David Rodgers

2022-03-31

Proof of $\mathrm{cos}\alpha +\mathrm{sin}\beta +\mathrm{cos}\gamma =4\cdot \mathrm{sin}\left(\frac{\alpha }{2}+45°\right)\cdot \mathrm{sin}\frac{\beta }{2}\cdot \mathrm{sin}\left(\frac{\gamma }{2}+45°\right)$
I tried to simplify right side:
$4\left(-\frac{1}{2}\left(\mathrm{cos}\left(\frac{\alpha }{2}+\frac{\gamma }{2}+\frac{\pi }{2}\right)-\mathrm{cos}\left(\frac{\alpha }{2}-\frac{\gamma }{2}\right)\right)\cdot \mathrm{sin}\left(\frac{\beta }{2}\right)=$
$=-2\cdot \mathrm{sin}\left(\beta \right)\cdot \mathrm{cos}\left(\frac{\alpha }{2}+\frac{\gamma }{2}+\frac{\pi }{2}\right)+2\mathrm{cos}\left(\frac{\alpha }{2}-\frac{\gamma }{2}\right)\cdot \mathrm{sin}\left(\frac{\beta }{2}\right)=$
$=-\mathrm{sin}\left(\frac{\beta }{2}+\frac{\alpha }{2}+\frac{\gamma }{2}+\frac{\pi }{2}\right)-\mathrm{sin}\left(\frac{\beta }{2}-\frac{\alpha }{2}-\frac{\gamma }{2}-\frac{\pi }{2}\right)+\mathrm{sin}\left(\frac{\beta }{2}+\frac{\alpha }{2}-\frac{\gamma }{2}\right)+\mathrm{sin}\left(\frac{\beta }{2}-\frac{\alpha }{2}+\frac{\gamma }{2}\right)$

German Ferguson

use that
$\mathrm{sin}\left(\frac{x}{2}+\frac{\pi }{4}\right)=\frac{1}{\sqrt{2}}\left(\left(\mathrm{sin}\left(\frac{x}{2}\right)+\mathrm{cos}\left(\frac{x}{2}\right)\right)$

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