Question

3π/θ＜2π,化简√1+sinθ-√1-sinθ. 详细步骤

Answer 1

题目应该是 3π/2<θ＜2π,化简√(1+sinθ)-√(1-sinθ) 吧！
已知：3π/2<θ＜2π，那么：3π/4<θ/2＜π
所以：sin(θ/2)>(√2)/2，cos(θ/2)<-(√2)/2
那么：sin(θ/2)+cos(θ/2)<0，sin(θ/2) -cos(θ/2)>0
所以：
√(1+sinθ)-√(1-sinθ)
=√[sin²(θ/2)+2sin(θ/2)cos(θ/2)+cos²(θ/2)] - √[sin²(θ/2)+2sin(θ/2)cos(θ/2)+cos²(θ/2)]
=√[sin(θ/2)+cos(θ/2)]² - √[sin(θ/2)-cos(θ/2)]²
=-[sin(θ/2)+cos(θ/2)]-[sin(θ/2)-cos(θ/2)]
=-2cos(θ/2)

追问

是的