数学题!!!!关于函数

f(x)＝cos²(x+π/12)+1/2sin2x
＝[1+cos(2x+π/6)]/2+1/2sin2x
＝1/2+1/2sin(π/3-2x)+1/2sin2x
＝1/2+sin(π/6)cos(π/6-2x)
＝1/2+1/2sin(2x+π/3)

(1)sin(2x+π/3)＝1→x＝kπ+π/12时，
所求最大值为：1；
sin(2x+π/3)＝-1→x＝kπ-5π/12时，
所求最小值为：0.

(2)单调递增时，
2kπ-π/2≤2x+π/3≤2kπ+π/2
→kπ-5π/12≤x≤kπ+π/12;
即x∈[kπ-5π/12, kπ+π/12];
单调递减时，
2kπ+π/2≤2x+π/3≤2kπ+3π/2
→kπ+π/12≤x≤kπ+7π/12,
即x∈[kπ+π/12, kπ+7π/12]。

函数f(x)=(1+cos(2x+pi/6))/2+1/2sin2x=0.5(1+sin(2x+pi/3))
-1<=sin(2x+pi/3)<=1=>0<=f(x)<=1
调增区间:2kpi-pi/2<=2x+pi/3<=2kpi+pi/2=>kpi-(5pi/12)<=x<=kpi+pi/12 k为整数

降次和倍角公式即可