(1)證明:△ABC中,利用餘弦定理可得cosC=a2+b2?c2
2ab
,
即a2+b2-c2=2ab?cosC.
再利用正弦定理可得sin2A+sin2B-sin2C=2sinAsinBcosC,
∴要證的等式成立.
(2)△ABC中,∵等式右邊=4sinA
2
sinB
cosC
=4sinA
cosπ?A?B
sinA+B
(sinA
cosB
+cosA
)
=2sin2A
sinB+2sinAsin2B
=(1-cosA)sinB+sinA(1-cosB)
=sinB+sinA-(sinBcosA+cosBsinA)=sinA+sinB-sin(A+B)
=sinA+sinB-sinC=左邊,
(1)證明:△ABC中,利用餘弦定理可得cosC=a2+b2?c2
2ab
,
即a2+b2-c2=2ab?cosC.
再利用正弦定理可得sin2A+sin2B-sin2C=2sinAsinBcosC,
∴要證的等式成立.
(2)△ABC中,∵等式右邊=4sinA
2
sinB
2
cosC
2
=4sinA
2
sinB
2
cosπ?A?B
2
=4sinA
2
sinB
2
sinA+B
2
=4sinA
2
sinB
2
(sinA
2
cosB
2
+cosA
2
sinB
2
)
=2sin2A
2
sinB+2sinAsin2B
2
=(1-cosA)sinB+sinA(1-cosB)
=sinB+sinA-(sinBcosA+cosBsinA)=sinA+sinB-sin(A+B)
=sinA+sinB-sinC=左邊,
∴要證的等式成立.