Question-200646

Question Number 200646 by Calculusboy last updated on 21/Nov/23

Answered by Frix last updated on 21/Nov/23

(((1+sin θ +i cos θ)/(1+sin θ −i cos θ)))^n =(sin θ +i cos θ)^n = =(cos ((π/2)−θ) +i sin ((π/2)−θ))^n =e^(i((π/2)−θ)n) = =cos (((π/2)−θ)n) +i sin (((π/2)−θ)n)

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

Commented by Calculusboy last updated on 22/Nov/23