Question

若tan2θ=-2√2，π/2<2θ<π,求[2cos^2(θ/2)-sinθ-1]/[√2sin(θ+π/4)]的值。

求详细解答过程！

Answer 1

tan2θ= 2tanθ/(1-tan^2(θ)) = -2根号2

π/2<2θ<π
π/4<θ<π/2
tanθ= 根号2

2cos^2(θ/2)-sinθ-1
=cosθ-sinθ
=根号2( cos45°cosθ-sin45°sinθ)
=根号2cos(θ+45°)

[2cos^2(θ/2)-sinθ-1]/[√2sin(θ+π/4)]
= 根号2cos(θ+45°)/ 根号2sin(θ+45)
=tan(45+θ)
= (1+tanθ)/(1-tanθ）
= -(3+2根号2)