求值域:y=sinx-cosx+2sinxcosx

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求值域:y=sinx-cosx+2sinxcosx
求值域:y=sinx-cosx+2sinxcosx

求值域:y=sinx-cosx+2sinxcosx
(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

令sinx-cosx=t,平方得,2sinxcosx=1-t^2,所以y=t+1-t^2=-(t-1/2)^2+5/4,因为t=sinx-cosx=√2sin(x-π/4),所以-√2<=t<=√2,所以t=-√2时,y(最小值)=-1-√2,t=1/2时,y(最大值)=5/4, 所以值域为[-1-√2,5/4]