Question

求数f(x)=sin(@/3+4x)+cos(4x-@/6)的最小正周期和递增区间

Answer 1

答：

f(x)=sin(π/3+4x)+cos(4x-π/6)

=sin(4x+π/3)+cos[-π/2+(π/3+4x)]

=sin(4x+π/3)+sin(π/3+4x)

=2sin(4x+π/3)

所以：f(x)的最小正周期T=2π/4=π/2

单调递增区间满足：2kπ-π/2<=4x+π/3<=2kπ+π/2

所以：单调递增区间为[kπ/2-5π/24，kπ/2+π/24]，k∈Z

Answer 2

f(x)=sin(Π/3+4x)+cos(4x-Π/6)
=1/2sin4x+√3/2cos4x+√3/2cos4x+1/2sin4x
=sin4x+√3cos4x
=2sin(4x+Π/3)
最小正周期T=Π/2
递增区间 2KΠ-Π/2<4X+Π/3<2KΠ+Π/2
KΠ/2-5Π/24k<X<KΠ/2+Π/24