SinA+SinB+SinC

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SinA+SinB+SinCSinA+SinB+SinCSinA+SinB+SinC在三角形ABC中证明(sinA)^2+(sinB)^2+(sinC)^2=2(1+cosAcosBcosC)由倍角公

SinA+SinB+SinC
SinA+SinB+SinC

SinA+SinB+SinC
在三角形ABC中证明(sinA)^2+(sinB)^2+(sinC)^2=2(1+cosAcosBcosC)
由倍角公式:
(sinA)^2+(sinB)^2+(sinC)^2
=(1-cos2A)/2+(1-cos2B)/2+(1-cos2C)/2
=3/2-1/2(cos2A+cos2B+cos2C) (对cos2A+cos2B用和差化积公式)
=3/2-1/2(2cos(A+B)cos(A-B)+2(cosC)^2-1)
=2-(cos(A+B)cos(A-B)+(cosC)^2)
=2-(-cos(A-B)cosC+(cosC)^2)
=2-cosC(cosC-cos(A-B)) (再用和差化积公式)
=2+2cosC[sin (C-A+B)/2*sin (C+A-B)/2]
=2+2cosC[sin (180-2A)/2*sin (180-2B)/2]
=2+2cosC[sin(90-A)*sin(90-B)]
=2+2cosCcosAcosB