原式=4+2根號(hào)2sinx+2根號(hào)2cosx+2sinxcosx=4+2根號(hào)2*（sinx+cosx）+2sinxcosx
設(shè)t=sinx+cosx,則有-根號(hào)2<=t<=根號(hào)2
t^2=1+2sinxcosx
故原式=4+2根號(hào)2t+t^2-1=t^2+2根號(hào)2 t+3=(t+ 根號(hào)2）^2+1
對(duì)稱軸是t=-根號(hào)2,則在[－根號(hào)2,根號(hào)2]上是單調(diào)增函數(shù).
故有最大值是在t=根號(hào)2時(shí)取得,即有y=9
有最小值是在t=-根號(hào)2時(shí)取得,即有y=1
即值域是[1,9]