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）.