作业帮求证：sin2x/[sinx+(cosx-1)][sinx-(cosx-1)]=(1+cosx)/sinx
来源：学生作业帮助网 编辑：作业帮 时间：2024/11/25 05:19:28
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求证：sin2x/[sinx+(cosx-1)][sinx-(cosx-1)]=(1+cosx)/sinx
求证：sin2x/[sinx+(cosx-1)][sinx-(cosx-1)]=(1+cosx)/sinx
求证：sin2x/[sinx+(cosx-1)][sinx-(cosx-1)]=(1+cosx)/sinx
原式=2sinxcosx/sin^2x-(cosx-1)^2=2sinxcosx/sin^2x-cos^2x+2cosx-1=2sinxcosx/-2cos^2x+2cosx=sinx/1-cosx=1+cosx/sinx
[sinx+(cosx-1)][sinx-(cosx-1)]=(sinx)^2 - (cosx-1)^2
sin2x = 2sinx cosx
代入后同乘以两侧的分母不就出来了？