If $0<(alpha),(beta),(gamma)<(pi)/2$ prove that $sin(alpha)+sin(beta)+sin(gamma)>sin(alpha+beta+gamma)$ ?

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Answer 1 · 2018-08-05T04:26:36

$sin(alpha)+sin(beta)+sin(gamma)-sin(alpha+beta+gamma)$

$=2sin((alpha+beta)/2)cos((alpha-beta)/2)-(sin(alpha+beta+gamma)-sin(gamma))$

$=2sin((alpha+beta)/2)cos((alpha-beta)/2)-(2cos(alpha/2+beta/2+gamma)sin((alpha+beta)/2))$

$=2sin((alpha+beta)/2)[cos((alpha-beta)/2)-cos(alpha/2+beta/2+gamma)]$

$=2sin((alpha+beta)/2)[2sin((alpha+gamma)/2)sin((beta+gamma)/2)]>0$

As it is given

$0 < alpha,beta and gamma< pi/2$ we have,

$((alpha+beta)/2),((alpha+gamma)/2)and((beta+gamma)/2)$ arr acute angles and sine of all must be $> 0$

Hence

$sin(alpha)+sin(beta)+sin(gamma) >sin(alpha+beta+gamma)$

If 0<(alpha),(beta),(gamma)<(pi)/20<(alpha),(beta),(gamma)<(pi)/2 prove that sin(alpha)+sin(beta)+sin(gamma)>sin(alpha+beta+gamma)sin(alpha)+sin(beta)+sin(gamma)>sin(alpha+beta+gamma)?