Answer to Question #122816 in Trigonometry for Ojugbele Daniel

Question

Prove that (1+costheta+isintheta)^n + (1+costheta-isintheta)^n =2^n+1 cos^n π/2 cosntheta/2

Answer 1

To prove $(1+cos\theta+isin\theta)^n+(1+cos\theta-isin\theta)^n=2^{n+1}cos^n\frac{\theta}{2}cos\frac{n\theta}{2}$

$we \space know\space that \space (1+cos\theta)=2cos^2\frac{\theta}{2}$ and $sin\theta=2sin\frac{\theta}{2}cos\frac{\theta}{2}$ .

L.H.S

$\space \space \space \space (1+cos\theta+isin\theta)^n+(1+cos\theta-isin\theta)^n\\ \space \\=(2cos^2\frac{\theta}{2}+i2sin\frac{\theta}{2}cos\frac{\theta}{2})^n+(2cos^2\frac{\theta}{2}-i2sin\frac{\theta}{2}cos\frac{\theta}{2})^n\\ \space \\=2^ncos^n\frac{\theta}{2}(cos\frac{\theta}{2}+isin\frac{\theta}{2})^n \space + 2^ncos^n\frac{\theta}{2}(cos\frac{\theta}{2}-isin\frac{\theta}{2})^n$

$=2^ncos^n\frac{\theta}{2}\bigg((cos\frac{\theta}{2}+isin\frac{\theta}{2})^n \space + (cos\frac{\theta}{2}-isin\frac{\theta}{2})^n\bigg)$

$Now \space using \space DeMoivre's \space Theorem \space which \space says \\(cos\alpha+isin\alpha)^n=cosn\alpha \space + isinn\alpha$

$=2^ncos^n\frac{\theta}{2}\bigg(cos\frac{n\theta}{2}+isin\frac{n\theta}{2}\space + cos\frac{n\theta}{2}-isin\frac{n\theta}{2}\bigg)$

$=2^ncos^n\frac{\theta}{2}(2cos\frac{n\theta}{2}).$

$=2^{n+1}cos^n\frac{\theta}{2}cos\frac{n\theta}{2}. \\=R.H.S$

Proved.