已知tan2θ=3/4(π/2＜θ＜π),求(2cos² ×θ/2+sinθ-1)/√2cos(θ+π/4)的值
问题描述：
已知tan2θ=3/4(π/2＜θ＜π),求(2cos² ×θ/2+sinθ-1)/√2cos(θ+π/4)的值
答
解
tan2θ=(2tanθ)/(1-tan^2θ)=3/4
即3(1-tan^2θ)=8tanθ
即3tan^2θ+8tanθ-3=0
(3tanθ-1)(tanθ+3)=0
∵π/2∴tanθ∴tanθ=-3
∴(2cos^2θ/2+sinθ-1)/(√2cos(θ+π/4))
=(cosθ+sinθ)/(cosθ-sinθ)
=(1+tanθ)/(1-tanθ)
=(1-3)/(1+3)
=-1/2