举报
微微笑乐悠悠
f(x)=cos(2x-π/3)+2sin(x-π/4)sin(x+π/4) =cos(2x-π/3)+cos[(x-π/4)-(x+π/4)]-cos[(x-π/4)+(x+π/4)] =cos(2x-π/3)-cos2x+cos(-π/2) =cos(2x-π/3)-cos2x =-2sin(2x-π/6)sin(-π/6) =sin(2x-π/6) 故最小正周期为π 令2x-π/6=kπ(k∈Z),解得:x=(k/2+1/12)π 这就是图像的对称轴方程 由x∈[-π/12,π/2]得2x-π/6∈[-π/3,5π/6] 由π/2∈[-π/3,5π/6] 函数最大值为1 最小值为sin(-π/3)=-√3/2