Question

已知函数f(x)=sinx平方+2倍根号3sin(x+派/4)cos(x-派/4)-cosx平方-根号3．

已知函数f(x)=sinx平方+2倍根号3sin(x+派/4)cos(x-派/4)-cosx平方-根号3．
1)求函数f(x)的最小正周期和单调递减区间
2)求f(x)在(-派/12,25派/36)上的值域

各位帮忙解下·万分感谢·

Answer 1

1、根据积化和差公式，
f(x)=(sinx)^2+2√3sin(x+π/4)cos(x-π/4)-(cosx)^2-√3
=-cos2x+√3[sin2x+sinπ/2)-√3
=2(√3sin2x/2-cos2x/2)
=2sin(2x-π/6),
最小正周期为2π/2=π，
2kπ+π/2<=2x-π/6<=2kπ+3π/2,
kπ+π/4<=x-π/12<=kπ+3π/4,
kπ+π/3<=x<=kπ+5π/6,
单调递减区间为：x∈[kπ+π/3,kπ+5π/6],(k∈Z),
2、单调递增区间为：x∈[kπ-5π/6,kπ+π/3],(k∈Z),
在区间（-π/12，25π/36）内，（-π/12，π/3）
为单调递增，最大值为2，最小为-√3，
（π/3，25π/36）内单调递减，最大为1，最小为2sin(2π/9),
故f(x)在（-π/12，25π/36）区间内值域：f(x)∈(-√3,2).

Answer 2

解：
f(x)=(sinx)^2+2√3sin(x+π/4)cos(x-π/4)-(cosx)^2-√3
=-((cosx)^2-(sinx)^2)+√3(sin(x+π/4+x-π/4)+sin(x+π/4-(x-π/4))-√3
=-cos(2x)+√3sin(2x)+√3-√3
=√3sin(2x)-cos(2x)
=2sin(2x-π/6)
(1)
最小正周期：2π/2=π
单调递减区间:2kπ+π/2≤2x-π/6≤2kπ+3π/2
即kπ+π/3≤x≤kπ+5π/6 (k∈n)
(2)
-π/3<2x-π/6<11π/9
sin(-π/3)<sin(2x-π/6)≤sin(π/2)
-√3/2<sin(2x-π/6)≤1
-√3<f(x)≤2