√[(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
分子分母相除大于0,所以二者同號,所以二者相乘也大于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,（上面的等式是一個倍角公式）
cosx