Olimpíadas ⇒ Elementos Da Matemática Volume 5 (Exercícios Gerais) Tópico resolvido

[BalekiEN]

Prove que:

[tex3]1+a.Cos(\theta) + a^{2}.Cos(2\theta )+...=\frac{1-aCos(\theta )}{1-2aCos(\theta )+a^{2}} [/tex3]

Minha tentativa:

[jedi]

[tex3]1+a.\cos(\theta)+a^2.c\cos(2\theta)+\dots=1+a\frac{(e^{i\theta}+e^{-i\theta})}{2}+a^2\frac{(e^{i2\theta}+e^{-i2\theta})}{2}+\dots[/tex3]

[tex3]=\frac{1+ae^{i\theta}+a^2e^{i2\theta}+\dots}{2}+\frac{1+ae^{-i\theta}+a^2e^{-i2\theta}+\dots}{2}[/tex3]

[tex3]=\frac{1}{2}.\frac{1}{1-ae^{i\theta}}+\frac{1}{2}.\frac{1}{1-ae^{-i\theta}}[/tex3]

[tex3]=\frac{1}{2}.\left(\frac{1-ae^{-i\theta}+1-ae^{i\theta}}{(1-ae^{i\theta}).(1-ae^{-i\theta})}\right)[/tex3]

[tex3]=\frac{1}{2}.\left(\frac{2-2a\cos(\theta)}{1-e^{i\theta}-e^{-i\theta}+a^2}\right)[/tex3]

[tex3]=\frac{1}{2}.\left(\frac{2-2a\cos(\theta)}{1-2a\cos(\theta)+a^2}\right)[/tex3]

[tex3]=\frac{1-a\cos(\theta)}{1-2a\cos(\theta)+a^2}[/tex3]