Question

已知sinα是方程5x^2-7x-6=0的根

已知sinα是方程5x^2-7x-6=0的根,求[cos(2π-α)tan^2(2π-α)]/[sin(π-α)sin(2π-α)tan(π-α)的值

Answer 1

解:方程5x^2-7x-6=0可因式分解为(x-2)(5x+3)=0
所以两个根分别为X1=2 X2=-3/5 因为 -1=<sinα<=1 所以sinα=-3/5
[cos(2π-α)tan^2(2π-α)]/[sin(π-α)sin(2π-α)tan(π-α)]=[cos(α)tan^2(α)]/[sin(α)sin(α)tan(α)
=1/sinα
=-5/3

追问

不好意思，我把题目打错了，是这样的：[cos(2π-α)cos(π+α)tan^2(2π-α)]/[sin(π-α)sin(2π-α)tan(π-α)]

追答

[cos(2π-α)cos(π+α)tan^2(2π-α)]/[sin(π-α)sin(2π-α)tan(π-α)]=[-cos(α)cos(α)tan^2(α)]/[sin(α)sin(α)tan(α)]=-1/tan(α)=-4/3或4/3