化简√(1+sinθ）-√（1-sinθ）

Question

如题所述

Answer 1

√(1+sinθ)-√(1-sinθ)
=√[1+2sin(θ/2)cos(θ/2)]-√[1-2sin(θ/2)cos(θ/2)]
=√[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)|

有没有给出θ的范围？有的话上面可以继续化简下去。追问

270·≤θ≤360·刚才忘了，Sorry.

追答

270°≤θ≤360°
135°≤θ/2≤180°
sin(θ/2)+cos(θ/2)≤0
sin(θ/2)-cos(θ/2)≥0

所以|sin(θ/2)+cos(θ/2)|+|sin(θ/2)-cos(θ/2)|
=-sin(θ/2)-cos(θ/2)+sin(θ/2)-cos(θ/2)
=-2cos(θ/2)

Answer 2

270·≤θ≤360· sinθ<0，原式为负值,cosθ>0
√(1+sinθ）-√（1-sinθ）
=-√ [ √(1+sinθ）-√（1-sinθ） ]^2
=-√ [ 1+sinθ）+（1-sinθ） - 2 * √(1+sinθ）√（1-sinθ） ]
=-√ [ 2 - 2 * √(1-sinθ^2） ]
=-√ [ 2 - 2 * √(cosθ^2） ]
=-√2* √( 1 - cosθ )
= - √2* √ (2 sin(θ/2)*sin(θ/2) )
= - 2 sin(θ/2)

Answer 3

√(1+sinθ）-√（1-sinθ）
=√[sin² θ/2+cos² θ/2+2 sin θ/2*cos θ/2]-√[sin² θ/2+cos² θ/2-2 sin θ/2*cos 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 |
又∵270°≤θ≤360°
135°≤θ/2≤180°
sin θ/2＞0＞cos θ/2
且sin θ/2+cos θ/2＜0
sin θ/2+cos θ/2＞0

上式= -(sin θ/2+cos θ/2) -(sin θ/2-cos θ/2)
= -2sin θ/2