If $0 < (α), (β), (γ) < \frac{π}{2}$ $0<(alpha),(beta),(gamma)<(pi)/2$ prove that $sin (α) + sin (β) + sin (γ) > sin (α + β + γ)$ $sin(alpha)+sin(beta)+sin(gamma)>sin(alpha+beta+gamma)$ ?

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If $0 < (α), (β), (γ) < \frac{π}{2}$ $0<(alpha),(beta),(gamma)<(pi)/2$ prove that $sin (α) + sin (β) + sin (γ) > sin (α + β + γ)$ $sin(alpha)+sin(beta)+sin(gamma)>sin(alpha+beta+gamma)$ ?

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Answer 1 · 2018-08-05T04:26:36

$sin (α) + sin (β) + sin (γ) - sin (α + β + γ)$ $sin(alpha)+sin(beta)+sin(gamma)-sin(alpha+beta+gamma)$

$= 2 sin (\frac{α + β}{2}) cos (\frac{α - β}{2}) - (sin (α + β + γ) - sin (γ))$ $=2sin((alpha+beta)/2)cos((alpha-beta)/2)-(sin(alpha+beta+gamma)-sin(gamma))$

$= 2 sin (\frac{α + β}{2}) cos (\frac{α - β}{2}) - (2 cos (\frac{α}{2} + \frac{β}{2} + γ) sin (\frac{α + β}{2}))$ $=2sin((alpha+beta)/2)cos((alpha-beta)/2)-(2cos(alpha/2+beta/2+gamma)sin((alpha+beta)/2))$

$= 2 sin (\frac{α + β}{2}) [cos (\frac{α - β}{2}) - cos (\frac{α}{2} + \frac{β}{2} + γ)]$ $=2sin((alpha+beta)/2)[cos((alpha-beta)/2)-cos(alpha/2+beta/2+gamma)]$

$= 2 sin (\frac{α + β}{2}) [2 sin (\frac{α + γ}{2}) sin (\frac{β + γ}{2})] > 0$ $=2sin((alpha+beta)/2)[2sin((alpha+gamma)/2)sin((beta+gamma)/2)]>0$

As it is given

$0 < α, β and γ < \frac{π}{2}$ $0 < alpha,beta and gamma< pi/2$ we have,

$(\frac{α + β}{2}), (\frac{α + γ}{2}) and (\frac{β + γ}{2})$ $((alpha+beta)/2),((alpha+gamma)/2)and((beta+gamma)/2)$ arr acute angles and sine of all must be $> 0$ $> 0$

Hence

$sin (α) + sin (β) + sin (γ) > sin (α + β + γ)$ $sin(alpha)+sin(beta)+sin(gamma) >sin(alpha+beta+gamma)$

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If $0 < (α), (β), (γ) < \frac{π}{2}$ $0<(alpha),(beta),(gamma)<(pi)/2$ prove that $sin (α) + sin (β) + sin (γ) > sin (α + β + γ)$ $sin(alpha)+sin(beta)+sin(gamma)>sin(alpha+beta+gamma)$ ?

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If $0 < (α), (β), (γ) < \frac{π}{2}$ $0<(alpha),(beta),(gamma)<(pi)/2$ prove that $sin (α) + sin (β) + sin (γ) > sin (α + β + γ)$ $sin(alpha)+sin(beta)+sin(gamma)>sin(alpha+beta+gamma)$ ?