设sinx+cosx=t属于[-√2,√2] => t^2=1+2sinxcosx =〉 sinxcosx=(t^2-1)/2
f(x)=(t^2-1)/2(1+t)=(t-1)/2属于[-(√2+1)/2,(√2-1)/2]
另外,分母不为零,所以1+sinx+cosxb不=0 ,既t≠-1
综上,值域属于[-(√2+1)/2,-1)∪(-1,(√2-1)/2]
设sinx+cosx=t属于[-√2,√2] => t^2=1+2sinxcosx =〉 sinxcosx=(t^2-1)/2
f(x)=(t^2-1)/2(1+t)=(t-1)/2属于[-(√2+1)/2,(√2-1)/2]
另外,分母不为零,所以1+sinx+cosxb不=0 ,既t≠-1
综上,值域属于[-(√2+1)/2,-1)∪(-1,(√2-1)/2]