已知函数 f(x)=1/2cos^2x-3sinxcosx-1/2sin^2x+1 x∈R求函数f(x)的最小正周期及在区间[_数学解答

f(x)=(1/2)(cosx)^2-3sinxcosx-(1/2)(sinx)^2 +1
==(1/2)[(cosx)^2-(sinx)^2]-(1/2)2sinxcosx -sin2x
=(1/2)(cos2x+sin2x)
=(√2/2)[cos2xcos(pi/4)+sin2xsin(pi/4)]
=(√2/2)cos(2x-pi/4)
1)周期T=2pi/2=pi
2)对称轴2x-pi/4=kpi--->x=kpi/2+pi/8
3)递减区间2kpi=<2x-pi/4=<2kpi+pi--->kpi+pi/8=递增区间2kpi+pi=<2x+pi/4=<2kpi--->kpi+5pi/8=