求证:sin2x/[sinx+(cosx-1)][sinx-(cosx-1)]=(1+cosx)/sinx
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![求证:sin2x/[sinx+(cosx-1)][sinx-(cosx-1)]=(1+cosx)/sinx](/uploads/image/z/1996146-18-6.jpg?t=%E6%B1%82%E8%AF%81%EF%BC%9Asin2x%2F%5Bsinx%2B%28cosx-1%29%5D%5Bsinx-%28cosx-1%29%5D%3D%281%2Bcosx%29%2Fsinx)
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?