot((π/4)+(((k−1)π)/6))−cot((π/4)+((kπ)/6))] =2[cot((π/4))−cot((π/4)+((13π)/6))] =2[1−cot(((29π)/(12)))] =2[1−cot(((5π)/(12)))] =2[1−(2−(√3))] =2((√3)−1) =2(√3)−2 (3)=Σ_(k=1) ^(13) [cot((π/4)+(((k−1)π)/6))−cot((π/4)+((kπ)/6))] =2[cot((π/4))−cot((π/4)+((13π)/6))] =2[1−cot(((5π)/(12)))] =2[1−(2−(√3))] =2((√3)−1)](https://www.tinkutara.com/question/Q220952.png)
$$=\underset{{k}=\mathrm{1}} {\overset{\mathrm{13}} {\sum}}\mathrm{2}\left[\mathrm{cot}\left(\frac{\pi}{\mathrm{4}}+\frac{\left({k}−\mathrm{1}\right)\pi}{\mathrm{6}}\right)−\mathrm{cot}\left(\frac{\pi}{\mathrm{4}}+\frac{{k}\pi}{\mathrm{6}}\right)\right] \\ $$$$=\mathrm{2}\left[\mathrm{cot}\left(\frac{\pi}{\mathrm{4}}\right)−\mathrm{cot}\left(\frac{\pi}{\mathrm{4}}+\frac{\mathrm{13}\pi}{\mathrm{6}}\right)\right] \\ $$$$=\mathrm{2}\left[\mathrm{1}−\mathrm{cot}\left(\frac{\pi}{\mathrm{4}}+\frac{\pi}{\mathrm{6}}\mathrm{2}\pi\right)\right] \\ $$$$=\mathrm{2}\left[\mathrm{1}−\mathrm{cot}\left(\frac{\mathrm{5}\pi}{\mathrm{12}}\right)\right] \\ $$$$\mathrm{cot}\left(\frac{\mathrm{5}\pi}{\mathrm{12}}\right)=\mathrm{2}−\sqrt{\mathrm{3}} \\ $$$$=\mathrm{2}\left[\mathrm{1}−\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)\right] \\ $$$$=\mathrm{2}\left(\sqrt{\mathrm{3}}−\mathrm{1}\right) \\ $$$$=\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{2} \\ $$$$\left(\mathrm{2}\right)=\mathrm{2}\underset{{k}=\mathrm{1}} {\overset{\mathrm{13}} {\sum}}\left[\mathrm{cot}\left(\frac{\pi}{\mathrm{4}}+\frac{\left({k}−\mathrm{1}\right)\pi}{\mathrm{6}}\right)−\mathrm{cot}\left(\frac{\pi}{\mathrm{4}}+\frac{{k}\pi}{\mathrm{6}}\right)\right] \\ $$$$=\mathrm{2}\left[\mathrm{cot}\left(\frac{\pi}{\mathrm{4}}\right)−\mathrm{cot}\left(\frac{\pi}{\mathrm{4}}+\frac{\mathrm{13}\pi}{\mathrm{6}}\right)\right] \\ $$$$=\mathrm{2}\left[\mathrm{1}−\mathrm{cot}\left(\frac{\mathrm{29}\pi}{\mathrm{12}}\right)\right] \\ $$$$=\mathrm{2}\left[\mathrm{1}−\mathrm{cot}\left(\frac{\mathrm{5}\pi}{\mathrm{12}}\right)\right] \\ $$$$=\mathrm{2}\left[\mathrm{1}−\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)\right] \\ $$$$=\mathrm{2}\left(\sqrt{\mathrm{3}}−\mathrm{1}\right) \\ $$$$=\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{2} \\ $$$$\left(\mathrm{3}\right)=\underset{{k}=\mathrm{1}} {\overset{\mathrm{13}} {\sum}}\left[\mathrm{cot}\left(\frac{\pi}{\mathrm{4}}+\frac{\left({k}−\mathrm{1}\right)\pi}{\mathrm{6}}\right)−\mathrm{cot}\left(\frac{\pi}{\mathrm{4}}+\frac{{k}\pi}{\mathrm{6}}\right)\right] \\ $$$$=\mathrm{2}\left[\mathrm{cot}\left(\frac{\pi}{\mathrm{4}}\right)−\mathrm{cot}\left(\frac{\pi}{\mathrm{4}}+\frac{\mathrm{13}\pi}{\mathrm{6}}\right)\right] \\ $$$$=\mathrm{2}\left[\mathrm{1}−\mathrm{cot}\left(\frac{\mathrm{5}\pi}{\mathrm{12}}\right)\right] \\ $$$$=\mathrm{2}\left[\mathrm{1}−\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)\right] \\ $$$$=\mathrm{2}\left(\sqrt{\mathrm{3}}−\mathrm{1}\right) \\ $$