求函数y=1+sinx+cosx+sinxcosx的值域

Question

求函数y=1+sinx+cosx+sinxcosx的值域

Answer 1

sinx+cosx=t
√2sin(x+∏/4)=t
-√2≤t≤√2
1+2sinxcosx=t²
sinxcosx=(t²-1)/2
y=1+sinx+cosx+sinxcosx
=1+t+(t²-1)/2
=t²/2+t+1/2
=1/2(t²+2t+1)
=1/2(t+1)²
t=-1 y=0
t=√2 y=1/2(√2+1)²
y∈{0,1/2(√2+1)²}