Prove that $\left(1 + \cos \theta + i \sin \theta\right)^n + \left(1 + \cos \theta - i \sin \theta\right)^n = 2^{n + 1} \cos^{n} \left(\dfrac{\theta}{2}\right) \cos \left(\dfrac{n \theta}{2}\right)$, $\;$ $n \in N$
Let $\;$ $z = \cos \theta + i \sin \theta$
$\because$ $\;$ $\left|z\right| = 1$ $\implies$ $\overline{z} = \dfrac{1}{z}$
$\therefore$ $\;$ From equation $(1a)$, $\;$ $\overline{z} = \dfrac{1}{z} = \cos \theta - i \sin \theta$ $\;\;\; \cdots \; (1)$
$\begin{aligned}
\therefore \; \left(1 + \cos \theta + i \sin \theta\right)^n + \left(1 + \cos \theta - i \sin \theta\right)^n & = \left(1 + z\right)^n + \left(1 + \dfrac{1}{z}\right)^n \\\\
& = \left(1 + z\right)^n + \dfrac{\left(1 + z\right)^n}{z^n} \\\\
& = \left(1 + z\right)^n \left(1 + \dfrac{1}{z^n}\right) \;\;\; \cdots \; (2)
\end{aligned}$
From equation $(1)$,
$\begin{aligned}
\dfrac{1}{z^n} & = \left(\cos \theta - i \sin \theta\right)^n \\\\
& = \cos \left(n \theta\right) - i \sin \left(n \theta\right) \;\; [\text{By De Moivre's theorem}]
\end{aligned}$
$\begin{aligned}
\therefore \; 1 + \dfrac{1}{z^n} & = 1 + \cos \left(n \theta\right) - i \sin \left(n \theta\right) \\\\
& = 2 \cos^2 \left(\dfrac{n \theta}{2}\right) - 2 i \sin \left(\dfrac{n \theta}{2}\right) \cos \left(\dfrac{n \theta}{2}\right) \\\\
& = 2 \cos \left(\dfrac{n \theta}{2}\right) \left[\cos \left(\dfrac{n \theta}{2}\right) - i \sin \left(\dfrac{n \theta}{2}\right)\right] \;\;\; \cdots \; (3a)
\end{aligned}$
$\begin{aligned}
Now, \; \left(1 + z\right) & = \left(1 + \cos \theta\right) + i \sin \theta \\\\
& = 2 \cos^2 \left(\dfrac{\theta}{2}\right) + 2 i \sin \left(\dfrac{\theta}{2}\right) \cos \left(\dfrac{\theta}{2}\right) \\\\
& = 2 \cos \left(\dfrac{\theta}{2}\right) \left[\cos \left(\dfrac{\theta}{2}\right) + i \sin \left(\dfrac{\theta}{2}\right)\right]
\end{aligned}$
$\begin{aligned}
\therefore \; \left(1 + z\right)^n & = \left\{2 \cos \left(\dfrac{\theta}{2}\right) \left[\cos \left(\dfrac{\theta}{2}\right) + i \sin \left(\dfrac{\theta}{2}\right)\right]\right\}^n \\\\
& = 2^n \cos^n \left(\dfrac{\theta}{2}\right) \left[\cos \left(\dfrac{n \theta}{2}\right) + i \sin \left(\dfrac{n \theta}{2}\right)\right] \;\;\; \cdots \; (3b)
\end{aligned}$
$\therefore$ $\;$ We have from equations $(3a)$ and $(3b)$,
$\begin{aligned}
\left(1 + z\right)^n \left(1 + \dfrac{1}{z^n}\right) & = 2^n \cos^n \left(\dfrac{\theta}{2}\right) \left[\cos \left(\dfrac{n \theta}{2}\right) + i \sin \left(\dfrac{n \theta}{2}\right)\right] \\
& \hspace{2.5cm} \times 2 \cos \left(\dfrac{n \theta}{2}\right) \left[\cos \left(\dfrac{n \theta}{2}\right) - i \sin \left(\dfrac{n \theta}{2}\right)\right] \\\\
& = 2^{n+1} \cos^n \left(\dfrac{\theta}{2}\right) \cos \left(\dfrac{n \theta}{2}\right) \\
& \hspace{1cm} \times \left[\cos \left(\dfrac{n \theta}{2}\right) + i \sin \left(\dfrac{n \theta}{2}\right)\right]\left[\cos \left(\dfrac{n \theta}{2}\right) - i \sin \left(\dfrac{n \theta}{2}\right)\right] \\\\
& = 2^{n+1} \cos^n \left(\dfrac{\theta}{2}\right) \cos \left(\dfrac{n \theta}{2}\right) \times \left[\cos^2 \left(\dfrac{n \theta}{2}\right) - i^2 \sin^2 \left(\dfrac{n \theta}{2}\right)\right] \\\\
& = 2^{n+1} \cos^n \left(\dfrac{\theta}{2}\right) \cos \left(\dfrac{n \theta}{2}\right) \times \left[\cos^2 \left(\dfrac{n \theta}{2}\right) + \sin^2 \left(\dfrac{n \theta}{2}\right)\right] \\\\
& = 2^{n+1} \cos^n \left(\dfrac{\theta}{2}\right) \cos \left(\dfrac{n \theta}{2}\right) \;\;\; \cdots \; (4)
\end{aligned}$
$\therefore$ $\;$ From equations $(2)$ and $(4)$ we have,
$\left(1 + \cos \theta + i \sin \theta\right)^n + \left(1 + \cos \theta - i \sin \theta\right)^n = 2^{n + 1} \cos^{n} \left(\dfrac{\theta}{2}\right) \cos \left(\dfrac{n \theta}{2}\right)$, $\;$ $n \in N$
Hence proved.