在三角形abc中 sin^A+sin^B+sin^ - 知识问答

在三角形abc中 sin^A+sin^B+sin^C

2个回答

(sinA)^2+(sinB)^2+(sinC)^2
=1-(cosA)^2+1-(cosB)^2+1-(cosC)^2
=3-[(cos2A+1)/2+(cosB+1)/2]-cosC^2
=2-[cos2A+cos2B]/2-(cosC)^2
=2-cos(A+B)cos(A-B)-(cosC)^2
=2+cosCcos(A-B)-(cosC)^2
=2+cosC[cos(A-B)-cosC]
=2+cosC[cos(A-B)+cos(A+B)]
=2+cosCcosAcosB＜2,
∴cosCcosAcosB＜0,
必有一角为钝角,所以为钝角三角形.

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