化简cos(α-π/2)/sin(α+5π/2)*sin(α-2π)*cos(2π-α),因为我刚学,所以要具体点的过程哦~

问题描述:

化简cos(α-π/2)/sin(α+5π/2)*sin(α-2π)*cos(2π-α),因为我刚学,所以要具体点的过程哦~
额……答案是sinα的平方

原式=(cos(-(π/2-α))/sin(2π+π/2+α))*sin(-(2π-α))*cos(2π-α)
=(cos(π/2-α)/sin(π/2+α))*(-sin(2π-α))*cosα
=(sinα/cosα)*sinα*cosα
=sin²α.