Prove that $sina+ sinb+ sinc- sin(a+b+c)= 4sin((a+b)/2) sin((b+c)/2) sin((c+a)/2)$ ?

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Answer 1 · 2018-05-25T05:11:23

$LHS=sina+ sinb+ sinc- sin(a+b+c)$

$=sina+ sinb+ sinc- sin((a+b+c)$

$=2sin((a+b)/2)cos((a-b)/2)- 2cos((a+b+2c)/2)sin((a+b)/2)]$

$=2sin((a+b)/2)[cos((a-b)/2)- cos((a+b+2c)/2)]$

$=2sin((a+b)/2)xx2sin((a+b+2c)/4-(a-b)/4)sin((a+b+2c)/4+(a-b)/4)$

$= 4sin((a+b)/2) sin((b+c)/2) sin((c+a)/2)=RHS$

Prove that sina+ sinb+ sinc- sin(a+b+c)= 4sin((a+b)/2) sin((b+c)/2) sin((c+a)/2)sina+ sinb+ sinc- sin(a+b+c)= 4sin((a+b)/2) sin((b+c)/2) sin((c+a)/2) ?