Question

求值域：y=sinx-cosx+2sinxcosx

Answer 1

(sinx-cosx)^2=(sinx)^2+(cosx)^2-2sinxcosx=1-2sinxcosx

y=sinx-cosx+1-(sinx-cosx)^2

t=sinx-cosx=√2sin(x-∏/4)

-√2≤t≤√2

y=t+1-t^2=-t^2+t+1=-(t-1/2)^2+5/4

t=1/2

ymax=5/4

t=-√2

ymin=-(-√2-1/2)^2+5/4=-1-√2

-1-√2≤y≤5/4