化简根号下[1-cos(a-pi)]\2-3pi

化简根号下[1-cos(a-pi)]\2-3pi
化简根号下[1-cos(a-pi)]\2
-3pi

化简根号下[1-cos(a-pi)]\2-3pi
应该是求 根号下{[1-cos(a-pi)]\2}吧?
[1-cos(a-pi)]\2
=[1-cos(pi-a)]/2
=[1+cosa]/2
=[1+2(cos(a/2))^2-1]/2
=(cos(a/2))^2
而 -3pi/2