已知θ∈（0,2分之π）,化简根号下2+sinθ—根号下1—sinθ

网友回答

是化简根号下1+sinθ—根号下1—sinθ吧?
因为θ∈（0,2分之π）,
所以θ/2∈（0,π/4）,
∴0根号下1+sinθ—根号下1—sinθ
=√(1+sinθ)-√(1—sinθ)
=√(sin(θ/2)+cos(θ/2))² -√(sin(θ/2)-cos(θ/2))²
= sin(θ/2)+cos(θ/2)-[ cos(θ/2) - sin(θ/2)]
=2 sin(θ/2).