在△ABC中,求证cos2A/a^2-cos2B/b^2=1/a^2-1/b^2
问题描述：
在△ABC中,求证cos2A/a^2-cos2B/b^2=1/a^2-1/b^2
答
根据二倍角公式
cos2A=1-2sin^2A,cos2B=1-2sin^2B
∴cos2A/a^2-cos2B/b^2
=(1-2sin^2A)/a^2-(1-2sin^2B)/b^1
=1/a^2-1/b^2-2(sin^2A/a^2-sin^B/b^2)
根据正弦定理
a/sinA=b/sinB
∴a^2/sin^2A=b^2/sin^2B
∴sin^2A/a^2-sin^B/b^2=0
∴1/a^2-1/b^2-2(sin^2A/a^2-sin^B/b^2)
=1/a^2-1/b^2
即cos2A/a^2-cos2B/b^2=1/a^2-1/b^2