在三角形ABC中,求证:SinA+SinB+SinC= 4CosA/2*CosB/2*CosC/2
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![在三角形ABC中,求证:SinA+SinB+SinC= 4CosA/2*CosB/2*CosC/2](/uploads/image/z/12618246-30-6.jpg?t=%E5%9C%A8%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E4%B8%AD%2C%E6%B1%82%E8%AF%81%EF%BC%9ASinA%2BSinB%2BSinC%3D+4CosA%2F2%2ACosB%2F2%2ACosC%2F2)
在三角形ABC中,求证:SinA+SinB+SinC= 4CosA/2*CosB/2*CosC/2
在三角形ABC中,求证:SinA+SinB+SinC= 4CosA/2*CosB/2*CosC/2
在三角形ABC中,求证:SinA+SinB+SinC= 4CosA/2*CosB/2*CosC/2
证明:
sinA+sinB+sinC
=2sin((A+B)/2)cos((A-B)/2)+2sin(C/2)cos(C/2)
=2sin((π-C)/2)cos((A-B)/2)+2sin(π-(A+B)/2)cos(C/2)
=2cos(C/2)cos((A-B)/2)+2cos((A+B)/2)cos(C/2)
=2cos(C/2)(cos((A-B)/2)+cos((A+B)/2))
=2cos(C/2)2cos(A/2)cos(B/2)
=4cos(A/2)cos(B/2)cos(C/2)