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}