在△ABC中,求证: (1)a2+b2c2=sin2A+sin2Bsin2C (2)a2+b2+c2=2(bccosA+cacosB+abcosC)

问题描述:

在△ABC中,求证:
(1)

a2+b2
c2
=
sin2A+sin2B
sin2C

(2)a2+b2+c2=2(bccosA+cacosB+abcosC)

证明:(1)由正弦定理可得a=2RsinA,b=2RsinB,c=2RsinC,

a2+b2
c2
=
4R2sin2A+4R2sin2B
4R2sin2C
=
sin2A+sin2B
sin2C

(2)由余弦定理可得2(bccosA+cacosB+abcosC)
=2bc•
b2+c2-a2
2bc
+2ac•
a2+c2-b2
2ac
+2ab•
a2+b2-c2
2ab
=a2+b2+c2
∴a2+b2+c2=2(bccosA+cacosB+abcosC)