Question

已知函数f(x)=2cos(x+三分之派）【sin（x+三分之派）-根号3cos（x+三分之派）】

Answer 1

f(x)=2cos(x+π/3)[sin(x+π/3)-√3cos(x+π/3)]，
=2cos(x+π/3)sin(x+π/3)-2√3cos^2(x+π/3)
=sin(2x+2π/3)-√3[2√3cos^2(x+π/3)-1]-√3
=sin(2x+2π/3)-√3cos(2x+2π/3)-√3
=2sin[(2x+2π/3)-π/3]-√3
=2sin(2x+π/3)-√3
最小正周期=2π/2=π
2x+π/3∈
[2Kπ+π/2,2Kπ+3π/2]单调递减区间
x∈
[Kπ+π/12,2Kπ+7π/12]单调递减区间

Answer 2

f(x)=2cos(x+π/3)[sin(x+π/3)-√3cos(x+π/3)]
=2cos(x+π/3)sin(x+π/3)-2√3cos²(x+π/3)
=sin(2x+2π/3)-√3[1+cos(2x+2π/3)]
=sin(2x+2π/3)-√3cos(2x+2π/3)-√3
=2[(1/2)sin(2x+2π/3)-(√3/2)cos(2x+2π/3)]-√3
=2sin(2x+2π/3-π/3)-√3
=2sin(2x+
π/3)-√3
最小正周期
Tmin=2π/2=π
当2kπ+π/2≤2x+π/3≤2kπ+3π/2
(k∈Z)时，函数单调递减，此时
kπ+π/12≤x≤kπ+7π/12
(k∈Z)
函数的单调递减区间为[kπ+π/12，kπ+7π/12]
(k∈Z)

Answer 3

f(x)=sin(2x+三分之派)+sin(2x-三分之派)+2cos方x-1
=sin2x*cos(3分之π)
+
cos2x*sin(3分之π)
+
sin2x*cos(3分之π)
-
cos2x*sin(3分之π)
+cos2x
=2sin2x*cos(3分之π)+cos2x
=2sin2x*
2分之1+cos2x
=sin2x+cos2x
=根号2*[sin2x*cos(4分之π)+
cos2x*sin(4分之π)]
=根号2*sin(2x
+
4分之π)
所以可知函数最小正周期t=2π/2=π