设f(α)=[2sin(π+α)cos(π-α)-cos(π+α)]/[(1+sin^2α+sin(π-α)-cos^2(π-α),求f(-23π/6)的值
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设f(α)=[2sin(π+α)cos(π-α)-cos(π+α)]/[(1+sin^2α+sin(π-α)-cos^2(π-α),求f(-23π/6)的值
设f(α)=[2sin(π+α)cos(π-α)-cos(π+α)]/[(1+sin^2α+sin(π-α)-cos^2(π-α),求f(-23π/6)的值
设f(α)=[2sin(π+α)cos(π-α)-cos(π+α)]/[(1+sin^2α+sin(π-α)-cos^2(π-α),求f(-23π/6)的值
f(α)=[2sin(π+α)cos(π-α)-cos(π+α)]/[(1+sin^2α+sin(π-α)-cos^2(π-α) = [2(-sinα)(-cosα)-(-cosα)]/[(1+sin^2α+sinα-cos^2α] = [2sinαcosα+cosα]/[2sin^2α+sinα] = cosα(2sinα+1)/sinα(2sinα+1) =cosα/sinα = cos(-17/6π)/sin(-17/6π) = cos(-5/6π)/sin(-5/6π) = cos(5/6π)/-sin(5/6π) = √3
设f(α)=[2sin(π+α)cos(π-α)-cos(π+α)]/[(1+sin^2α+sin(π-α)-cos^2(π-α),求f(-23π/6)的值
设f(α)=[2sin(π+α)cos(π-α)-cos(π+α)]/[(1+sin^2α+sin(π-α)-cos^2(π-α),求f(-23π/6)的值
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