∵cosx=cos²(x/2)-sin²(x/2)
=[cos(x/2)+sin(x/2)][cos(x/2)-sin(x/2)]
1-sinx=sin²(x/2)-2sin(x/2)cos(x/2)+cos²(x/2)
=[cos(x/2)-sin(x/2)]²
∴左边
=[cos(x/2)+sin(x/2)]/[cos(x/2)-sin(x/2)]
=[1+tan(x/2)]/[1-tan(x/2)]
=[1+tanπ/4*tan(x/2)]/[tanπ/4-tan(x/2)]
=1/tan(π/4-x/2)
=cot(π/4-x/2)
追问
这步是怎么得到?
根据两角差正切公式
tan(a-b)=(tana-tanb)/(1-tanatanb)
得:
(tanπ/4-x/2)=(tanπ/4-x/2)/(1-tanπ/4tanx/2)
两边取倒数就给以了
您的两角差正切公式是不是写错了?应该是:
对分母是加号,不好意思,手头在判作业,分神了
(tanπ/4-x/2)=(tanπ/4-x/2)/(1+tanπ/4tanx/2)