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
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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
额……答案是sinα的平方
![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](/uploads/image/z/7825207-31-7.jpg?t=%E5%8C%96%E7%AE%80cos%28%CE%B1-%CF%80%2F2%29%2Fsin%28%CE%B1%2B5%CF%80%2F2%29%2Asin%28%CE%B1-2%CF%80%29%2Acos%282%CF%80-%CE%B1%29%2C%E5%9B%A0%E4%B8%BA%E6%88%91%E5%88%9A%E5%AD%A6%2C%E6%89%80%E4%BB%A5%E8%A6%81%E5%85%B7%E4%BD%93%E7%82%B9%E7%9A%84%E8%BF%87%E7%A8%8B)
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.
=(cos(π/2-α)/sin(π/2+α))*(-sin(2π-α))*cosα
=(sinα/cosα)*sinα*cosα
=sin²α.
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
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