egin{equation*}
extbf{已知}x_1,x_2<pi,x_{n+1}=x_n+left{ egin{aligned}
sin x_n &,x_n>x_{n+1}\
cos x_n&,x_nle x_{n+1}\
end{aligned}
ight.
end{equation*}
证明:$ x_n<dfrac{3pi}{2}$
假设存在$n_0,x_{n_0}<dfrac{3pi}{2},x_{n_0+1}gedfrac{3pi}{2},ecause x_{n_0+1}-x_{n_0}le1, herefore x_{n_0}ge x_{n_0+1}-1ge dfrac{3pi}{2}-1>pi,$
$ herefore pi<x_{n_0}<dfrac{3pi}{2}$但此时由$x_{n_0+1}$ 的定义知道$x_{n_0+1}<x_{n_0}$ 与假设矛盾.所以假设不成立.