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Question Number 135557 by bemath last updated on 14/Mar/21

(1/(1−cos θ−i sin θ)) =? i=(√(−1))

$$\frac{\mathrm{1}}{\mathrm{1}−\mathrm{cos}\:\theta−{i}\:\mathrm{sin}\:\theta}\:=? \\ $$$${i}=\sqrt{−\mathrm{1}} \\ $$

Answered by mathmax by abdo last updated on 14/Mar/21

(1/(1−cosθ−isinθ))=(1/(2sin^2 ((θ/2))−2isin((θ/2))cos((θ/2)))) =(1/(−2isin((θ/2)){cos((θ/2))+isin((θ/2))})) =−(1/(2isin((θ/2)) e^((iθ)/2) )) =−(e^(−((iθ)/2)) /(2isin((θ/2))))=−(1/(2isin((θ/2)))){cos((θ/2))−isin((θ/2))} =(i/2)cotan((θ/2))+(1/2)

$$\frac{\mathrm{1}}{\mathrm{1}−\mathrm{cos}\theta−\mathrm{isin}\theta}=\frac{\mathrm{1}}{\mathrm{2sin}^{\mathrm{2}} \left(\frac{\theta}{\mathrm{2}}\right)−\mathrm{2isin}\left(\frac{\theta}{\mathrm{2}}\right)\mathrm{cos}\left(\frac{\theta}{\mathrm{2}}\right)} \\ $$$$=\frac{\mathrm{1}}{−\mathrm{2isin}\left(\frac{\theta}{\mathrm{2}}\right)\left\{\mathrm{cos}\left(\frac{\theta}{\mathrm{2}}\right)+\mathrm{isin}\left(\frac{\theta}{\mathrm{2}}\right)\right\}}\:=−\frac{\mathrm{1}}{\mathrm{2isin}\left(\frac{\theta}{\mathrm{2}}\right)\:\mathrm{e}^{\frac{\mathrm{i}\theta}{\mathrm{2}}} } \\ $$$$=−\frac{\mathrm{e}^{−\frac{\mathrm{i}\theta}{\mathrm{2}}} }{\mathrm{2isin}\left(\frac{\theta}{\mathrm{2}}\right)}=−\frac{\mathrm{1}}{\mathrm{2isin}\left(\frac{\theta}{\mathrm{2}}\right)}\left\{\mathrm{cos}\left(\frac{\theta}{\mathrm{2}}\right)−\mathrm{isin}\left(\frac{\theta}{\mathrm{2}}\right)\right\} \\ $$$$=\frac{\mathrm{i}}{\mathrm{2}}\mathrm{cotan}\left(\frac{\theta}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Commented by bemath last updated on 14/Mar/21

thank you

$${thank}\:{you} \\ $$

Commented by mathmax by abdo last updated on 15/Mar/21

you are welcome

$$\mathrm{you}\:\mathrm{are}\:\mathrm{welcome} \\ $$

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