Question

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

Answer 1

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