已知cos（π/6+a）cos（π/3-a）=1/4,a属于（π/3,π/2）.求.sina和tana-1/tana分别为多少.

Question

已 知cos（π/6+a）cos（π/3-a）=1/4,a属于（π/3,π/2）.求.sina和tana-1/tana分别为多少.

Answer 1

2cos(π/6+a)cos(π/3-a)=-1/2,
积化和差得cos(π/2)+cos(2a-π/6)=-1/2,
∴cos(2a-π/6)=-1/2,
a∈（π/3,π/2）,u=2a-π/6的值域是（π/2,5π/6）,
∴2a-π/6=2π/3,a=5π/12.
∴sin2a=1/2,
tana-1/tana
=sina/cosa-cosa/sina
=[(sina)^2-(cosa)^2]/[sinacosa]
=-2cos2a/sin2a
=-2/tan2a
=-2/(-1/√3)
=2√3.