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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

$${jension}\:{inequality} \\ $$

Answered by Kunal12588 last updated on 21/May/19

cos(α)+cos(β)+cos(γ)  =cos(α)+2cos(((β+γ)/2))cos(((β−γ)/2))  =cos(α)+2cos((π/2)−(α/2))cos(((β−γ)/2))  =1−2sin^2 ((α/2))+2sin((α/2))cos(((β−γ)/2))  =1−2sin((α/2)){sin (α/2) − cos ((β−γ)/2)}  =1−2sin((α/2)){sin((π/2)−((β+γ)/2))−cos ((β−γ)/2)}  =1−2sin((α/2)){cos ((β+γ)/2) − cos ((β−γ)/2)}  =1−2sin((α/2))(−2sin((β/2))sin((γ/2)))  =1+4sin((α/2))sin((β/2))sin((γ/2))  =1+4a

$${cos}\left(\alpha\right)+{cos}\left(\beta\right)+{cos}\left(\gamma\right) \\ $$$$={cos}\left(\alpha\right)+\mathrm{2}{cos}\left(\frac{\beta+\gamma}{\mathrm{2}}\right){cos}\left(\frac{\beta−\gamma}{\mathrm{2}}\right) \\ $$$$={cos}\left(\alpha\right)+\mathrm{2}{cos}\left(\frac{\pi}{\mathrm{2}}−\frac{\alpha}{\mathrm{2}}\right){cos}\left(\frac{\beta−\gamma}{\mathrm{2}}\right) \\ $$$$=\mathrm{1}−\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{\alpha}{\mathrm{2}}\right)+\mathrm{2}{sin}\left(\frac{\alpha}{\mathrm{2}}\right){cos}\left(\frac{\beta−\gamma}{\mathrm{2}}\right) \\ $$$$=\mathrm{1}−\mathrm{2}{sin}\left(\frac{\alpha}{\mathrm{2}}\right)\left\{{sin}\:\frac{\alpha}{\mathrm{2}}\:−\:{cos}\:\frac{\beta−\gamma}{\mathrm{2}}\right\} \\ $$$$=\mathrm{1}−\mathrm{2}{sin}\left(\frac{\alpha}{\mathrm{2}}\right)\left\{{sin}\left(\frac{\pi}{\mathrm{2}}−\frac{\beta+\gamma}{\mathrm{2}}\right)−{cos}\:\frac{\beta−\gamma}{\mathrm{2}}\right\} \\ $$$$=\mathrm{1}−\mathrm{2}{sin}\left(\frac{\alpha}{\mathrm{2}}\right)\left\{{cos}\:\frac{\beta+\gamma}{\mathrm{2}}\:−\:{cos}\:\frac{\beta−\gamma}{\mathrm{2}}\right\} \\ $$$$=\mathrm{1}−\mathrm{2}{sin}\left(\frac{\alpha}{\mathrm{2}}\right)\left(−\mathrm{2}{sin}\left(\frac{\beta}{\mathrm{2}}\right){sin}\left(\frac{\gamma}{\mathrm{2}}\right)\right) \\ $$$$=\mathrm{1}+\mathrm{4}{sin}\left(\frac{\alpha}{\mathrm{2}}\right){sin}\left(\frac{\beta}{\mathrm{2}}\right){sin}\left(\frac{\gamma}{\mathrm{2}}\right) \\ $$$$=\mathrm{1}+\mathrm{4}{a} \\ $$

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