怎样求函数2cos(x/2)-3sin(x/3)的最小正周期

Answer 1

2cos(x/2)-3sin(x/3)
令x=x+T
2cos((x+T)/2)-3sin((x+T)/3)
=2(cos(x/2)cos(T/2)-sin(x/2)sin(T/2))-3(sin(x/3)cos(T/3)+cos(x/3)sin(T/3)).(1)
T为周期,原式和(1)式应该相等.
即:
cos(T/2)=1.(a)
sin(T/2)=0.(b)
cos(T/3)=1.(c)
sin(T/3)=0.(d)
由(a)(b):T/2=2k∏
=> T=4k∏ (k 为使得T为正数的整数,下同)
由(c)(d):T/3=2k'∏
=> T=6k'∏
因此当k=3,k'=2时T=12∏
周期是12∏