θ∈[4/5π,2/3π],所以sinθ＞0＞cosθ
根号下的1–sin2θ–根号下1+sin2θ
=√(sinθ-cosθ)²-√(sinθ+cosθ)²
=sinθ-cosθ+sinθ+cosθ
=2sinθ