x-x³/6＜sinx＜x
1.令 f(x) = x - sinx ,f(0)=0,
f '(x) = 1 - cosx ≥ 0
=> f(x) 單增,當x>0時,f(x) > f(0) = 0
=> f(x) > 0,即 sinx < x
2.令 g(x) = sinx - (x-x³/6),g(0)=0,
g '(x) = cosx - 1 + x²/2,g '(0)=0
當x>0時,g ''(x) = x - sinx > 0 => g '(x) 單增
=> 當x>0時,g '(x) > g '(0) = 0 => g (x) 單增
=> 當x>0時,g(x) > g(0) = 0
即 當x>0時,x-x³/6＜sinx