The circular curve defeats the equality form of the vector-valued mean value theorem

Given: The curve $f(t)=(\cos t,\sin t)$ on $[0,2\pi ]$.

The functions $\sin$ and $\cos$ are differentiable on $\mathbb{R}$, with $(\sin t{)}^{\mathrm{\prime }}=\cos t$ and $(\cos t{)}^{\mathrm{\prime }}=-\sin t$; also $\sin 0=0$ and $\cos 0=1$ (The derivatives of sine and cosine are cosine and minus sine).

Let $m\in \mathbb{N}$ with $m\ge 1$, let $A\subseteq \mathbb{R}$, and let $c\in A$ be a limit point of $A$. A map $g:A\to {\mathbb{R}}^m$ is differentiable at $c$ if and only if every component is differentiable there, and its derivative then has the component derivatives as coordinates (The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral).

Let $m\in \mathbb{N}$ with $m\ge 1$, let $A$ be a subspace of a metric space, and let $g:A\to {\mathbb{R}}^m$. The map $g$ is continuous at a point of $A$ if and only if every component is continuous there (A vector-valued function has a limit, or is continuous, if and only if each of its components does; with the algebra of continuous vector-valued functions).

For a real-valued function on $A\subseteq \mathbb{R}$, differentiability at a limit point $c\in A$ implies continuity there (A function differentiable at $c$ is continuous at $c$).

For every real $t$, ${\sin }^{2}t+{\cos }^{2}t=1$ (Parity and the Pythagorean identity for sine and cosine).

Both sine and cosine have period $2\pi$ (The zero sets of sine and cosine and the least positive common period 2 pi).

On ${\mathbb{R}}^2$, $\parallel (u,v){\parallel }_2=\sqrt{u^2+v^2}$ (The $p$-norms $\parallel x{\parallel }_p$ for rational $p\ge 1$, and $\parallel x{\parallel }_{\mathrm{\infty }}$).

The number $\pi$ is positive (Pi as twice the smallest positive zero of cosine).