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Question Number 46085 by peter frank last updated on 20/Oct/18

Commented by maxmathsup by imad last updated on 21/Oct/18

$$letS=\sum _{n=1}^{\mathrm{\infty }}{cos}^n\theta cos\left(n\theta \right)=Re\left(\sum _{n=1}^{\mathrm{\infty }}{e}^{in\theta }{cos}^n\theta \right)but$$$$\sum _{n=1}^{\mathrm{\infty }}{e}^{in\theta }{cos}^n\theta =\sum _{n=1}^{\mathrm{\infty }}{e}^{in\theta }{\left(\frac{e^{i\theta }+e^{-i\theta }}{2}\right)}^n$$$$=\sum _{n=1}^{\mathrm{\infty }}\frac{e^{in\theta }}{2^n}\left(\sum _{k=0}^n{C}_n^k{e}^{ik\theta }{e}^{-i(n-k)\theta }\right)$$$$=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{2^n}\left(\sum _{k=0}^n{C}_n^k{e}^{2ik\theta }\right)=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{2^n}\sum _{k=0}^n{C}_n^k{\left(e^{2i\theta }\right)}^k$$$$=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{2^n}{(1+e^{2i\theta })}^n=\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{1+e^{2i\theta }}{2}\right)}^n=\frac{1}{1-\frac{1+e^{2i\theta }}{2}}-1$$$$=\frac{2}{1-e^{2i\theta }}-1=\frac{2-1+e^{2i\theta }}{1-e^{2i\theta }}=\frac{1+e^{2i\theta }}{1-e^{2i\theta }}=\frac{1+cos\left(2\theta \right)+isin\left(2\theta \right)}{1-cos\left(2\theta \right)-isin\left(2\theta \right)}$$$$=\frac{2{cos}^2\left(\theta \right)+2isin\theta cos\theta }{2{sin}^2\theta -2isin\theta cos\theta }=\frac{cos\theta e^{i\theta }}{-isin\theta e^{i\theta }}=icotan\theta \Rightarrow S=0$$$$$$

Commented by peter frank last updated on 21/Oct/18

$$\mathrm{thanks}\mathrm{sir}$$

Answered by tanmay.chaudhury50@gmail.com last updated on 21/Oct/18

$$p=cos\theta cos\theta +{cos}^2\theta cos2\theta +{cos}^3\theta cos3\theta +...$$$$q=cos\theta sin\theta +{cos}^2\theta sin2\theta +{cos}^3\theta sin3\theta +...$$$$now{e}^{i\alpha }=cos\alpha +isin\alpha$$$$p+iq=cos\theta e^{i\theta }+{cos}^2\theta e^{i2\theta }+{cos}^3\theta e^{i3\theta }+...$$$$p+iq=t+t^2+t^3+...\mathrm{\infty }$$$$hereiamediting...$$$$sumoftermsuptoinfinity$$$$S_{\mathrm{\infty }}=\frac{t}{1-t}=\frac{cos\theta e^{i\theta }}{1-cos\theta e^{i\theta }}$$$$=\frac{cos\theta (cos\theta +isin\theta )}{1-cos\theta (cos\theta +isin\theta )}$$$$$$$$$$$$=\frac{cos\theta (cos\theta +isn\theta )}{1-cos\theta (cos\theta +isin\theta )}$$$$=\frac{{cos}^2\theta +isin\theta cos\theta }{1-{cos}^2\theta -isin\theta cos\theta }$$$$=\frac{cos\theta }{sin\theta }\times \frac{cos\theta +isin\theta }{sin\theta -icos\theta }$$$$=cot\theta \times \frac{(cos\theta +isin\theta )(sin\theta +icos\theta )}{(sin\theta -icos\theta )(sin\theta +icos\theta )}$$$$=cot\theta \times \frac{cos\theta sin\theta +{icos}^2\theta +{isin}^2\theta -sin\theta cos\theta }{{sin}^2\theta +{cos}^2\theta }$$$$=cot\theta \times i$$$$sop+iq=0+icot\theta$$$$sop=0$$$$q=cot\theta$$$$$$$$\mathit{s}\mathit{o}\mathit{r}\mathit{e}\mathit{q}\mathit{u}\mathit{i}\mathit{r}\mathit{e}\mathit{d}\mathit{a}\mathit{n}\mathit{s}\mathit{w}\mathit{e}\mathit{r}\mathit{f}\mathit{o}\mathit{r}\mathit{t}\mathit{h}\mathit{e}\mathit{g}\mathit{i}\mathit{v}\mathit{e}\mathit{n}\mathit{s}\mathit{e}\mathit{r}\mathit{i}\mathit{e}\mathit{s}\mathit{i}\mathit{s}\mathit{z}\mathit{e}\mathit{r}\mathit{o}$$$$$$

Commented by peter frank last updated on 21/Oct/18

$$\mathrm{okay}\mathrm{sir}$$

Commented by peter frank last updated on 21/Oct/18

$$\mathrm{sorry}\mathrm{sir}\mathrm{first}\mathrm{line}\mathrm{with}\mathrm{red}\mathrm{colour}.$$

Commented by tanmay.chaudhury50@gmail.com last updated on 21/Oct/18

$$plscheck...ihavecorrected$$

Commented by peter frank last updated on 21/Oct/18

$$\mathrm{thank}\mathrm{you}\mathrm{sir}.$$

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