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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

$j e n s i o n i n e q u a l i t y$
	Answered by Kunal12588 last updated on 21/May/19
	$cos(α)+cos(β)+cos(γ) =cos(α)+2cos(((β+γ)/2))cos(((β−γ)/2)) =cos(α)+2cos((π/2)−(α/2))cos(((β−γ)/2)) =1−2sin^2 ((α/2))+2sin((α/2))cos(((β−γ)/2)) =1−2sin((α/2)){sin (α/2) − cos ((β−γ)/2)} =1−2sin((α/2)){sin((π/2)−((β+γ)/2))−cos ((β−γ)/2)} =1−2sin((α/2)){cos ((β+γ)/2) − cos ((β−γ)/2)} =1−2sin((α/2))(−2sin((β/2))sin((γ/2))) =1+4sin((α/2))sin((β/2))sin((γ/2)) =1+4a$
	$c o s (α) + c o s (β) + c o s (γ)$ $= c o s (α) + 2 c o s (\frac{β + γ}{2}) c o s (\frac{β - γ}{2})$ $= c o s (α) + 2 c o s (\frac{π}{2} - \frac{α}{2}) c o s (\frac{β - γ}{2})$ $= 1 - 2 {s i n}^{2} (\frac{α}{2}) + 2 s i n (\frac{α}{2}) c o s (\frac{β - γ}{2})$ $= 1 - 2 s i n (\frac{α}{2}) {s i n \frac{α}{2} - c o s \frac{β - γ}{2}}$ $= 1 - 2 s i n (\frac{α}{2}) {s i n (\frac{π}{2} - \frac{β + γ}{2}) - c o s \frac{β - γ}{2}}$ $= 1 - 2 s i n (\frac{α}{2}) {c o s \frac{β + γ}{2} - c o s \frac{β - γ}{2}}$ $= 1 - 2 s i n (\frac{α}{2}) (- 2 s i n (\frac{β}{2}) s i n (\frac{γ}{2}))$ $= 1 + 4 s i n (\frac{α}{2}) s i n (\frac{β}{2}) s i n (\frac{γ}{2})$ $= 1 + 4 a$
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