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Question Number 142900 by aliibrahim1 last updated on 06/Jun/21

Commented by qaz last updated on 07/Jun/21

$$\ln \left(2\sin \frac{\mathrm{x}}{2}\right)=-\underset{\mathrm{n}=1}{\sum ^{\mathrm{\infty }}}\frac{\cos \left(\mathrm{nx}\right)}{\mathrm{n}}$$

Commented by mathmax by abdo last updated on 07/Jun/21

$$\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{\cos \left(\mathrm{nx}\right)}{\mathrm{n}}=\mathrm{Re}\left(\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{\mathrm{e}}^{\mathrm{inx}}}{\mathrm{n}}\right)\mathrm{let}\mathrm{w}\left(\mathrm{x}\right)=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{\mathrm{e}}^{\mathrm{inx}}}{\mathrm{n}}\Rightarrow$$$$\mathrm{w}\left(\mathrm{x}\right)=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{\left({\mathrm{e}}^{\mathrm{ix}}\right)}^{\mathrm{n}}}{\mathrm{n}}=-\log (1-{\mathrm{e}}^{\mathrm{ix}})$$$$=-\log (1-\mathrm{cosx}-\mathrm{isinx})=-\log ({2\sin }^2\left(\frac{\mathrm{x}}{2}\right)-2\mathrm{isin}\left(\frac{\mathrm{x}}{2}\right)\cos \left(\frac{\mathrm{x}}{2}\right))$$$$=-\log (-2\mathrm{isin}\left(\frac{\mathrm{x}}{2}\right){\mathrm{e}}^{\frac{\mathrm{ix}}{2}})=-\log (-\mathrm{i})-\log \left(2\sin \left(\frac{\mathrm{x}}{2}\right)\right)-\frac{\mathrm{ix}}{2}$$$$=-\log \left({\mathrm{e}}^{-\frac{\mathrm{i}\pi }{2}}\right)-\log \left(2\sin \left(\frac{\mathrm{x}}{2}\right)\right)-\frac{\mathrm{ix}}{2}=\frac{\mathrm{i}\pi }{2}-\frac{\mathrm{ix}}{2}-\log \left(2\sin \left(\frac{\mathrm{x}}{2}\right)\right)$$$$=\mathrm{i}\left(\frac{\pi -\mathrm{x})}{2}\right)-\log \left(2\sin \left(\frac{\mathrm{x}}{2}\right)\right)\Rightarrow \sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{\cos \left(\mathrm{nx}\right)}{\mathrm{n}}=-\log \left(2\sin \left(\frac{\mathrm{x}}{2}\right)\right)\Rightarrow$$$$\log \left(\sin \left(\frac{\mathrm{x}}{2}\right)\right)=-\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{\cos \left(\mathrm{nx}\right)}{\mathrm{n}}$$

Commented by aliibrahim1 last updated on 19/Jun/21

$$sorrytookmeawhiletlreplyiwasofflinebutthankyousoomuchsir$$

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