Kinsley Moon

2022-01-24

Why is $\frac{\mathrm{cos}\left(\alpha +\beta \right)+\mathrm{cos}\left(-\alpha \right)}{\mathrm{sin}\left(\alpha +\beta \right)+\mathrm{sin}\left(-\alpha \right)}$ independent of $\alpha$?

Prince Huang

$\mathrm{cos}\left(\alpha +\beta \right)+\mathrm{cos}\left(-\alpha \right)=2\mathrm{cos}\frac{\alpha +\alpha +\beta }{2}\mathrm{cos}\frac{\alpha +\beta -\alpha }{2}=2\mathrm{cos}\left\{\frac{\beta }{2}\right\}\mathrm{cos}\left(\alpha +\frac{\beta }{2}\right)$
$\mathrm{sin}\left(\alpha +\beta \right)+\mathrm{sin}\left(-\alpha \right)=2\mathrm{sin}\frac{\alpha +\beta -\alpha }{2}\mathrm{cos}\frac{\alpha +\beta -\left(-\alpha \right)}{2}=2\mathrm{sin}\frac{\beta }{2}\mathrm{cos}\left(\alpha +\frac{\beta }{2}\right)$

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