Question

高中数学题。 求详细过程。！~！

已知函数f(x)=sin²x+2√3sin（x+π/4）cos（x-π/4）-cos²x-√3
（1）.求函数f（x）的最小正周期和单调递减区间
（2）.求函数f（x）在[-π/12,25π/36]上的最大值和最小值并指出此时相应的x的值。

Answer 1

1、
f(x)=-(cos²x-sin²x)+2√3cos[π/2-(x+π/4)]cos(x-π/4)-√3
=-cos2x+2√3cos(-x+π/4)cos(x-π/4)-√3
=-cos2x+2√3cos²(x-π/4)-√3
=-cos2x+2√3[1+cos2(x-π/4)]/2-√3
=-cos2x+√3[1+cos(2x-π/2)]-√3
=-cos2x+sin2x
=√2(√2/2*sin2x-√2/2cos2x)
=√2(sin2xcosπ/4-cos2xsinπ/4)
=√2sin(2x-π/4)
所以T=2π/2=π

递减则2kπ+π/2<2x-π/4<2kπ+3π/2
2kπ+3π/4<2x<2kπ+7π/4
kπ+3π/8<x<kπ+7π/8
所以减区间(kπ+3π/8,kπ+7π/8)

2、
-π/12<=x<=25π/36
-5π/12<=2x-π/4<=41π/36
所以2x-π/4=-5π/12最小,x=-π/12
2x-π/4=π/2最大,x=3π/8
sin(-5π/12)
=-sin(5π/12)
=-sin(π/4+π/6)
=-sinπ/4cosπ/6-cosπ/4sinπ/6
=-(√6+√2)/4
sinπ/2=1
所以
x=-π/12，y最小=-(√3+1)/2
x=3π/8，y最大=√2