若根號1-sinx/1 sinx=sinx-1/cosx,則x的取值範圍是
同題.
選項爲：
A.2kπ≤x≤2kπ π/2 B.2kπ π/2網友廻答：
- (sinx-1)/cosx>=0
分[fèn]子小於等於0，分母小於0，所以x在二三象限
（題目最好打上括[kuò]號）
網友廻答：
- √[(1-sinx)/(1 sinx)]=(sinx-1)/cosx,
將式中所有的1都換成sin²(x/2) cos²(x/2),sinx都換成2sin(x/2)cos(x/2),cosx換成cos²(x/2)-sin²(x/2),得
√{[sin²(x/2)-2sin(x/2)cos(x/2) cos²(x/2)]/[sin²(x/2) 2sin(x/2)cos(x/2) cos²(x/2)]}
= -[sin²(x/2)-2sin(x/2)cos(x/2) cos²(x/2)]/[cos²(x/2)-sin²(x/2)],
√{[sin(x/2)-cos(x/2)]²/[sin(x/2) cos(x/2)]²}=
[sin(x/2)-cos(x/2)]²/{[sin(x/2)-cos(x/2)]*[sin(x/2) cos(x/2)]},
| [sin(x/2)-cos(x/2)]/[sin(x/2) cos(x/2)] |=[sin(x/2)-cos(x/2)]/[sin(x/2) cos(x/2)],
上式實際上是|a|=a,說明a>0,所以
[sin(x/2)-cos(x/2)]/[sin(x/2) cos(x/2)]>0
分[fèn]子分母相除大於0,所以二者同號,所以二者相乘也大於0,即
[sin(x/2)-cos(x/2)]*[sin(x/2) cos(x/2)]>0,變形
[sin(x/2)]²-[cos(x/2)]²>0
-cosx>0,（上麪的等式是一個倍角公式）
cosx
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