Question and Answers Forum

All Questions      Topic List

Trigonometry Questions

Previous in All Question      Next in All Question      

Previous in Trigonometry      Next in Trigonometry      

Question Number 34117 by tanmay.chaudhury50@gmail.com last updated on 30/Apr/18

Commented by abdo imad last updated on 01/May/18

$$letputf\left(x\right)=\frac{sin\theta }{x^2-2xcos\theta +1}polesoff\left(x\right)?$$$$x^2-2xcos\theta +1=0$$$${\mathrm{\Delta }}^{{}^{\prime }}={cos}^2\theta -1=-{sin}^2\theta ={\left(isin\theta \right)}^2$$$$x_1=cos\theta +isin\theta =e^{i\theta }and{x}_2=e^{-i\theta }\Rightarrow$$$$f\left(x\right)=\frac{sin\theta }{(x-e^{i\theta })(x-e^{-i\theta })}=sin\theta (\frac{1}{x-e^{i\theta }}-\frac{1}{x-e^{-i\theta }}).\frac{1}{2isin\theta }$$$$=\frac{1}{2i}(\frac{-1}{e^{i\theta }-x}+\frac{1}{e^{-i\theta }-x})$$$$=\frac{1}{2i}(\frac{-e^{-i\theta }}{1-{xe}^{-i\theta }}+\frac{e^{i\theta }}{1-x{e}^{i\theta }})$$$$=\frac{1}{2i}(e^{i\theta }\sum _{n=0}^{\mathrm{\infty }}{x}^n{e}^{in\theta }-e^{-i\theta }\sum _{n=0}^{\mathrm{\infty }}{x}^n{e}^{-in\theta })$$$$=\frac{1}{2i}(\sum _{n=0}^{\mathrm{\infty }}{e}^{i(n+1)\theta }{x}^n-\sum _{n=0}^{\mathrm{\infty }}{e}^{-i(n+1)\theta }{x}^n)$$$$=\frac{1}{2i}\sum _{n=0}^{\mathrm{\infty }}(e^{i(n+1)\theta }-e^{-i(n+1)\theta })x^n$$$$=\sum _{n=0}^{\mathrm{\infty }}sin\left((n+1)\theta \right)x^n$$$$=sin\theta +sin\left(2\theta \right)x^2+sin\left(3\theta \right)x^3+....$$$$$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com