a^2·sin2B+b^2·sin2A
=4R^2((sinA)^2sin2B+(sinB)^2sin2A)
=8R^2sinAsinB(sinAcosB+cosBsinA)
=8R^2sinAsinBsin(A+B)
=8R^2sinAsinBsinC
2absinC
=8R^2sinAsinBsinC
所以a^2·sin2B+b^2·sin2A=2absinC能跟我再详细地解释下每步的意思,我刚学这个,不是很懂。谢谢了。再细点:a^2·sin2B+b^2·sin2A=(2RsinA)^2*sin2B+(2RsinB)^2*sin2A【用正弦定理】=4R^2((sinA)^2sin2B+(sinB)^2sin2A)=4R^2((sinA)^2*sinBcosB+(sinB)^2*2sinAcosA)【用2倍角的正弦公式】=8R^2sinAsinB(sinAcosB+cosBsinA)=8R^2sinAsinBsin(A+B)【用两角和的正弦公式】=8R^2sinAsinBsinC2absinC=8R^2sinAsinBsinC所以a^2·sin2B+b^2·sin2A=2absinC