Question Number 82519 by peter frank last updated on 22/Feb/20
Commented by mathmax by abdo last updated on 23/Feb/20
$${A}_{\theta} =\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}^{{n}} }{cos}\left({n}\theta\right)\:\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}} \:{cos}\left({n}\theta\right) \\ $$$${Re}\left(\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−\frac{\mathrm{1}}{\mathrm{2}}{e}^{{i}\theta} \right)^{{n}} \right)\:\:{but}\:\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\frac{\mathrm{1}}{\mathrm{2}}{e}^{{i}\theta} \right)^{{n}} \:=\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}{e}^{{i}\theta} } \\ $$$$=\frac{\mathrm{2}}{\mathrm{2}+{cos}\theta\:+{isin}\theta}\:=\frac{\mathrm{2}\left(\mathrm{2}+{cos}\theta\:−{isin}\theta\right)}{\left(\mathrm{2}+{cos}\theta\right)^{\mathrm{2}} \:+{sin}^{\mathrm{2}} \theta}\:=\frac{\mathrm{2}\left(\mathrm{2}+{cos}\theta\right)}{\mathrm{4}+\mathrm{4}{cos}\theta\:+\mathrm{1}}−\frac{{isin}\theta}{\mathrm{4}+\mathrm{4}{cos}\theta\:+\mathrm{1}} \\ $$$$\Rightarrow{A}_{\theta} =\frac{\mathrm{4}+\mathrm{2}{cos}\theta}{\mathrm{5}+\mathrm{4}{cos}\theta} \\ $$
Answered by mr W last updated on 22/Feb/20
$${z}=\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)\left(\mathrm{cos}\:\theta+{i}\:\mathrm{sin}\:\theta\right)=−\frac{{e}^{{i}\theta} }{\mathrm{2}} \\ $$$${z}^{{k}} =\left(−\mathrm{1}\right)^{{k}} \frac{\mathrm{1}}{\mathrm{2}^{{k}} }\left(\mathrm{cos}\:{k}\theta+{i}\:\mathrm{sin}\:{k}\theta\right)=\left(−\frac{{e}^{{i}\theta} }{\mathrm{2}}\right)^{{k}} \\ $$$$\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{k}} \frac{\mathrm{1}}{\mathrm{2}^{{k}} }\:\mathrm{cos}\:{k}\theta+{i}\:\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{k}} \frac{\mathrm{1}}{\mathrm{2}^{{k}} }\:\mathrm{sin}\:{k}\theta=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\frac{{e}^{{i}\theta} }{\mathrm{2}}\right)^{{k}} \\ $$$$\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{k}} \frac{\mathrm{1}}{\mathrm{2}^{{k}} }\:\mathrm{cos}\:{k}\theta+{i}\:\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{k}} \frac{\mathrm{1}}{\mathrm{2}^{{k}} }\:\mathrm{sin}\:{k}\theta=\frac{\mathrm{1}}{\mathrm{1}+\frac{{e}^{{i}\theta} }{\mathrm{2}}} \\ $$$$\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{k}} \frac{\mathrm{1}}{\mathrm{2}^{{k}} }\:\mathrm{cos}\:{k}\theta+{i}\:\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{k}} \frac{\mathrm{1}}{\mathrm{2}^{{k}} }\:\mathrm{sin}\:{k}\theta=\frac{\mathrm{2}}{\mathrm{2}+\mathrm{cos}\:\theta+{i}\:\mathrm{sin}\:\theta} \\ $$$$\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{k}} \frac{\mathrm{1}}{\mathrm{2}^{{k}} }\:\mathrm{cos}\:{k}\theta+{i}\:\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{k}} \frac{\mathrm{1}}{\mathrm{2}^{{k}} }\:\mathrm{sin}\:{k}\theta=\frac{\mathrm{2}\left(\mathrm{2}+\mathrm{cos}\:\theta−{i}\:\mathrm{sin}\:\theta\right)}{\left(\mathrm{2}+\mathrm{cos}\:\theta\right)^{\mathrm{2}} +\mathrm{sin}^{\mathrm{2}} \:\theta} \\ $$$$\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{k}} \frac{\mathrm{1}}{\mathrm{2}^{{k}} }\:\mathrm{cos}\:{k}\theta+{i}\:\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{k}} \frac{\mathrm{1}}{\mathrm{2}^{{k}} }\:\mathrm{sin}\:{k}\theta=\frac{\mathrm{2}\left(\mathrm{2}+\mathrm{cos}\:\theta\right)−{i}\:\mathrm{2}\:\mathrm{sin}\:\theta\left(\mathrm{2}+\mathrm{cos}\:\theta\right)}{\mathrm{5}+\mathrm{4}\:\mathrm{cos}\:\theta} \\ $$$$\Rightarrow\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{k}} \frac{\mathrm{1}}{\mathrm{2}^{{k}} }\:\mathrm{cos}\:{k}\theta=\frac{\mathrm{2}\left(\mathrm{2}+\mathrm{cos}\:\theta\right)}{\mathrm{5}+\mathrm{4}\:\mathrm{cos}\:\theta} \\ $$$$\Rightarrow\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{k}} \frac{\mathrm{1}}{\mathrm{2}^{{k}} }\:\mathrm{sin}\:{k}\theta=\frac{−\mathrm{2}\:\mathrm{sin}\:\theta\left(\mathrm{2}+\mathrm{cos}\:\theta\right)}{\mathrm{5}+\mathrm{4}\:\mathrm{cos}\:\theta} \\ $$
Commented by peter frank last updated on 22/Feb/20
$${thank}\:{you}\:{very}\:{much} \\ $$