Question

已知f(x)=1-2sin^2(x+22.5°)+2sin(x+22.5°)*cos(x+22.5°)

Answer 1

1)

f(x)=1-2sin^2(x+22.5°)+2sin(x+22.5°)*cos(x+22.5°)

=1-1+[2cos^2(x+22.5°)-1]+sin(2x+45°)

=1-1+cos(2x+45°)+sin(2x+45°)

=√2sin(2x+90°)

函数f(x)的最小正周期 ∏,

2)

函数f(x)的单调递增区间[(2k-1) ∏,k ∏]