If $x = \cos \alpha + i \sin \alpha$; $\;$ $y = \cos \beta + i \sin \beta$, $\;$ then prove that $\;$ $x^m y^n + \dfrac{1}{x^m y^n} = 2 \cos \left(m \alpha + n \beta\right)$
Given: $\;$ $x = \cos \alpha + i \sin \alpha$; $\;\;$ $y = \cos \beta + i \sin \beta$
$\implies$ $\dfrac{1}{x} = \cos \alpha - i \sin \alpha$; $\;$ $\dfrac{1}{y} = \cos \beta - i \sin \beta$
Now, $\;$ $x^m = \left(\cos \alpha + i \sin \alpha\right)^m = \cos \left(m \alpha\right) + i \sin \left(m \alpha
\right)$
$y^n = \left(\cos \beta + i \sin \beta\right)^n = \cos \left(n \beta\right) + i \sin \left(n \beta\right)$
$\begin{aligned}
\therefore \; x^m y^n & = \left[\cos \left(m \alpha\right) + i \sin \left(m \alpha\right)\right] \left[\cos \left(n \beta\right) + i \sin \left(n \beta\right)\right] \\\\
& = \left[\cos \left(m \alpha\right) \cos \left(n \beta\right) + i^2 \sin \left(m \alpha\right) \sin \left(n \beta\right)\right] \\
& \hspace{2cm} + i \left[\sin \left(m \alpha\right) \cos \left(n \beta\right) + \sin \left(n \beta\right) \cos \left(m \alpha\right)\right] \\\\
& = \left[\cos \left(m \alpha\right) \cos \left(n \beta\right) - \sin \left(m \alpha\right) \sin \left(n \beta\right)\right] \\
& \hspace{2cm} + i \left[\sin \left(m \alpha\right) \cos \left(n \beta\right) + \sin \left(n \beta\right) \cos \left(m \alpha\right)\right] \\\\
& = \cos \left(m \alpha + n \beta\right) + i \sin \left(m \alpha + n \beta\right)
\end{aligned}$
$\dfrac{1}{x^m} = \left(\cos \alpha - i \sin \alpha\right)^m = \cos \left(m \alpha\right) - i \sin \left(m \alpha\right)$
$\dfrac{1}{y^n} = \left(\cos \beta - i \sin \beta\right)^n = \cos \left(n \beta\right) - i \sin \left(n \beta\right)$
$\begin{aligned}
\therefore \; \dfrac{1}{x^m y^n} & = \left[\cos \left(m \alpha\right) - i \sin \left(m \alpha\right)\right] \left[\cos \left(n \beta\right) - i \sin \left(n \beta\right)\right] \\\\
& = \left[\cos \left(m \alpha\right) \cos \left(n \beta\right) + i^2 \sin \left(m \alpha\right) \sin \left(n \beta\right)\right] \\
& \hspace{2cm} - i \left[\sin \left(m \alpha\right) \cos \left(n \beta\right) + \sin \left(n \beta\right) \cos \left(m \alpha\right)\right] \\\\
& = \left[\cos \left(m \alpha\right) \cos \left(n \beta\right) - \sin \left(m \alpha\right) \sin \left(n \beta\right)\right] \\
& \hspace{2cm} - i \left[\sin \left(m \alpha\right) \cos \left(n \beta\right) + \sin \left(n \beta\right) \cos \left(m \alpha\right)\right] \\\\
& = \cos \left(m \alpha + n \beta\right) - i \sin \left(m \alpha + n \beta\right)
\end{aligned}$
$\begin{aligned}
\therefore \; x^m y^n + \dfrac{1}{x^m y^n} & = \cos \left(m \alpha + n \beta\right) + i \sin \left(m \alpha + n \beta\right) \\
& \hspace{2cm} + \cos \left(m \alpha + n \beta\right) - i \sin \left(m \alpha + n \beta\right) \\\\
& = 2 \cos \left(m \alpha + n \beta\right)
\end{aligned}$
Hence proved.