已知α是第四象限角,化简sin(kπ+α)√[1+cos(kπ+α)]/[1-cos(kπ+α)],k属于z

题目

已知α是第四象限角,化简sin(kπ+α)√[1+cos(kπ+α)]/[1-cos(kπ+α)],k属于z

答案

根据半角公式cot²(A/2)=(1+cosA)/(1-cosA)
即[1+cos(kπ+α)]/[(1-cos(kπ+α)]=cot²[(kπ+α)/2]
sin(kπ+α)√[1+cos(kπ+α)]/[1-cos(kπ+α)]
=sin(kπ+α)√{cos²[(kπ+α)/2]/sin²[(kπ+α)/2]}
=2sin[(kπ+a)/2]cos[(kπ+a)/2]√{cos²[(kπ+α)/2]/sin²[(kπ+α)/2]}
=2cos[(kπ+a)/2]√cos²[(kπ+α)/2]
=±2cos²[(kπ+a)/2]
=±2sin²(a/2)