Question Number 54700 by peter frank last updated on 09/Feb/19
$$\mathrm{F}{i}\mathrm{nd}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{to}\:\mathrm{infinity} \\ $$$$\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}\:}\mathrm{cos}\:\theta+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{cos}\:\mathrm{2}\theta−\frac{\mathrm{1}}{\mathrm{8}}\mathrm{cos}\:\mathrm{3}\theta+………. \\ $$$$ \\ $$
Commented by maxmathsup by imad last updated on 09/Feb/19
$${we}\:{have}\:{S}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}^{{n}} }{cos}\left({n}\theta\right)\:\Rightarrow{S}={Re}\left(\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}^{{n}} }\:{e}^{{in}\theta} \right)\:{but} \\ $$$$\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}^{{n}} }\:{e}^{{in}\theta} \:=\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\frac{\mathrm{1}}{\mathrm{2}}\:{e}^{{i}\theta} \right)^{{n}} \:\:\left(\:{we}\:{have}\:\mid−\frac{\mathrm{1}}{\mathrm{2}}\:{e}^{{i}\theta} \mid<\mathrm{1}\:\right) \\ $$$$\:=\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}{e}^{{i}\theta} }\:=\frac{\mathrm{2}}{\mathrm{2}\:+{cos}\theta\:+{isin}\theta}\:=\frac{\mathrm{2}\left(\mathrm{2}+{cos}\theta−{isin}\theta\right)}{\left(\mathrm{2}+{cos}\theta\right)^{\mathrm{2}} \:+{sin}^{\mathrm{2}} \theta}\:\Rightarrow \\ $$$${S}\:=\frac{\mathrm{2}\left(\mathrm{2}+{cos}\theta\right)}{\mathrm{4}+\mathrm{4}{cos}\theta\:+\mathrm{1}}\:=\frac{\mathrm{4}\:+\mathrm{2}{cos}\theta}{\mathrm{5}\:+\mathrm{4}{cos}\theta} \\ $$$$ \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 09/Feb/19
$${p}=\mathrm{1}−\frac{{cos}\theta}{\mathrm{2}}+\frac{{cos}\mathrm{2}\theta}{\mathrm{4}}−\frac{{cos}\mathrm{3}\theta}{\mathrm{8}}+… \\ $$$${q}=−\frac{{sin}\theta}{\mathrm{2}}+\frac{{sin}\mathrm{2}\theta}{\mathrm{4}}−\frac{{sin}\mathrm{3}\theta}{\mathrm{8}}+… \\ $$$${p}+{iq}=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\left({cos}\theta+{isin}\theta\right)+\frac{\mathrm{1}}{\mathrm{4}}\left({cos}\mathrm{2}\theta+{isin}\mathrm{2}\theta\right)−\frac{\mathrm{1}}{\mathrm{8}}\left({cos}\mathrm{3}\theta+{isin}\mathrm{3}\theta\right)+… \\ $$$${p}+{iq}=\mathrm{1}−\frac{{e}^{{i}\theta} }{\mathrm{2}}+\frac{{e}^{{i}\mathrm{2}\theta} }{\mathrm{2}^{\mathrm{2}} }−\frac{{e}^{{i}\mathrm{3}\theta} }{\mathrm{2}^{\mathrm{3}} }+… \\ $$$${p}+{iq}=\mathrm{1}−{a}+{a}^{\mathrm{2}} −{a}^{\mathrm{3}} +… \\ $$$${p}+{iq}=\frac{\mathrm{1}}{\mathrm{1}+{a}}=\frac{\mathrm{1}}{\mathrm{1}+\frac{{e}^{{i}\theta} }{\mathrm{2}}}=\frac{\mathrm{2}}{\mathrm{2}+{e}^{{i}\theta} }=\frac{\mathrm{2}}{\mathrm{2}+{cos}\theta+{isin}\theta} \\ $$$${p}+{iq}=\frac{\mathrm{2}\left(\mathrm{2}+{cos}\theta−{isin}\theta\right)}{\left(\mathrm{2}+{cos}\theta\right)^{\mathrm{2}} +\left({sin}\theta\right)^{\mathrm{2}} } \\ $$$${p}+{iq}=\frac{\mathrm{4}+\mathrm{2}{cos}\theta}{\mathrm{4}+\mathrm{4}{cos}\theta+\mathrm{1}}+\frac{−\mathrm{2}{isin}\theta}{\mathrm{5}+\mathrm{4}{cos}\theta} \\ $$$${so}\:{required}\:{ans}\:{is}=\frac{\mathrm{4}+\mathrm{2}{cos}\theta}{\mathrm{5}+\mathrm{4}{cos}\theta} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$\mathrm{2} \\ $$
Commented by peter frank last updated on 09/Feb/19
$${thank}\:{you}\:{sir} \\ $$
Commented by peter frank last updated on 09/Feb/19
$${thank}\:{you} \\ $$
Commented by peter frank last updated on 09/Feb/19
$${sorry}\:{sir}\:{ellaborate}\: \\ $$$${second}\:{line} \\ $$$$ \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 09/Feb/19
$${p}={the}\:{sum}\:{of}\:{series}\:{of}\:{cos}\theta \\ $$$${q}={sum}\:{of}\:{series}\:{of}\:{sin}\theta \\ $$$${the}\:{method}\:{is}\:{find}\:{p}+{iq}\:{and}\:{convert}\:{it}\:{to}\:{series}\:{of}\:{e}^{{i}\theta} \\ $$$${then}\:{separate}\:{real}\:{snd}\:{imaginary}\:{part} \\ $$$${real}\:{part}\rightarrow{p}\:\:\:{imaginary}\:{part}\rightarrow{q} \\ $$