Answer to Question #122813 in Trigonometry for Ojugbele Daniel

Question

Prove that the general value of theta which satisfies the equation (costheta+isintheta)×(cos2theta+isintheta).... (Cosntheta+isintheta)=1 is 4mπ over n(n+1) where m is any integer

Answer 1

(cos$\theta$ +isin$\theta$ )=eⁱ$^\theta$

(cos2$\theta$ +isin2$\theta$ )=eⁱ²$^\theta$

(cosn$\theta$ +isinn$\theta$ )=eⁱⁿ$^\theta$

so

(cos$\theta$ +isin$\theta$ )×(cos2$\theta$ +isin2$\theta$ ).... (Cosn$\theta$ +isinn$\theta$ =1

(eⁱ$^\theta$)x(eⁱ²$^\theta$).....(eⁱⁿ$^\theta$)=1

eⁱ$^\theta$^(1+2+...+n)=1

eⁱ$^\theta$^{n(n+1)/2}=1

now we write eⁱ$^\theta$  in terms of cos and sin

cos[n(n+1)/2]$\theta$ + i sin[n(n+1)/2]$\theta$ = 1

for value of $\theta$

cos [n(n+1)/2]$\theta$ = 1

[n(n+1)/2 ]$\theta$ = cos^-1

[n(n+1)/2]$\theta$ = 2mπ

$\theta$ =4mπ/n(n+1