Show that 1-1/2t^2<=cost<=1 for 0<=t<=1.?

Consider f(t) = cost - 1 + t²/2.
Since this is differentiable for all R, let's apply the Lagrange mean value theorem to f(t) for 0 < t ≤ k:

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

However, ξ > sinξ for ξ > 0.
⇒ (ξ - sinξ) k > 0 for k > 0 ,
then cosk - 1 + k²/2 > 0
⇒ 1 - t²/2 < cost for t > 0.