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Question Number 144702 by mathmax by abdo last updated on 28/Jun/21

$$\mathrm{let}\mathrm{g}\left(\mathrm{x}\right)=\log (\mathrm{cosx}+2\mathrm{sinx})$$$$\mathrm{developp}\mathrm{f}\mathrm{at}\mathrm{fourier}\mathrm{serie}$$$$$$

Answered by mathmax by abdo last updated on 28/Jun/21

$$\mathrm{we}\mathrm{have}\mathrm{cosx}+2\mathrm{sinx})=\sqrt{5}(\frac{1}{\sqrt{5}}\mathrm{cosx}+\frac{2}{\sqrt{5}}\mathrm{sinx})$$$$\mathrm{let}\cos \alpha =\frac{1}{\sqrt{5}}\mathrm{and}\sin \alpha =\frac{2}{\sqrt{5}}\Rightarrow \tan \alpha =2\Rightarrow \alpha =\arctan 2\Rightarrow$$$$\mathrm{cosx}+2\mathrm{sinx}=\sqrt{5}\cos (\mathrm{x}-\alpha )\Rightarrow \mathrm{g}\left(\mathrm{x}\right)=\frac{1}{2}\log 5+\log \left(\cos (\mathrm{x}-\alpha )\right)$$$$\mathrm{the}\mathrm{problem}\mathrm{is}\mathrm{turned}\mathrm{to}\mathrm{developp}\phi \left(\mathrm{t}\right)=\log \left(\mathrm{cost}\right)$$$$\mathrm{we}\mathrm{have}\phi \left(\mathrm{t}\right)=\log \left(\frac{{\mathrm{e}}^{\mathrm{it}}+{\mathrm{e}}^{-\mathrm{t}}}{2}\right)=\log ({\mathrm{e}}^{\mathrm{it}}+{\mathrm{e}}^{-\mathrm{it}})-\log \left(2\right)$$$$=\mathrm{it}+\log (1+{\mathrm{e}}^{-2\mathrm{it}})-\log 2$$$$\mathrm{we}[\mathrm{have}{\log }^{{}^{\prime }}(1+\mathrm{u})=\frac{1}{1+\mathrm{u}}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\mathrm{u}}^{\mathrm{n}}\Rightarrow$$$$\log (1+\mathrm{u})=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}{\mathrm{u}}^{\mathrm{n}+1}}{\mathrm{n}+1}=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}-1}\frac{{\mathrm{u}}^{\mathrm{n}}}{\mathrm{n}}$$$$\Rightarrow \log (1+{\mathrm{e}}^{-2\mathrm{it}})=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}-1}\frac{{\mathrm{e}}^{-2\mathrm{int}}}{\mathrm{n}}$$$$\Rightarrow \phi \left(\mathrm{t}\right)=-\log 2+\mathrm{it}+\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}-1}}{\mathrm{n}}(\cos \left(2\mathrm{nt}\right)+\mathrm{isin}\left(2\mathrm{nt}\right))$$$$=-\log 2+\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}-1}}{\mathrm{n}}\cos \left(2\mathrm{nt}\right)+\mathrm{i}(\mathrm{t}+\mathrm{\Sigma }(...)\mathrm{sint})$$$$\phi \left(\mathrm{t}\right)\mathrm{real}\Rightarrow \phi \left(\mathrm{t}\right)=-\log 2+\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}-1}}{\mathrm{n}}\cos \left(2\mathrm{nt}\right)\Rightarrow$$$$\log (\mathrm{cosx}+2\mathrm{sinx})=\frac{1}{2}\log 5+\phi (\mathrm{x}-\alpha )$$$$=\frac{1}{2}\log 5-\log 2+\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}-1}}{\mathrm{n}}\cos \left(2\mathrm{n}(\mathrm{x}-\alpha )\right)$$$$=\frac{1}{2}\log 5-\log 2+\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}-1}}{\mathrm{n}}\cos (2\mathrm{nx}-2\mathrm{narctan}\left(2\right))$$

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