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)
