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Question Number 144464 by mathmax by abdo last updated on 25/Jun/21

$$\mathrm{f}\left(\mathrm{x}\right)=\frac{2}{{(1+\mathrm{sinx})}^2}$$$$\mathrm{developp}\mathrm{f}\mathrm{at}\mathrm{fourier}\mathrm{serie}$$

Answered by mathmax by abdo last updated on 26/Jun/21

$$\mathrm{let}\mathrm{f}\left(\mathrm{a}\right)=\frac{2}{\mathrm{a}+\mathrm{sinx}}\Rightarrow {\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)=-\frac{2}{{(\mathrm{a}+\mathrm{sinx})}^2}\Rightarrow -{\mathrm{f}}^{{}^{\prime }}\left(1\right)=\frac{2}{{(1+\mathrm{sinx})}^2}$$$$\mathrm{we}\mathrm{have}\mathrm{f}\left(\mathrm{a}\right)=\frac{2}{\mathrm{a}+\frac{{\mathrm{e}}^{\mathrm{ix}}-{\mathrm{e}}^{-\mathrm{ix}}}{2\mathrm{i}}}=\frac{4\mathrm{i}}{2\mathrm{ia}+{\mathrm{e}}^{\mathrm{ix}}-{\mathrm{e}}^{-\mathrm{ix}}}$$$${=}_{{\mathrm{e}}^{\mathrm{ix}}=\mathrm{z}}\frac{4\mathrm{i}}{2\mathrm{ia}+\mathrm{z}-{\mathrm{z}}^{-1}}=\frac{4\mathrm{iz}}{2\mathrm{iaz}+{\mathrm{z}}^2-1}=\frac{4\mathrm{iz}}{{\mathrm{z}}^2+2\mathrm{iaz}-1}$$$${\mathrm{\Delta }}^{{}^{\prime }}=-{\mathrm{a}}^2+1=1-{\mathrm{a}}^2\Rightarrow {\mathrm{z}}_1=-\mathrm{ia}+\sqrt{1-{\mathrm{a}}^2}$$$${\mathrm{z}}_2=-\mathrm{ia}-\sqrt{1-{\mathrm{a}}^2}\Rightarrow \mathrm{f}\left(\mathrm{a}\right)=\frac{4\mathrm{iz}}{(\mathrm{z}-{\mathrm{z}}_1)(\mathrm{z}-{\mathrm{z}}_2)}$$$$=4\mathrm{iz}(\frac{1}{\mathrm{z}-{\mathrm{z}}_1}-\frac{1}{\mathrm{z}-{\mathrm{z}}_2}).\frac{1}{{\mathrm{z}}_1-{\mathrm{z}}_2}=\frac{4\mathrm{i}}{2\sqrt{1-{\mathrm{a}}^2}}(\frac{\mathrm{z}}{\mathrm{z}-{\mathrm{z}}_1}-\frac{\mathrm{z}}{\mathrm{z}-{\mathrm{z}}_2})$$$$\mid \frac{\mathrm{z}}{{\mathrm{z}}_1}\mid =1\mathrm{and}\mid \frac{\mathrm{z}}{{\mathrm{z}}_2}\mid =1\Rightarrow \mathrm{f}\left(\mathrm{a}\right)=\frac{2\mathrm{i}}{\sqrt{1-{\mathrm{a}}^2}}(\frac{\mathrm{z}}{{\mathrm{z}}_1(\frac{\mathrm{z}}{{\mathrm{z}}_1}-1)}-\frac{\mathrm{z}}{{\mathrm{z}}_2(\frac{\mathrm{z}}{{\mathrm{z}}_2}-1)})$$$$=\frac{2\mathrm{iz}}{\sqrt{1-{\mathrm{a}}^2}}(\frac{1}{1-\frac{\mathrm{z}}{{\mathrm{z}}_2}}-\frac{1}{1-\frac{\mathrm{z}}{{\mathrm{z}}_1}})=\frac{2\mathrm{iz}}{\sqrt{1-{\mathrm{a}}^2}}(\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{\mathrm{z}}^{\mathrm{n}}}{{\mathrm{z}}_2^{\mathrm{n}}}-\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{\mathrm{z}}^{\mathrm{n}}}{{\mathrm{z}}_1^{\mathrm{n}}})$$$${\mathrm{z}}_1={\mathrm{e}}^{\mathrm{iarctan}(-\frac{\mathrm{a}}{\sqrt{1-{\mathrm{a}}^2}})}\mathrm{and}{\mathrm{z}}_2={\mathrm{e}}^{\mathrm{iarctan}\left(\frac{\mathrm{a}}{\sqrt{1-{\mathrm{a}}^2}}\right)}\Rightarrow$$$$\Rightarrow \frac{1}{{\mathrm{z}}_1^{\mathrm{n}}}={\mathrm{e}}^{\mathrm{inarctan}\left(\frac{\mathrm{a}}{\sqrt{1-{\mathrm{a}}^2}}\right)}+{\mathrm{e}}^{-\mathrm{inarctan}\left(\frac{\mathrm{a}}{\sqrt{1-{\mathrm{a}}^2}}\right)}=2\mathrm{Re}(....)$$$$=2\cos \left(\mathrm{narctan}\left(\frac{\mathrm{a}}{\sqrt{1-{\mathrm{a}}^2}}\right)\right)\Rightarrow$$$$\mathrm{f}\left(\mathrm{a}\right)=\frac{2\mathrm{iz}}{\sqrt{1-{\mathrm{a}}^2}}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}2\cos \left(\mathrm{narctan}\left(\frac{\mathrm{a}}{\sqrt{1-{\mathrm{a}}^2}}\right)\right){\mathrm{e}}^{\mathrm{inx}}$$$$=\frac{4\mathrm{i}}{\sqrt{1-{\mathrm{a}}^2}}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\cos \left(\mathrm{narctan}\left(\frac{\mathrm{a}}{\sqrt{1-{\mathrm{a}}^2}}\right)\right)(\cos (\mathrm{n}+1)\mathrm{x}+\mathrm{isin}(\mathrm{n}+1)\mathrm{x})$$$$=4\mathrm{i}(....)-\frac{4}{\sqrt{1-{\mathrm{a}}^2}}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\cos \left(\mathrm{narctan}\left(\frac{\mathrm{a}}{\sqrt{1-{\mathrm{a}}^2}}\right)\right)\sin (\mathrm{n}+1)\mathrm{x}$$$$\mathrm{f}\left(\mathrm{a}\right)\mathrm{is}\mathrm{reaal}\Rightarrow$$$$\mathrm{f}\left(\mathrm{a}\right)=\frac{-4}{\sqrt{1-{\mathrm{a}}^2}}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\cos \left(\mathrm{narctan}\left(\frac{\mathrm{a}}{\sqrt{1-{\mathrm{a}}^2}}\right)\right)\sin (\mathrm{n}+1)\mathrm{x}$$$$\mathrm{rest}\mathrm{to}\mathrm{calculate}{\mathrm{f}}^{{}^{\prime }}\left(\mathrm{a}\right)....\mathrm{be}\mathrm{continued}$$

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