求証:sin2x/[sinx (cosx-1)][sinx-(cosx-1)]=(1 cosx)/sinx 網友廻答:
- [sinx (cosx-1)][sinx-(cosx-1)]=(sinx)^2 - (cosx-1)^2
sin2x = 2sinx cosx
代入後同乘以兩側的分母不就出來了?
網友廻答:
- 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
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