Question Number 60056 by bhanukumarb2@gmail.com last updated on 17/May/19
Commented by maxmathsup by imad last updated on 17/May/19
$${we}\:{see}\:{that}\:\:{S}\:=\sum_{{k}=\mathrm{0}} ^{\infty} \:\frac{{sin}\left({k}\theta\right){cos}^{{k}} \theta}{{k}!}\:\Rightarrow{S}\:={Im}\left(\sum_{{k}=\mathrm{0}} ^{\infty} \:\frac{{e}^{{ik}\theta} \:{cos}^{{k}} \theta}{{k}!}\right)\:{but} \\ $$$$\sum_{{k}=\mathrm{0}} ^{\infty} \:\:\frac{{e}^{{ik}\theta} \:{cos}^{{k}} \theta}{{k}!}\:=\sum_{{k}=\mathrm{0}} ^{\infty} \:\:\frac{\left({e}^{{i}\theta} \:{cos}\theta\right)^{{k}} }{{k}!}\:={e}^{\left({e}^{\left.{i}\theta{cos}\theta\right)} \right.} \:\:\:\:={e}^{\left.{cos}\left(\theta\:{cos}\theta\right)\:+{isin}\left(\theta\:{cos}\theta\right)\right)} \\ $$$$={e}^{{cos}\left(\theta\:{cos}\theta\right)} \left\{\:{cos}\left({sin}\left(\theta{cos}\theta\right)\:+{i}\:{sin}\left({sin}\left(\theta\:{cos}\theta\right)\right)\:\right\}\:\Rightarrow\right. \\ $$$${S}\:={e}^{{cos}\left(\theta\:{cos}\theta\right)} \:{sin}\left({sin}\left(\theta\:{cos}\theta\right)\right)\:. \\ $$
Answered by tanmay last updated on 17/May/19
![p=cosθcosθ+((cos2θcos^2 θ)/(2!))+((cos3θcos^3 θ)/(3!))+... q=sinθcosθ+((sin2θcos^2 θ)/(2!))+((sin3θcos^3 θ)/(3!))+.. p+iq =e^(iθ) cosθ+e^(i2θ) ×((cos^2 θ)/(2!))+((e^(i3θ) ×cos^3 θ)/(3!))+... t=e^(iθ) cosθ p+iq=t+(t^2 /(2!))+(t^3 /(3!))+... =e^t −1 =e^(e^(iθ) cosθ) −1 =e^(cosθ(cosθ+isinθ)) −1 =e^(cos^2 θ+isinθcosθ) −1 =e^(cos^2 θ) ×(e^(isinθcosθ) )−1 =e^(cos^2 θ) ×e^((isin2θ)/2) −1 =e^(cos^2 θ) [cos(((sin2θ)/2))+isin(((sin2θ)/2))]−1 =e^(cos^2 θ) [cos(((sin2θ)/2))]−1+ie^(cos^2 θ) sin(((sin2θ)/2)) So p=e^(cos^2 θ) [cos(((sin2θ)/2))]−1←real part q=e^(cos^2 θ) sin(((sin2θ)/2))←imaginary part So required answer is q=e^(cos^2 θ) sin(((sin2θ)/2)) pls check....](https://www.tinkutara.com/question/Q60061.png)
$${p}={cos}\theta{cos}\theta+\frac{{cos}\mathrm{2}\theta{cos}^{\mathrm{2}} \theta}{\mathrm{2}!}+\frac{{cos}\mathrm{3}\theta{cos}^{\mathrm{3}} \theta}{\mathrm{3}!}+… \\ $$$${q}={sin}\theta{cos}\theta+\frac{{sin}\mathrm{2}\theta{cos}^{\mathrm{2}} \theta}{\mathrm{2}!}+\frac{{sin}\mathrm{3}\theta{cos}^{\mathrm{3}} \theta}{\mathrm{3}!}+.. \\ $$$${p}+{iq} \\ $$$$={e}^{{i}\theta} {cos}\theta+{e}^{{i}\mathrm{2}\theta} ×\frac{{cos}^{\mathrm{2}} \theta}{\mathrm{2}!}+\frac{{e}^{{i}\mathrm{3}\theta} ×{cos}^{\mathrm{3}} \theta}{\mathrm{3}!}+… \\ $$$${t}={e}^{{i}\theta} {cos}\theta \\ $$$${p}+{iq}={t}+\frac{{t}^{\mathrm{2}} }{\mathrm{2}!}+\frac{{t}^{\mathrm{3}} }{\mathrm{3}!}+… \\ $$$$={e}^{{t}} −\mathrm{1} \\ $$$$={e}^{{e}^{{i}\theta} {cos}\theta} −\mathrm{1} \\ $$$$={e}^{{cos}\theta\left({cos}\theta+{isin}\theta\right)} −\mathrm{1} \\ $$$$={e}^{{cos}^{\mathrm{2}} \theta+{isin}\theta{cos}\theta} −\mathrm{1} \\ $$$$={e}^{{cos}^{\mathrm{2}} \theta} ×\left({e}^{{isin}\theta{cos}\theta} \right)−\mathrm{1} \\ $$$$={e}^{{cos}^{\mathrm{2}} \theta} ×{e}^{\frac{{isin}\mathrm{2}\theta}{\mathrm{2}}} −\mathrm{1} \\ $$$$={e}^{{cos}^{\mathrm{2}} \theta} \left[{cos}\left(\frac{{sin}\mathrm{2}\theta}{\mathrm{2}}\right)+{isin}\left(\frac{{sin}\mathrm{2}\theta}{\mathrm{2}}\right)\right]−\mathrm{1} \\ $$$$={e}^{{cos}^{\mathrm{2}} \theta} \left[{cos}\left(\frac{{sin}\mathrm{2}\theta}{\mathrm{2}}\right)\right]−\mathrm{1}+{ie}^{{cos}^{\mathrm{2}} \theta} {sin}\left(\frac{{sin}\mathrm{2}\theta}{\mathrm{2}}\right) \\ $$$$\boldsymbol{{S}}{o}\:{p}={e}^{{cos}^{\mathrm{2}} \theta} \left[{cos}\left(\frac{{sin}\mathrm{2}\theta}{\mathrm{2}}\right)\right]−\mathrm{1}\leftarrow{real}\:{part} \\ $$$${q}={e}^{{cos}^{\mathrm{2}} \theta} {sin}\left(\frac{{sin}\mathrm{2}\theta}{\mathrm{2}}\right)\leftarrow{imaginary}\:{part} \\ $$$$\boldsymbol{{S}}{o}\:\boldsymbol{{required}}\:\boldsymbol{{answer}}\:\boldsymbol{{is}} \\ $$$$\boldsymbol{{q}}=\boldsymbol{{e}}^{\boldsymbol{{cos}}^{\mathrm{2}} \theta} {sin}\left(\frac{{sin}\mathrm{2}\theta}{\mathrm{2}}\right) \\ $$$${pls}\:{check}…. \\ $$$$ \\ $$
Commented by bhanukumarb2@gmail.com last updated on 17/May/19
$${right}\:{sir}\:{thanku} \\ $$