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)#