Question

化简：[(1+sinθ+cosθ)(sinθ/2-cosθ/2)]/(√2+2cosθ) (0<θ<π)=

Answer 1

原式=[(1+sinθ)+cosθ](sin θ/2 - cos θ/2) / √2(1+cosθ)
=[(sin θ/2 + cos θ/2)² + (cos θ/2 + sin θ/2)(cos θ/2 - sin θ/2)](sin θ/2 - cos θ/2) / √2×2cos² θ/2
(∵0<θ<π ，∴0<θ/2<π/2)
=[(cos θ/2 + sin θ/2)(2cos θ/2)](sin θ/2 - cos θ/2) / (2cos θ/2)
=(cos θ/2 + sin θ/2)(sin θ/2 - cos θ/2)
=-(cos² θ/2 - sin² θ/2)
=-cosθ

Answer 2

原式=[（2sinθ/2cosθ/2+2cosθ/2^2)(sinθ/2-cosθ/2)]/(2cosθ/2)=2cosθ/2*(cosθ/2+sinθ/2)(sinθ/2-cosθ/2)/2cosθ/2=-cosθ