Using a cosine curve, where is the largest osculating circle found for 0 <= t <= pi

Curvature of circle with radius r:
k = 1/r

So the larger the circle (i.e. radius), then the smaller the curvature.

We need to find point on curve where curvature is smallest (i.e. closest to 0)
Curvature at a certain point on curve:
k = |y''| / (1+(y')²)^(3/2)

y = cos(x)
y' = -sin(x)
y'' = -cos(x)

k = |-cos(x)| / (1+sin²x)^(3/2)

Now k is smallest when numerator is smallest, and denominator is largest.
|-cos(x)| is smallest when cos(x) = 0 ------------> x = π/2
(1+sin²x)^(3/2) is largest when sin(x) = ±1 ----> x = π/2

Since numerator is smallest and denominator is largest for same value of x, curvature is smallest when x = π/2

NOTE: when x = π/2, then curvature is:
k = |-cos(π/2)| / (1+sin²(π/2))^(3/2) = 0/(1+1)^(3/2) = 0

Now we cannot actually have a circle with curvature = 0, since k = 1/r is never = 0, but k does approach 0 as r approaches infinity. So as x approaches π/2 (from the left or right), osculating circle gets larger and larger, until it degenerates into a straight line at x = π/2.