Dania Mueller

2022-07-01

I have been trying to simplify this expression:
$\mathrm{cos}\left(\theta \right)-\frac{1}{\mathrm{cos}\left(\theta \right)+i\mathrm{sin}\left(\theta \right)}$

Kaydence Washington

Expert

$cos\left(\theta \right)-\frac{1}{\mathrm{cos}\left(\theta \right)+i\mathrm{sin}\left(\theta \right)}=cos\left(\theta \right)-\frac{1}{{e}^{i\theta }}=$
$cos\left(\theta \right)-{e}^{-i\theta }=\frac{1}{2}{e}^{i\theta }+\frac{1}{2}{e}^{-i\theta }-{e}^{-i\theta }=$
$=\frac{1}{2}{e}^{i\theta }-\frac{1}{2}{e}^{-i\theta }=i\left(\frac{1}{2i}{e}^{i\theta }-\frac{1}{2i}{e}^{-i\theta }\right)=i\mathrm{sin}\left(\theta \right)$

Gaaljh

Expert

HINT:
${e}^{-i\theta }=\mathrm{cos}\left(-\theta \right)+i\mathrm{sin}\left(-\theta \right)=\mathrm{cos}\theta -i\mathrm{sin}\theta$
$\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\mathrm{cos}\theta -i\mathrm{sin}\theta =\frac{1}{\mathrm{cos}\theta +i\mathrm{sin}\theta }$
which is also evident from $\left(\mathrm{cos}\theta +i\mathrm{sin}\theta \right)\left(\mathrm{cos}\theta -i\mathrm{sin}\theta \right)=\cdots =1$

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