sin2A-sin2B+sin2C=4cosAsinBcosC,A+B+C=180°

guawang2022-10-04 11:39:541条回答

已提交，审核后显示！提交回复

xxfly 共回答了19个问题 | 采纳率78.9%: sin2A-sin2B+sin2C
=2sinAcosA-sin2B+2sinCcosC 1
=2sinAcosA-sin(360°-2A-2C)+2sinCcosC
=2sinAcosA-sin2(180°-A-C)+2sinCcosC
=sin2A+sin(2A+2C)+sin2C
=sin2A+sin2Acos2C+sin2Ccos2A+sin2C
=(sin2A+sin2C)(cos2C+1)
4cosAsinBcosC
=4cosAsin(180°-A-C)cosC
=4cosAsin(A+C)cosC
=2sinAcosA*2cos^2C+2sinCcosC*2cos^2C 2
=(sin2A+sin2C)(cos2C+1)
所以左边等于右边; 1年前