(sin²a-sin²b)/(sina·cosa-sinb·cosb) 如何化简,

来源：学生作业帮助网编辑：六六作业网时间：2024/12/23 19:20:04

(sin²a-sin²b)/(sina·cosa-sinb·cosb)如何化简,(sin²a-sin²b)/(sina·cosa-sinb·cosb)如何化简,

(sin²a-sin²b)/(sina·cosa-sinb·cosb) 如何化简,
(sin²a-sin²b)/(sina·cosa-sinb·cosb) 如何化简,

(sin²a-sin²b)/(sina·cosa-sinb·cosb) 如何化简,
(sin²a-sin²b)/(sina·cosa-sinb·cosb)
=[(1-cos2a)/2-(1-cos2b)/2]/[(sin2a)/2-(sin2b)/2]
=(cos2b-cos2a)/(sin2a-sin2b)
【cos2x=1-2sinx^2 sin2x=sinxcosx】