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Question Number 134475 by EDWIN88 last updated on 04/Mar/21
If 1+cos x+cos 2x+cos 3x+cos 4x+...+∞ = 3 for 0<x≤(π/2) find the value of sin x+sin 2x+sin 3x+...+∞
$$\mathrm{If}\:\mathrm{1}+\mathrm{cos}\:\mathrm{x}+\mathrm{cos}\:\mathrm{2x}+\mathrm{cos}\:\mathrm{3x}+\mathrm{cos}\:\mathrm{4x}+…+\infty\:=\:\mathrm{3} \\ $$$$\mathrm{for}\:\mathrm{0}<\mathrm{x}\leqslant\frac{\pi}{\mathrm{2}} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{sin}\:\mathrm{x}+\mathrm{sin}\:\mathrm{2x}+\mathrm{sin}\:\mathrm{3x}+…+\infty \\ $$
Commented by Dwaipayan Shikari last updated on 04/Mar/21
Σ_(n=0) ^∞ cos(nx)=(1/2)Σ_(n=0) ^∞ e^(inx) +(1/2)Σ_(n=0) ^∞ e^(−inx) =(1/2).(1/(1−e^(ix) ))+(1/2).(e^(ix) /(e^(ix) −1)) =(1/2) Σ_(n=1) ^∞ ((cos(nx))/n)=−log(4sin^2 (x/2)) −Σ_(n=1) ^∞ sin(nx)=−(d/dx)log(4sin^2 (x/2)) Σ_(n=1) ^∞ sin(nx)=((2sinx)/(2−2cosx))=((sinx)/(1−cosx))
$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{cos}\left({nx}\right)=\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{e}^{{inx}} +\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{e}^{−{inx}} =\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{1}}{\mathrm{1}−{e}^{{ix}} }+\frac{\mathrm{1}}{\mathrm{2}}.\frac{{e}^{{ix}} }{{e}^{{ix}} −\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\: \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{cos}\left({nx}\right)}{{n}}=−{log}\left(\mathrm{4}{sin}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}\right) \\ $$$$−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{sin}\left({nx}\right)=−\frac{{d}}{{dx}}{log}\left(\mathrm{4}{sin}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}\right) \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{sin}\left({nx}\right)=\frac{\mathrm{2}{sinx}}{\mathrm{2}−\mathrm{2}{cosx}}=\frac{{sinx}}{\mathrm{1}−{cosx}} \\ $$
Commented by benjo_mathlover last updated on 04/Mar/21
what the answer sir?
$$\mathrm{what}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{sir}? \\ $$
Commented by Dwaipayan Shikari last updated on 04/Mar/21
If 1+cosx+cos^2 x+cos^3 x+...=(1/(1−cosx))=3 ⇒cosx=(2/3)⇒sinx=((√5)/3) Then sinx+sin^2 x+...=((sinx)/(1−sinx))=((√5)/(3−(√5)))=((√5)/4)((√5)+3) =(1/4)(3(√5)+5) :)
$${If}\:\: \\ $$$$\mathrm{1}+{cosx}+{cos}^{\mathrm{2}} {x}+{cos}^{\mathrm{3}} {x}+…=\frac{\mathrm{1}}{\mathrm{1}−{cosx}}=\mathrm{3} \\ $$$$\Rightarrow{cosx}=\frac{\mathrm{2}}{\mathrm{3}}\Rightarrow{sinx}=\frac{\sqrt{\mathrm{5}}}{\mathrm{3}} \\ $$$${Then}\:{sinx}+{sin}^{\mathrm{2}} {x}+…=\frac{{sinx}}{\mathrm{1}−{sinx}}=\frac{\sqrt{\mathrm{5}}}{\mathrm{3}−\sqrt{\mathrm{5}}}=\frac{\sqrt{\mathrm{5}}}{\mathrm{4}}\left(\sqrt{\mathrm{5}}+\mathrm{3}\right) \\ $$$$\left.=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{3}\sqrt{\mathrm{5}}+\mathrm{5}\right)\:\::\right) \\ $$
Commented by EDWIN88 last updated on 04/Mar/21
The available answer (a) 2((√5)−3) (c) (1/4)(3(√5)−5) (b) (1/2)(2(√3)+2) (d) (1/4)(3(√5)+5) (e) (1/2)(2(√3)−2)
$$\mathrm{The}\:\mathrm{available}\:\mathrm{answer}\: \\ $$$$\left(\mathrm{a}\right)\:\mathrm{2}\left(\sqrt{\mathrm{5}}−\mathrm{3}\right)\:\:\:\:\:\:\left(\mathrm{c}\right)\:\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{3}\sqrt{\mathrm{5}}−\mathrm{5}\right) \\ $$$$\left(\mathrm{b}\right)\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{2}\right)\:\:\:\left(\mathrm{d}\right)\:\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{3}\sqrt{\mathrm{5}}+\mathrm{5}\right) \\ $$$$\left(\mathrm{e}\right)\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{2}\right) \\ $$
Commented by Dwaipayan Shikari last updated on 04/Mar/21
May be error in your question sir!
$${May}\:{be}\:{error}\:{in}\:{your}\:{question}\:{sir}! \\ $$
Commented by EDWIN88 last updated on 04/Mar/21
yes sir .i think it wrong
$$\mathrm{yes}\:\mathrm{sir}\:.\mathrm{i}\:\mathrm{think}\:\mathrm{it}\:\mathrm{wrong} \\ $$

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