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若sinα是方程6x=1-√x的根,试求tan(π-α)tan(2π-α)cos(5π-α)/cos(3π/2+α)的值
人气:276 ℃ 时间:2019-11-06 16:14:19
解答
化简:tan(π-α)tan(2π-α)cos(5π-α)/cos(3π/2+α) = -tanα
解根号方程:6x+√x-1=0
x=1/3(舍去负根)
所以:sinα = 1/3
因α没有限制,所以α可能锐角可能钝角
若α锐角
tanα = √2/4
tan(π-α)tan(2π-α)cos(5π-α)/cos(3π/2+α) = -√2/4
若α钝角
tanα = -√2/4
tan(π-α)tan(2π-α)cos(5π-α)/cos(3π/2+α) = √2/4
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