Question Number 158005 by mkam last updated on 30/Oct/21
Commented by mkam last updated on 30/Oct/21
$${how}\:{can}\:{it}\:{prove}\:{this}\:{help}\:{me}\:{please}\:? \\ $$
Commented by mkam last updated on 30/Oct/21
$$?????\:\: \\ $$
Commented by benhamimed last updated on 30/Oct/21
![z=1+sin θ +icos θ =1+cos ((π/2)−θ)+isin ((π/2)−θ) =1+e^(i((π/2)−θ)) =e^(i((π/4)−(θ/2))) (e^(−i((π/4)−(θ/2))) +e^(i((π/4)−(θ/2))) ) =2cos ((π/4)−(θ/2)).e^(i((π/4)−(θ/2))) [((1+sin θ +icos θ)/(1+sin θ −icos θ))]^n =[(z/z^− )]^n =[e^(i((π/2)−θ)) ]^n =e^(i(((nπ)/2)−nθ)) =cos (((nπ)/2)−nθ)+isin (((nπ)/2)−nθ)](Q158084.png)
$${z}=\mathrm{1}+\mathrm{sin}\:\theta\:+{i}\mathrm{cos}\:\theta \\ $$$$=\mathrm{1}+\mathrm{cos}\:\left(\frac{\pi}{\mathrm{2}}−\theta\right)+{i}\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}−\theta\right) \\ $$$$=\mathrm{1}+{e}^{{i}\left(\frac{\pi}{\mathrm{2}}−\theta\right)} \\ $$$$={e}^{{i}\left(\frac{\pi}{\mathrm{4}}−\frac{\theta}{\mathrm{2}}\right)} \left({e}^{−{i}\left(\frac{\pi}{\mathrm{4}}−\frac{\theta}{\mathrm{2}}\right)} +{e}^{{i}\left(\frac{\pi}{\mathrm{4}}−\frac{\theta}{\mathrm{2}}\right)} \right) \\ $$$$=\mathrm{2cos}\:\left(\frac{\pi}{\mathrm{4}}−\frac{\theta}{\mathrm{2}}\right).{e}^{{i}\left(\frac{\pi}{\mathrm{4}}−\frac{\theta}{\mathrm{2}}\right)} \\ $$$$\left[\frac{\mathrm{1}+\mathrm{sin}\:\theta\:+{i}\mathrm{cos}\:\theta}{\mathrm{1}+\mathrm{sin}\:\theta\:−{i}\mathrm{cos}\:\theta}\right]^{{n}} =\left[\frac{{z}}{\overset{−} {{z}}}\right]^{{n}} \\ $$$$=\left[{e}^{{i}\left(\frac{\pi}{\mathrm{2}}−\theta\right)} \right]^{{n}} ={e}^{{i}\left(\frac{{n}\pi}{\mathrm{2}}−{n}\theta\right)} \\ $$$$=\mathrm{cos}\:\left(\frac{{n}\pi}{\mathrm{2}}−{n}\theta\right)+{i}\mathrm{sin}\:\left(\frac{{n}\pi}{\mathrm{2}}−{n}\theta\right) \\ $$