紧急!紧急!紧急!三角函数!为什么(sin2t-cos(2t+∏/6))^2=3sin^2(2t-∏/6)

Question

紧急!紧急!紧急!三角函数!
为什么(sin2t-cos(2t+∏/6))^2=3sin^2(2t-∏/6)

Answer 1

左边
=[sin2t - squr3/2·cos2t+sin2t/2]^2
=[3/2·sin2t-squr3/2·cos2t]^2
=[squr3·(squr3/2·sin2t-cos2t/2)]^2
=3·[squr3/2·sin2t-cos2t/2]^2
=3·[sin(2t-∏/6)]^2