The period of a (non-linear) simple pendulum is $$ T(\theta_0) = \sqrt{8}/\omega_0 \int_0^{\theta_0} \frac{1}{\sqrt{\cos\theta-\cos\theta_0}}d\theta. $$
Using elliptic functions, we can show that the term of order $1$ in $\theta_0$ is $2\pi/\omega_0$, which is precisely the period of the linear simple pendulum.
On the other hand, writing $$ \cos\theta-\cos\theta_0 \simeq \sin\theta_0(\theta_0-\theta) $$ leads to $$ T(\theta_0) \simeq \sqrt{5}/2/\omega_0\simeq 5.65/\omega_0 $$ which is far from the result $2\pi/\omega_0$.
Is it due to the fact I neglected terms of order $>1$?
So if I would take the Taylor-Lagrange expansion $$ \cos\theta-\cos\theta_0=(\theta_0-\theta)\sin\theta_0-1/2(\theta_0-\theta)^2\cos \xi, \quad \xi\in (0,\theta_0), $$ could I compute $\xi(\theta_0)$ so that the first term of $T(\theta_0)$ in $\theta_0$ is $2\pi/\omega_0$ ?
The issue is that $$\frac{\cos\theta-\cos\theta_0}{\theta-\theta_0}\approx -\sin\theta_0$$ only if $\theta$ and $\theta_0$ are close, otherwise that is not a good approximation, so you cannot recover the exact first term of the wanted Taylor series from it, since the value of the actual integrand function in a right neighbourhood of the origin is fairly different from the value of the approximated integrand function.
On the other hand $$\frac{\cos\theta-\cos\theta_0}{\theta^2-\theta_0^2}\approx -\frac{\sin\theta_0}{2\theta_0}$$ leads to the correct outcome since $\int_{0}^{T}\frac{dt}{\sqrt{T^2-t^2}}=\frac{\pi}{2}$ for any $T>0$.