由正弦定理我们可以知道
a/sinA=b/sinB=c/sinC=2R
代入余弦定理的左边a^2+b^2-2abcosC=4R^2(sinA)^2+4R^2(sinB)^2-8R^2sinAsinBcosC
所以(a^2+b^2-2abcosC)/4R^2
=(sinA)^2+(sinB)^2-2sinAsinBcosC
=(1-cos2A)/2+(1-cos2B)/2-2sinAsinBcosC
=1-(cos2A+cos2B)/2+(cos(A+B)/2-cos(A-B)/2)cosC
=1-cos(A+B)cos(A-B)+(cos(A+B)/2-cos(A-B)/2)cosC
=1+cosCcos(A-B)+(sinC/2-cos(A-B)/2)cosC
哦,太繁了,接下来的你自己算吧.只要证明到余弦定理的右边c^2就可以了
a/sinA=b/sinB=c/sinC=2R
代入余弦定理的左边a^2+b^2-2abcosC=4R^2(sinA)^2+4R^2(sinB)^2-8R^2sinAsinBcosC
所以(a^2+b^2-2abcosC)/4R^2
=(sinA)^2+(sinB)^2-2sinAsinBcosC
=(1-cos2A)/2+(1-cos2B)/2-2sinAsinBcosC
=1-(cos2A+cos2B)/2+(cos(A+B)/2-cos(A-B)/2)cosC
=1-cos(A+B)cos(A-B)+(cos(A+B)/2-cos(A-B)/2)cosC
=1+cosCcos(A-B)+(sinC/2-cos(A-B)/2)cosC
哦,太繁了,接下来的你自己算吧.只要证明到余弦定理的右边c^2就可以了
