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已知sinα是方程5x^2-7x-6=0的根,且α为第三象限角 求:
sin(α+3/2*π)*sin(3/2*π-α)*tan^2(2π-α)*tan(π-α)÷cos(π-α)*cos(π/2+α)
人气:102 ℃ 时间:2019-11-02 15:33:40
解答
5x^2-7x-6=0(5x+3)(x-2)=0x1=-3/5,x2=2sinα是方程5x^2-7x-6=0的根sinα≠2∴sinα=-3/5sin(α+3/2*π)*sin(3/2*π-α)*tan^2(2π-α)*tan(π-α)÷【cos(π-α)*cos(π/2+α)】= -cosα * (-cos...
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