设-3π

题目

设-3π

答案

1-cos(a-π)=2*[sin(a-π)/2]^2
√[1-cos(a-π)]/2=√｛2*{sin[(a-π)/2]}^2/2｝
=√{sin[(a-π)/2]}^2
=|sin[(a-π)/2]|
因为-3π所以-2π<(a-π)/2<-7π/4
在一象限,所以sin(a-π)/2>0
所以 √[1-cos(a-π)]/2=sin[(a-π)/2]
=sin(a/2-π/2)
=-cos(a/2)