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# Show that, (1+cos theta + i*sin theta)^n + (1+cos theta - i*sin theta)^n = 2^(n+1) * (cos theta/2)^n * cos (n*theta/2) ?

Jan 27, 2018

#### Explanation:

Let $1 + \cos \theta + i \sin \theta = r \left(\cos \alpha + i \sin \alpha\right)$, here $r = \sqrt{{\left(1 + \cos \theta\right)}^{2} + {\sin}^{2} \theta} = \sqrt{2 + 2 \cos \theta}$
= $\sqrt{2 + 4 {\cos}^{2} \left(\frac{\theta}{2}\right) - 2} = 2 \cos \left(\frac{\theta}{2}\right)$

and $\tan \alpha = \sin \frac{\theta}{1 + \cos \theta} = = \frac{2 \sin \left(\frac{\theta}{2}\right) \cos \left(\frac{\theta}{2}\right)}{2 {\cos}^{2} \left(\frac{\theta}{2}\right)} = \tan \left(\frac{\theta}{2}\right)$ or $\alpha = \frac{\theta}{2}$

then $1 + \cos \theta - i \sin \theta = r \left(\cos \left(- \alpha\right) + i \sin \left(- \alpha\right)\right) = r \left(\cos \alpha - i \sin \alpha\right)$

and we can write ${\left(1 + \cos \theta + i \sin \theta\right)}^{n} + {\left(1 + \cos \theta - i \sin \theta\right)}^{n}$ using DE MOivre's theorem as

${r}^{n} \left(\cos n \alpha + i \sin n \alpha + \cos n \alpha - i \sin n \alpha\right)$

= $2 {r}^{n} \cos n \alpha$

= $2 \cdot {2}^{n} {\cos}^{n} \left(\frac{\theta}{2}\right) \cos \left(\frac{n \theta}{2}\right)$

= ${2}^{n + 1} {\cos}^{n} \left(\frac{\theta}{2}\right) \cos \left(\frac{n \theta}{2}\right)$

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