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Question Number 15322 by tawa tawa last updated on 09/Jun/17

Commented by mrW1 last updated on 09/Jun/17

$${\sin }^2{\mathrm{x}}_{\mathrm{k}}=\frac{4}{2015}$$$${\cos }^2{\mathrm{x}}_{\mathrm{k}}=1-\frac{4}{2015}=\frac{2011}{2015}$$$$\sin {2\mathrm{x}}_{\mathrm{k}}=2\left(\sin {\mathrm{x}}_{\mathrm{k}}\cos {\mathrm{x}}_{\mathrm{k}}\right)=2\sqrt{\frac{4\times 2011}{{2015}^2}}=\frac{4\sqrt{2011}}{2015}$$$$\underset{\mathrm{k}=1}{\sum ^{2015}}\sin {2\mathrm{x}}_{\mathrm{k}}=2015\times \frac{4\sqrt{2011}}{2015}=4\sqrt{2011}\approx 179.38$$$$$$$$\mathrm{I}\mathrm{believe}\mathrm{this}\mathrm{is}\mathrm{the}\mathrm{maximum}.$$

Commented by tawa tawa last updated on 09/Jun/17

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

Commented by tawa tawa last updated on 09/Jun/17

$$\mathrm{i}\mathrm{really}\mathrm{appreciate}\mathrm{the}\mathrm{workings}\mathrm{sir}.\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

Commented by tawa tawa last updated on 09/Jun/17

$$\mathrm{But}\mathrm{sir},\mathrm{\Sigma }{\sin }^2{\mathrm{x}}_{\mathrm{k}}=4\left(\mathrm{k}\mathrm{ranging}\mathrm{from}1\mathrm{to}2015\right)$$$$\mathrm{Does}\mathrm{that}\mathrm{means}\mathrm{that}\mathrm{\Sigma }{\sin }^2{\mathrm{x}}_{\mathrm{k}}=\frac{4}{2015}??,\mathrm{How}\mathrm{sir}?$$$$$$$$\mathrm{Does}\mathrm{that}\mathrm{means},{\mathrm{x}}_1={\mathrm{x}}_2=......={\mathrm{x}}_{2015}?\mathrm{is}\mathrm{a}\mathrm{constant}\mathrm{sequence}.$$$$$$$$\mathrm{Does}\mathrm{that}\mathrm{means},{\sin }^2{\mathrm{x}}_1={\sin }^2{\mathrm{x}}_2=...={\sin }^2{\mathrm{x}}_{\mathrm{n}}?$$$$$$$$\mathrm{please}\mathrm{explain}\mathrm{to}\mathrm{me}\mathrm{sir}.$$

Commented by mrW1 last updated on 09/Jun/17

$$\mathrm{Maximum}\mathrm{or}\mathrm{minimum}\mathrm{occurs}\mathrm{in}$$$$\mathrm{such}\mathrm{symmetric}\mathrm{cases}\mathrm{when}\mathrm{all}$$$$\mathrm{variables}\mathrm{are}\mathrm{equal}.\mathrm{that}\mathrm{means}$$$${\mathrm{x}}_1={\mathrm{x}}_2=...={\mathrm{x}}_{\mathrm{k}}=...={\mathrm{x}}_{2015}=\alpha$$$$\underset{\mathrm{k}=1}{\sum ^{2015}}{\sin }^2{\mathrm{x}}_{\mathrm{k}}=\underset{\mathrm{k}=1}{\sum ^{2015}}{\sin }^2\alpha =2015\times {\sin }^2\alpha =4$$$$\Rightarrow {\sin }^2{\mathrm{x}}_{\mathrm{k}}={\sin }^2\alpha =\frac{4}{2015}$$$${\cos }^2{\mathrm{x}}_{\mathrm{k}}={\cos }^2\alpha =1-{\sin }^2\alpha =1-\frac{4}{2015}=\frac{2011}{2015}$$$$$$$$\underset{\mathrm{k}=1}{\sum ^{2015}}\sin {2\mathrm{x}}_{\mathrm{k}}=\underset{\mathrm{k}=1}{\sum ^{2015}}\sin 2\alpha =2015\times \sin 2\alpha$$$$=2015\times 2\times \sin \alpha \cos \alpha$$$$=2015\times 2\times \sqrt{{\sin }^2\alpha {\cos }^2\alpha }$$$$=2015\times 2\times \sqrt{\frac{4}{2015}\times \frac{2011}{2015}}$$$$=2015\times 2\times \frac{2\sqrt{2011}}{2015}$$$$=4\sqrt{2011}$$$$$$$$\mathrm{note}:\mathrm{what}\mathrm{I}\mathrm{did}\mathrm{is}\mathrm{not}\mathrm{a}\mathrm{strict}\mathrm{proof},$$$$\mathrm{though}\mathrm{I}\mathrm{believe}\mathrm{it}\mathrm{is}\mathrm{correct}.$$

Commented by tawa tawa last updated on 09/Jun/17

$$\mathrm{I}\mathrm{appreciate}\mathrm{your}\mathrm{time}\mathrm{sir}.\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

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