Question Number 60452 by ANTARES VY last updated on 21/May/19
Commented by bhanukumarb2@gmail.com last updated on 21/May/19
$${jension}\:{inequality} \\ $$
Answered by Kunal12588 last updated on 21/May/19
$${cos}\left(\alpha\right)+{cos}\left(\beta\right)+{cos}\left(\gamma\right) \\ $$$$={cos}\left(\alpha\right)+\mathrm{2}{cos}\left(\frac{\beta+\gamma}{\mathrm{2}}\right){cos}\left(\frac{\beta−\gamma}{\mathrm{2}}\right) \\ $$$$={cos}\left(\alpha\right)+\mathrm{2}{cos}\left(\frac{\pi}{\mathrm{2}}−\frac{\alpha}{\mathrm{2}}\right){cos}\left(\frac{\beta−\gamma}{\mathrm{2}}\right) \\ $$$$=\mathrm{1}−\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{\alpha}{\mathrm{2}}\right)+\mathrm{2}{sin}\left(\frac{\alpha}{\mathrm{2}}\right){cos}\left(\frac{\beta−\gamma}{\mathrm{2}}\right) \\ $$$$=\mathrm{1}−\mathrm{2}{sin}\left(\frac{\alpha}{\mathrm{2}}\right)\left\{{sin}\:\frac{\alpha}{\mathrm{2}}\:−\:{cos}\:\frac{\beta−\gamma}{\mathrm{2}}\right\} \\ $$$$=\mathrm{1}−\mathrm{2}{sin}\left(\frac{\alpha}{\mathrm{2}}\right)\left\{{sin}\left(\frac{\pi}{\mathrm{2}}−\frac{\beta+\gamma}{\mathrm{2}}\right)−{cos}\:\frac{\beta−\gamma}{\mathrm{2}}\right\} \\ $$$$=\mathrm{1}−\mathrm{2}{sin}\left(\frac{\alpha}{\mathrm{2}}\right)\left\{{cos}\:\frac{\beta+\gamma}{\mathrm{2}}\:−\:{cos}\:\frac{\beta−\gamma}{\mathrm{2}}\right\} \\ $$$$=\mathrm{1}−\mathrm{2}{sin}\left(\frac{\alpha}{\mathrm{2}}\right)\left(−\mathrm{2}{sin}\left(\frac{\beta}{\mathrm{2}}\right){sin}\left(\frac{\gamma}{\mathrm{2}}\right)\right) \\ $$$$=\mathrm{1}+\mathrm{4}{sin}\left(\frac{\alpha}{\mathrm{2}}\right){sin}\left(\frac{\beta}{\mathrm{2}}\right){sin}\left(\frac{\gamma}{\mathrm{2}}\right) \\ $$$$=\mathrm{1}+\mathrm{4}{a} \\ $$