Question

If $2\cos \theta =x+\frac{1}{x}$ and $2\cos \varphi =y+\frac{1}{y}$, then prove that:$x^m{y}^n+\frac{1}{x^m{y}^n}=2\cos \left(m\theta +n\varphi \right)$

Answer 1

let $x=\cos \theta =e^{i\theta }$

and $y-\frac{1}{y}=2\cos \varphi$

$y=\cos \varphi =e^i{y}^{\varphi }$

now $x^m{y}^m+\frac{1}{x^m{y}^m}={{(e}^{i\theta }{e}^{i\phi })}^m+\frac{1}{{{(e}^{i\theta }{e}^{i\phi })}^m}={(e}^{i\theta m}{e}^{i\phi m})+\frac{1}{{(e}^{i\theta m}{e}^{i\phi m})}=e^{im(\theta +\phi )}+\frac{1}{e^{im(\theta +\phi )}}=e^{im(\theta +\phi )}{+e}^{-im(\theta +\phi )}$

as we know

$\cos A=\frac{e^{ia}{e}^{-ia}}{2}$

using

$2\cos (m\theta +m\varphi )$