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Question Number 82519 by peter frank last updated on 22/Feb/20

Commented by mathmax by abdo last updated on 23/Feb/20

$$A_{\theta }=\sum _{n=0}^{\mathrm{\infty }}\frac{{(-1)}^n}{2^n}cos\left(n\theta \right)=\sum _{n=0}^{\mathrm{\infty }}{(-\frac{1}{2})}^{n}cos\left(n\theta \right)$$$$Re\left(\sum _{n=0}^{\mathrm{\infty }}{(-\frac{1}{2}{e}^{i\theta })}^n\right)but\sum _{n=0}^{\mathrm{\infty }}{(-\frac{1}{2}{e}^{i\theta })}^n=\frac{1}{1+\frac{1}{2}{e}^{i\theta }}$$$$=\frac{2}{2+cos\theta +isin\theta }=\frac{2(2+cos\theta -isin\theta )}{{(2+cos\theta )}^2+{sin}^2\theta }=\frac{2(2+cos\theta )}{4+4cos\theta +1}-\frac{isin\theta }{4+4cos\theta +1}$$$$\Rightarrow A_{\theta }=\frac{4+2cos\theta }{5+4cos\theta }$$

Answered by mr W last updated on 22/Feb/20

$$z=(-\frac{1}{2})(\cos \theta +i\sin \theta )=-\frac{e^{i\theta }}{2}$$$$z^k={(-1)}^k\frac{1}{2^k}(\cos k\theta +i\sin k\theta )={(-\frac{e^{i\theta }}{2})}^k$$$$\underset{k=0}{\sum ^{\mathrm{\infty }}}{(-1)}^k\frac{1}{2^k}\cos k\theta +i\underset{k=0}{\sum ^{\mathrm{\infty }}}{(-1)}^k\frac{1}{2^k}\sin k\theta =\underset{k=0}{\sum ^{\mathrm{\infty }}}{(-\frac{e^{i\theta }}{2})}^k$$$$\underset{k=0}{\sum ^{\mathrm{\infty }}}{(-1)}^k\frac{1}{2^k}\cos k\theta +i\underset{k=0}{\sum ^{\mathrm{\infty }}}{(-1)}^k\frac{1}{2^k}\sin k\theta =\frac{1}{1+\frac{e^{i\theta }}{2}}$$$$\underset{k=0}{\sum ^{\mathrm{\infty }}}{(-1)}^k\frac{1}{2^k}\cos k\theta +i\underset{k=0}{\sum ^{\mathrm{\infty }}}{(-1)}^k\frac{1}{2^k}\sin k\theta =\frac{2}{2+\cos \theta +i\sin \theta }$$$$\underset{k=0}{\sum ^{\mathrm{\infty }}}{(-1)}^k\frac{1}{2^k}\cos k\theta +i\underset{k=0}{\sum ^{\mathrm{\infty }}}{(-1)}^k\frac{1}{2^k}\sin k\theta =\frac{2(2+\cos \theta -i\sin \theta )}{{(2+\cos \theta )}^2+{\sin }^2\theta }$$$$\underset{k=0}{\sum ^{\mathrm{\infty }}}{(-1)}^k\frac{1}{2^k}\cos k\theta +i\underset{k=0}{\sum ^{\mathrm{\infty }}}{(-1)}^k\frac{1}{2^k}\sin k\theta =\frac{2(2+\cos \theta )-i2\sin \theta (2+\cos \theta )}{5+4\cos \theta }$$$$\Rightarrow \underset{k=0}{\sum ^{\mathrm{\infty }}}{(-1)}^k\frac{1}{2^k}\cos k\theta =\frac{2(2+\cos \theta )}{5+4\cos \theta }$$$$\Rightarrow \underset{k=0}{\sum ^{\mathrm{\infty }}}{(-1)}^k\frac{1}{2^k}\sin k\theta =\frac{-2\sin \theta (2+\cos \theta )}{5+4\cos \theta }$$

Commented by peter frank last updated on 22/Feb/20

$$thankyouverymuch$$

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