If $n$ is an integer, then show that : $(1 + \cos heta + i\sin heta)^n + (1 + \cos heta - i\sin heta)^n = 2^{n+1}\cos^n\left(\dfrac{ heta}{2} ight)\cdot\cos\left(\dfrac{n heta}{2} ight)$.

Question

If $n$ is an integer, then show that : $(1 + \cos	heta + i\sin	heta)^n + (1 + \cos	heta - i\sin	heta)^n = 2^{n+1}\cos^n\left(\dfrac{	heta}{2}ight)\cdot\cos\left(\dfrac{n	heta}{2}ight)$.

Accepted Answer

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> [!TLDR]
> Writing each base as $2\cos	frac	heta2(\cos	frac	heta2 \pm i\sin	frac	heta2)$ and using De Moivre's theorem gives the required identity.
>
> Using the half-angle identities $1 + \cos	heta = 2\cos^2	frac	heta2$ and $\sin	heta = 2\sin	frac	heta2\cos	frac	heta2$:
>
> $1 + \cos	heta + i\sin	heta = 2\cos^2	frac	heta2 + 2i\sin	frac	heta2\cos	frac	heta2 = 2\cos	frac	heta2\left(\cos	frac	heta2 + i\sin	frac	heta2ight)$.
>
> Similarly,
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> $1 + \cos	heta - i\sin	heta = 2\cos	frac	heta2\left(\cos	frac	heta2 - i\sin	frac	heta2ight)$.
>
> Raising to the $n$th power and applying De Moivre's theorem:
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> $(1 + \cos	heta + i\sin	heta)^n = 2^n\cos^n	frac	heta2\left(\cos	frac{n	heta}{2} + i\sin	frac{n	heta}{2}ight)$,
> …

Q.If $n$ is an integer, then show that : $(1 + \cos\theta + i\sin\theta)^n + (1 + \cos\theta - i\sin\theta)^n = 2^{n+1}\cos^n\left(\dfrac{\theta}{2}\right)\cdot\cos\left(\dfrac{n\theta}{2}\right)$ .

Concept understanding — Complex Number Transform

The Intuition: Why "Transform" a Number?

The Precise Statement

A Concrete Example

Why This Matters

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