Patatiniuh

2022-07-01

Let $z=\mathrm{cos}x+i\mathrm{sin}x$. Show that
$1+z=2\mathrm{cos}\frac{1}{2}x\left(\mathrm{cos}\frac{1}{2}x+i\mathrm{sin}\frac{1}{2}x\right)$

Miguidi4y

Expert

Since
$\begin{array}{rcl}\mathrm{cos}x& =& 2{\mathrm{cos}}^{2}\frac{x}{2}-1\\ \mathrm{sin}x& =& 2\mathrm{sin}\frac{x}{2}\mathrm{cos}\frac{x}{2},\end{array}$
we have
$\begin{array}{rcl}1+z& =& 1+\mathrm{cos}x+i\mathrm{sin}x\\ & =& 1+\left(2{\mathrm{cos}}^{2}\frac{x}{2}-1\right)+i\left(2\mathrm{sin}\frac{x}{2}\mathrm{cos}\frac{x}{2}\right)\\ & =& 2{\mathrm{cos}}^{2}\frac{x}{2}+2i\mathrm{sin}\frac{x}{2}\mathrm{cos}\frac{x}{2}\\ & =& 2\mathrm{cos}\frac{x}{2}\left(\mathrm{cos}\frac{x}{2}+i\mathrm{sin}\frac{x}{2}\right)\end{array}$

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