△ABC中,求证;cosA+cosB+cosC=1+4sinA/2 * sinB/2 * sinC/2

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△ABC中,求证;cosA+cosB+cosC=1+4sinA/2 * sinB/2 * sinC/2
△ABC中,求证;cosA+cosB+cosC=1+4sinA/2 * sinB/2 * sinC/2

△ABC中,求证;cosA+cosB+cosC=1+4sinA/2 * sinB/2 * sinC/2
很简单
因为A+B+C=派 所以C=派-A-B,即C/2=派/2-(A+B)/2
cosA+cosB+cosC=cosA+cosB+cos(派-(A+B))
=cosA+cosB-cos(A+B)
=2cos(A+B)/2*cos(A-B)/2+1-2cos^2(A+B)/2
=1+2cos(A+B)/2*(cos(A-B)/2-cos(A+B)/2)
=1+4cos(A+B)/2*(sinA/2+sinB/2)
=1+4sinA/2*sinB/2*sinC/2