设函数f（x）=x/2+sinx的所有正的极小值点从小到大排成的数列为xn 1）求数列xn的通项公

f(x)=x/2 + sin(x),
f'(x) = 1/2 + cos(x),令0=f'(x),有,-1/2 = cos(x),x=2kπ+4π/3或x=2kπ-4π/3,k=0,1,-1,2,-2,...,n,-n,...
f''(x) = - sin(x),x为极小值点,则-sin(x)=f''(x)>=0,因此,x=2kπ+4π/3,k=0,1,-1,2,-2,...,n,-n,...
x(n)=2(n-1)π+4π/3,
s(n)=n(n-1)π+4nπ/3,
n=3m,s(n)=3m(3m-1)π+4mπ,sin[s(n)] = sin(0)=0,n=3m,m=1,2,...
n=3m-1,s(n)=(3m-1)(3m-2)π + 4(3m-1)π/3 = (3m-1)(3m-2)π + 4mπ - 4π/3,
sin[s(n)] = sin(-4π/3)=sin(2π/3)=sin(π/3)=3^(1/2)/2,n=3m-1,m=1,2,...
n=3m-2,s(n) = (3m-2)(3m-3)π +4(3m-2)π/3 = (3m-2)(3m-3)π + 4mπ - 8π/3,
sin[s(n)] = sin(-8π/3)=sin(-2π/3)=-sin(2π/3)=-sin(π/3)=-3^(1/2)/2,n=3m-2,m=1,2,...
综合,有,
n=3m-2时,sin[s(n)]=-3^(1/2)/2,
n=3m-1时,sin[s(n)]= 3^(1/2)/2,
n=3m时,sin[s(n)]=0,
m=1,2,...

f（x）取极小值时，sinx=-1
即x=2kπ+3π/2,k∈Z
极小值通式为：kπ+3π/4-1
k>=0时，满足kπ+3π/4-1>0,即k=0时为Xn的第一项X1
Xn=(n-1)π+3π/4-1=nπ-π/4-1
Sn=n(n+1)π/2-nπ/4-n=n(2n+1)π/4-n
sinSn=sin[n(2n+1)π/4-n]