Question

已知函数f(x)=1-2sin^2(x+π/8)+2sin(x+π/8)*cos(x+π/8)

求函数最小值与单调增区间
tan4/π才应该=b/a=1吧，为啥是√2sin(2x+π/2)呢

Answer 1

f(x)=1-2sin^2(x+π/8)+2sin(x+π/8)*cos(x+π/8)
=cos(2x+π/4)+sin(2x+π/4)
=√2sin(2x+π/4+π/4)
=√2cos2x
f(x)=√2cos2x的最小值即cos2x=-1时
f(x)的最小值为 -√2
递增区间为[-π/2+kπ，kπ] (k∈Z)