向量a=（cosx+2sinx,sinx）向量b=（cosx-sinx,2cosx） f（x)=向量a*向量b 求f（x）的单调区间

a=(cosx+2sinx,sinx),b=(cosx-sinx,2cosx)
f(x)=a·b=(cosx+2sinx)(cosx-sinx)+2sinx*cosx
=(cosx)^2+sinxcosx-2(sinx)^2+2sinxcosx
=(cosx)^2-2(sinx)^2+3sinxcosx
=(1+cos2x)/2-2(1-cos2x)/2+(3/2)sin2x
=-1/2-(3/2)cos2x+(3/2)sin2x
=-1/2+(3√2/2)[√2/2)sin2x-(√2/2)cos2x]
=-1/2+(3√2/2)sin(2x-π/4)
函数sinx在-π/2+2kπ=所以此函数的递增区间满足-π/8+kπ=-π/8+kπ=