Question

已知向量m=(sinwx,-根号3coswx),n=(sinwx,cos(wx+π/2))(w>0)函数f(x)=mn的最小正周期为π

Answer 1

1.cos(wx+π/2)=-sinwx
则f(x)=mn=(sinwx)×(sinwx)+(-√3coswx)×(-sinwx)=sin²wx+√3coswxsinwx
=(1-cos2wx)/2
+
(√3/2)sin2wx
=(√3/2)sin2wx
-
(1/2)cos2wx
+
1/2
=sin(2wx
-
π/6)
+
1/2
f(x)最小正周期=2π/2w
∵f(x)的最小正周期为π
∴2π/2w=π
w=1
2.向左平移π/12各单位:f(x)=sin(2x
-
π/6)=sin[2(x
-
π/6)]=sin[2(x
-
π/6
+
π/12)]=sin[2(x
-
π/12)]=sin(2x
-
π/6)
图像上个点的横左边伸长到原来的4倍:
f(X)=sin[2(x/4)
-
π/6]=sin(x/2
-
π/6)
∵函数y=sinx的递减区间是[2kπ+π/2,2kπ+3π/2](k∈Z)
∴令x/2-π/6∈[2kπ+π/2,2kπ+3π/2]
(k∈Z)
x/2∈[2kπ+2π/3,2kπ+5π/3]
(k∈Z)
x∈[4kπ+4π/3,4kπ+10π/3]
(k∈Z)
g(x)的单调减区间是:x∈[4kπ+4π/3,4kπ+10π/3]
(k∈Z)