【急】數學三角函數題

√(1-sinθ)+√2/2*√(1-cosθ)
=√(1-2sin(θ/2)cos(θ/2))+√2/2*√(1-(1-2sin^2(θ/2))
=|sin(θ/2)-cos(θ/2)|+|sin(θ/2)|=4/5
1).cos(θ/2)-sin(θ/2)>0, sin(θ/2)>0
-->cos(θ/2)=4/5, cos(3θ/2)=4cos^3(θ/2)-3cos(θ/2)=-44/125
2).cos(θ/2)-sin(θ/2)<0, sin(θ/2)<0
cos(θ/2)=-4/5, cos(3θ/2)=4(-4/5)^3-3(-4/5)=44/125
3).cos(θ/2)-sin(θ/2)>0, sin(θ/2)<0
或是cos(θ/2)-sin(θ/2)<0, sin(θ/2)>0
此時cos(θ/2)跟sin(θ/2)無解

~根號一般我是用貼的

2010-06-30 22:51:21 補充：
原本我想可能題目有θ的範圍忘了列出
所以需要分別討論
如果意見裡天助大的說明
3).cos(θ/2)-sin(θ/2)>0, sin(θ/2)<0
或是cos(θ/2)-sin(θ/2)<0, sin(θ/2)>0

cos(θ/2)-sin(θ/2)>0, sin(θ/2)<0
cos(θ/2)=t, sin(θ/2)=-√(1-t^2)
-->t+√(1-t^2)+√(1-t^2)=4/5
解出t代回去~太麻煩 懶得做了

3).cos(θ/2)-sin(θ/2)>0, sin(θ/2)<0
或是cos(θ/2)-sin(θ/2)<0, sin(θ/2)>0
此時cos(θ/2)跟sin(θ/2) "都有解"