来自陈东奎的问题
已知2sin(π-α)-cos(π+α)=1(0<α<π),求cos(2π-α)+sin(π+α)的值
已知2sin(π-α)-cos(π+α)=1(0<α<π),求cos(2π-α)+sin(π+α)的值
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2020-12-28 05:38
已知2sin(π-α)-cos(π+α)=1(0<α<π),求cos(2π-α)+sin(π+α)的值
已知2sin(π-α)-cos(π+α)=1(0<α<π),求cos(2π-α)+sin(π+α)的值
原式等价于2sinα+cosα=1
移项,两边平方
4(1-cos²α)=4sin²α=(1-cosα)²
解得cosα=-3/5(或cosα=1,与0<α<π矛盾,舍去)
于是sinα=4/5
cos(2π-α)+sin(π+α)=cosα-sinα=-7/5