mean value theorem?

a)
只須證 0 < x ≤ π 時 x > sinx.
由 f(x) = sinx 在 [0,x] 上連續且可導,根據 Lagrange mean value theorem , 必有 ξ ∈ (0,x) 使得
(sinx - sin0) / (x - 0) = (sinξ)' = cosξ < 1
sinx / x < 1
x > sinx.

b)
Consider f(x) = cosx - 1 + x²/2.
Since this is differentiable for all R, let's apply the Lagrange mean value theorem to f(x) for 0 < x ≤ k:

f(k) - f(0) = f '(ξ) (k - 0) for 0 < ξ ≤ k
⇒ cosk - 1 + k²/2 = (ξ - sinξ)k.

However, by a) , ξ > sinξ for ξ > 0.
⇒ (ξ - sinξ) k > 0 for k > 0 ,
then cosk - 1 + k²/2 > 0
⇒ cosx > 1 - x²/2 for x > 0.