Question

已知sinα是方程5x²-7x-6=o的根,

求[sin(α+ 3π/2 )sin(3π/2 -α)tan²(2π-α)tan(π-α)] / [cos(π/2 -α)cos(π/2 +α)]的值

Answer 1

解：sin(α+ 3π/2 )=-cosα
sin(3π/2 -α)=-cosα
tan(2π-α)=-tanα
tan(π-α)=-tanα
cos(π/2-α)=sinα
cos(π/2+α)=-sinα
原式=
[(cosα)^2*(-tanα)^3]/[-(sinα)^2]
=[(cosα)^2*(tanα)^3]/[(sinα)^2]
=tanα
5x²-7x-6=0
(5x+3)(x-2)=0
∵sinα∈[-1,1]
∴sinα=-3/5
cosα=±4/5
tanα=±3/4
答：原式=±3/4