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Question Number 53781 by maxmathsup by imad last updated on 25/Jan/19

$${calculateA}_n=\sum _{n=1}^{\mathrm{\infty }}{x}^{n}cos\left(n\theta \right)and{B}_n=\sum _{n=1}^{\mathrm{\infty }}{x}^{n}sin\left(n\theta \right)$$

Commented by maxmathsup by imad last updated on 26/Jan/19

$$wehave{A}_n+i{B}_n=\sum _{n=1}^{\mathrm{\infty }}{x}^n{e}^{in\theta }=\sum _{n=0}^{\mathrm{\infty }}{\left(x{e}^{i\theta }\right)}^n-1$$$$soif\mid x\mid <1weget{A}_n+{iB}_n=\frac{1}{1-{xe}^{i\theta }}-1=\frac{1}{1-xcos\theta -ixsin\theta }-1$$$$=\frac{1-xcos\theta +ixsin\theta }{{(1-xcos\theta )}^2+x^2{sin}^2\theta }-1$$$$=\frac{1-xcos\theta }{1-2xcos\theta +x^2}-1+i\frac{xsin\theta }{x^2-2xcos\theta +1}$$$$=\frac{1-xcos\theta -1+2xcos\theta -x^2}{x^2-2xcos\theta +1}+i\frac{xsin\theta }{x^2-2xcos\theta +1}\Rightarrow$$$$A_n=\frac{-x^2+xcos\theta +1}{x^2-2xcos\theta +1}and{B}_n=\frac{xsin\theta }{x^2-2xcos\theta +1}$$$$$$

Answered by Smail last updated on 26/Jan/19

$$A_n+{iB}_n=\underset{n=1}{\sum ^{\mathrm{\infty }}}{x}^n{e}^{in\theta }$$$$=\underset{n=0}{\sum ^{\mathrm{\infty }}}{\left({xe}^{i\theta }\right)}^n-1=\frac{1}{1-{xe}^{i\theta }}-1=\frac{{xe}^{i\theta }}{1-{xe}^{i\theta }}$$$$=\frac{{xe}^{i\theta }}{1-xcos\theta -xisin\theta }$$$$=\frac{{xe}^{i\theta }(1-{xe}^{-i\theta })}{(1-{xe}^{i\theta })(1-{xe}^{-i\theta })}=\frac{x(e^{i\theta }-x)}{1-x(e^{i\theta }+e^{-i\theta })+x^2}$$$$=\frac{x(cos\theta -x+isin\theta )}{1+x^2-2xcos\theta }$$$$A_n+{iB}_n=\frac{x(cos\theta -x)}{1+x^2-2xcos\theta }+i\frac{xsin\theta }{1+x^2-2xcos\theta }$$$$A_n=\frac{x(cos\theta -x)}{{(x-cos\theta )}^2+{sin}^2\theta }$$$$B_n=\frac{xsin\theta }{{(x-cos\theta )}^2+{sin}^2\theta }$$

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