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Prove: $\mathrm{sin}\alpha +\mathrm{sin}\beta +\mathrm{sin}\gamma =4\mathrm{cos}\frac{\alpha }{2}\mathrm{cos}\frac{\beta }{2}\mathrm{cos}\frac{\gamma }{2}$ when $\alpha +\beta +\gamma =\pi$
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You may go this way:
$\mathrm{cos}\frac{\gamma }{2}=\mathrm{cos}\frac{\pi -\alpha -\beta }{2}=\mathrm{sin}\frac{\alpha +\beta }{2}=\mathrm{sin}\frac{\alpha }{2}\mathrm{cos}\frac{\beta }{2}+\mathrm{cos}\frac{\alpha }{2}\mathrm{sin}\frac{\beta }{2}$
so the right hand side becomes
$4\mathrm{cos}\frac{\alpha }{2}\mathrm{cos}\frac{\beta }{2}\mathrm{sin}\frac{\alpha }{2}\mathrm{cos}\frac{\beta }{2}+4\mathrm{cos}\frac{\alpha }{2}\mathrm{cos}\frac{\beta }{2}\mathrm{cos}\frac{\alpha }{2}\mathrm{sin}\frac{\beta }{2}$
Recalling the duplication formula for the sine we get
$2\mathrm{sin}\alpha {\mathrm{cos}}^{2}\frac{\beta }{2}+2\mathrm{sin}\beta {\mathrm{cos}}^{2}\frac{\alpha }{2}$
and we can recall
$2{\mathrm{cos}}^{2}\frac{\delta }{2}=1+\mathrm{cos}\delta$
to get
$\mathrm{sin}\alpha +\mathrm{sin}\alpha \mathrm{cos}\beta +\mathrm{sin}\beta +\mathrm{sin}\beta \mathrm{cos}\alpha =\mathrm{sin}\alpha +\mathrm{sin}\beta +\mathrm{sin}\left(\alpha +\beta \right)=\mathrm{sin}\alpha +\mathrm{sin}\beta +\mathrm{sin}\gamma$