已知sinα+cosα=1/5(0<α<π) ⑴求(2sinαcosα+2sin²α)/(cosα-sinαtanα)的值;
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![已知sinα+cosα=1/5(0<α<π) ⑴求(2sinαcosα+2sin²α)/(cosα-sinαtanα)的值;](/uploads/image/z/2758178-2-8.jpg?t=%E5%B7%B2%E7%9F%A5sin%CE%B1%2Bcos%CE%B1%3D1%2F5%280%EF%BC%9C%CE%B1%EF%BC%9C%CF%80%EF%BC%89+%E2%91%B4%E6%B1%82%EF%BC%882sin%CE%B1cos%CE%B1%2B2sin%26%23178%3B%CE%B1%EF%BC%89%2F%EF%BC%88cos%CE%B1-sin%CE%B1tan%CE%B1%29%E7%9A%84%E5%80%BC%EF%BC%9B)
已知sinα+cosα=1/5(0<α<π) ⑴求(2sinαcosα+2sin²α)/(cosα-sinαtanα)的值;
已知sinα+cosα=1/5(0<α<π) ⑴求(2sinαcosα+2sin²α)/(cosα-sinαtanα)的值;
已知sinα+cosα=1/5(0<α<π) ⑴求(2sinαcosα+2sin²α)/(cosα-sinαtanα)的值;
cosα-sinα=-√5/5
所以平方得:1-sin2a=1/5,sin2a=4/5
因为0