已知函数f(x)=cos^4x+2sinxcosx-sin^4x

　　求(1)f(x)单调区间.(2)若x∈[0,π/2],求f(x)的最大值和最小值

　　【解】①f(x)=cos^4x+2sinxcosx-sin^4x=(cos^4x-sin^4x)+2sinxcosx

　　=(cos^2x+sin^2x)(cos^2x-sin^2x)+sin2x

　　=cos^2x-sin^2x+sin2x=cos2x+sin2x

　　=√2sin(2x+π/4)

　　2kπ-π/2≤2x+π/4≤2kπ+π/2,k∈Z.

　　kπ-3π/8≤x≤kπ+π/8,k∈Z.

　　递增区间是[kπ-3π/8,kπ+π/8],k∈Z.

　　2kπ+π/2≤2x+π/4≤2kπ+3π/2,k∈Z.

　　kπ+π/8≤x≤kπ+5π/8,k∈Z.

　　递减区间是[kπ+π/8,kπ+5π/8],k∈Z.

　　②x∈[0,π/2],则2x+π/4∈[π/4,5π/4],

　　∴sin(2x+π/4)∈[-√2/2,1].

　　√2sin(2x+π/4)∈[-1,√2].

　　f(x)的最大值是√2（x=π/8时）,最小值是-1（x=π/2时）.