求下列函数的值域:(1)y=4-3sin(x-π/3) (2)y=cos^2x-sinx.

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求下列函数的值域:(1)y=4-3sin(x-π/3) (2)y=cos^2x-sinx.

(1)y=4-3sin(x-π/3)
x-π/3是整体,所以-1≤sin(x-π/3)≤1
所以-3≤3sin(x-π/3)≤3 所以y=4-3sin(x-π/3) 的值域为[1,7]
(2)y=cos^2x-sinx
y=(2cos^2x-1)/2+1/2-Sinx
=cosx-sinx+1/2
=√2cos(x+π/4 )+1/2 -1≤cos(x+π/4)≤1
所以-√2+1/2≤y≤√2 +1/2