∵sinxcosx
=[（sinx+cosx）^2-1]/2
=(1+sinx+cosx)(sinx+cosx-1)/2
∴y=sinxcosx/(1+sinx+cosx)
=(sinx+cosx-1)/2
又1+sinx+cosx≠0即sinx+cosx≠-1
且sinx+cosx=√2sin(x+π/4)∈[-√2,√2].
∴y=sinxcosx/(1+sinx+cosx)的值域?yàn)?br/>[(-√2-1)/2,-1)∪(-1,(√2-1)/2].
參考:
設(shè)sinx+cosx=t屬于[-根號(hào)2,根號(hào)2] => t^2=1+2sinxcosx =〉 sinxcosx=(t^2-1)/2
f(x）=（t^2-1）/2(1+t)=(t-1)/2屬于[-(√2+1)/2,(√2-1)/2]
另外,分母不為零,所以1+sinx+cosxb不=0 ,既t不=-1
綜上,值域?qū)儆赱-(√2+1)/2,-1）并上（-1,(√2-1)/2]