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Question Number 68206 by turbo msup by abdo last updated on 07/Sep/19

$$findS\left(\theta \right)=\sum _{n=0}^{\mathrm{\infty }}\frac{{sin}^3\left(n\theta \right)}{n!}$$

Answered by Smail last updated on 07/Sep/19

$${sin}^2\left(n\theta \right)=\frac{3}{4}sin\left(n\theta \right)-\frac{1}{4}sin\left(3n\theta \right)$$$$Thus,$$$$S\left(\theta \right)=\frac{1}{4}\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{3sin\left(n\theta \right)-sin\left(3n\theta \right)}{n!}$$$$=\frac{3}{4}\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{sin\left(n\theta \right)}{n!}-\frac{1}{4}\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{sin\left(3n\theta \right)}{n!}$$$$=\frac{3}{4}Im\left(\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{e^{in\theta }}{n!}\right)-\frac{1}{4}Im\left(\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{e^{3in\theta }}{n!}\right)$$$$=\frac{3}{4}Im\left(e^{e^{i\theta }}\right)-\frac{1}{4}Im\left(e^{e^{3i\theta }}\right)$$$$Im\left(e^{e^{i\theta }}\right)=Im\left(e^{cos\theta }{e}^{isin\theta }\right)=e^{cos\theta }sin\left(sin\theta \right)$$$$Im\left(e^{e^{3i\theta }}\right)=e^{cos3\theta }sin\left(sin\left(3\theta \right)\right)$$$$S\left(\theta \right)=\frac{1}{4}(3e^{cos\theta }sin\left(sin\theta \right)-e^{cos3\theta }sin\left(sin\left(3\theta \right)\right))$$

Commented by mathmax by abdo last updated on 07/Sep/19

$$thankyousir.$$

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