[proofplan]
We transfer the algebraic polynomial $p$ on $[-1,1]$ to a trigonometric polynomial by writing $x = \cos \theta$ and setting $q(\theta) = p(\cos \theta)$. The derivative of this composite contains exactly the endpoint factor: $q'(\theta) = -\sin \theta\,p'(\cos \theta)$. Since $q$ is a trigonometric polynomial of degree at most $n$, the trigonometric Bernstein inequality bounds $|q'|$ by $n\|q\|_\infty$, and the equality $\|q\|_\infty = \|p\|_{L^\infty([-1,1])}$ transfers the estimate back to $p$.
[/proofplan]
[step:Pull back the polynomial along the cosine parametrisation]
Fix $p \in \mathcal{P}_n$ and $x \in (-1,1)$. Since $x \in (-1,1)$, define $\theta_x \in (0,\pi)$ by
\begin{align*}
\theta_x := \arccos x.
\end{align*}
Thus $\cos \theta_x = x$ and $\sin \theta_x > 0$.
Define the function $q: \mathbb{R} \to \mathbb{C}$ by $q(\theta) := p(\cos \theta)$ for $\theta \in \mathbb{R}$.
Because $p$ is a polynomial and $\cos: \mathbb{R} \to [-1,1]$ is smooth, the function $q$ is differentiable on $\mathbb{R}$. By the chain rule, for every $\theta \in \mathbb{R}$,
\begin{align*}
q'(\theta) = -\sin \theta\, p'(\cos \theta).
\end{align*}
In particular,
\begin{align*}
|q'(\theta_x)| = \sin \theta_x\, |p'(x)|.
\end{align*}
[guided]
The point of introducing $q$ is that the interval $[-1,1]$ is exactly the image of the cosine map. Since $x \in (-1,1)$, there is a unique angle $\theta_x \in (0,\pi)$ with $\cos \theta_x = x$, namely
\begin{align*}
\theta_x := \arccos x.
\end{align*}
The restriction $\theta_x \in (0,\pi)$ is important because it gives $\sin \theta_x > 0$, so no absolute value ambiguity appears when we later identify $\sin \theta_x$ with $\sqrt{1-x^2}$.
Now define the function $q: \mathbb{R} \to \mathbb{C}$ by $q(\theta) := p(\cos \theta)$ for $\theta \in \mathbb{R}$. This is a differentiable function because $p: \mathbb{C} \to \mathbb{C}$ is a polynomial and $\cos: \mathbb{R} \to [-1,1]$ is smooth. Applying the chain rule to the composite $\theta \mapsto p(\cos \theta)$ gives, for every $\theta \in \mathbb{R}$,
\begin{align*}
q'(\theta) = p'(\cos \theta)(-\sin \theta).
\end{align*}
Equivalently,
\begin{align*}
q'(\theta) = -\sin \theta\, p'(\cos \theta).
\end{align*}
Evaluating at $\theta = \theta_x$ and using $\cos \theta_x = x$ gives
\begin{align*}
q'(\theta_x) = -\sin \theta_x\, p'(x).
\end{align*}
Taking absolute values and using $\sin \theta_x > 0$, we obtain
\begin{align*}
|q'(\theta_x)| = \sin \theta_x\, |p'(x)|.
\end{align*}
This identity is the mechanism behind the endpoint factor in Bernstein's inequality.
[/guided]
[/step]
[step:Verify that the pullback is a trigonometric polynomial of degree at most $n$]
Write
\begin{align*}
p(z) = \sum_{k=0}^{n} a_k z^k
\end{align*}
with coefficients $a_0,\dots,a_n \in \mathbb{C}$. Let $i \in \mathbb{C}$ denote the imaginary unit, so $i^2 = -1$. Using
\begin{align*}
\cos \theta = \frac{e^{i\theta} + e^{-i\theta}}{2},
\end{align*}
we have, for each $k \in \{0,\dots,n\}$,
\begin{align*}
(\cos \theta)^k = 2^{-k}\sum_{j=0}^{k} \binom{k}{j} e^{i(k-2j)\theta}.
\end{align*}
Therefore $q$ is a finite linear combination of functions $\theta \mapsto e^{im\theta}$ with integers $m$ satisfying $|m| \leq n$. Hence $q$ is a trigonometric polynomial of degree at most $n$.
[guided]
We now check that the pullback has the exact form required for the trigonometric Bernstein estimate. Since $p \in \mathcal{P}_n$, there are coefficients $a_0,\dots,a_n \in \mathbb{C}$ such that
\begin{align*}
p(z) = \sum_{k=0}^{n} a_k z^k.
\end{align*}
Let $i \in \mathbb{C}$ denote the imaginary unit, so $i^2 = -1$. [Euler's formula](/theorems/2014) gives
\begin{align*}
\cos \theta = \frac{e^{i\theta} + e^{-i\theta}}{2}.
\end{align*}
For each integer $k \in \{0,\dots,n\}$, the [binomial theorem](/theorems/750) applied to the two summands $e^{i\theta}$ and $e^{-i\theta}$ gives
\begin{align*}
(\cos \theta)^k = 2^{-k}\sum_{j=0}^{k} \binom{k}{j} e^{i(k-2j)\theta}.
\end{align*}
Thus every monomial $(\cos \theta)^k$ is a finite linear combination of exponential modes $\theta \mapsto e^{im\theta}$, where $m = k-2j$ for some $j \in \{0,\dots,k\}$. Since $0 \leq k \leq n$, each such integer satisfies $|m| \leq k \leq n$. Therefore $q(\theta)=p(\cos \theta)$ is a finite linear combination of modes $\theta \mapsto e^{im\theta}$ with $|m| \leq n$, which is precisely the assertion that $q$ is a trigonometric polynomial of degree at most $n$.
[/guided]
[/step]
[step:Apply the trigonometric Bernstein estimate and transfer the norm]
We apply the standard [Bernstein inequality for trigonometric polynomials](/theorems/6877) to $q$: every trigonometric polynomial $r: \mathbb{R} \to \mathbb{C}$ of degree at most $n$ satisfies $|r'(\theta)| \leq n\|r\|_{L^\infty(\mathbb{R})}$ for every $\theta \in \mathbb{R}$. Its hypotheses are satisfied because the previous step shows that $q$ is a trigonometric polynomial of degree at most $n$. Thus, for every $\theta \in \mathbb{R}$,
\begin{align*}
|q'(\theta)| \leq n \|q\|_{L^\infty(\mathbb{R})}.
\end{align*}
Since $\cos(\mathbb{R}) = [-1,1]$, the range of $q$ is exactly the range of $p$ restricted to $[-1,1]$. Consequently,
\begin{align*}
\|q\|_{L^\infty(\mathbb{R})} = \|p\|_{L^\infty([-1,1])}.
\end{align*}
Evaluating the trigonometric Bernstein estimate at $\theta = \theta_x$ gives
\begin{align*}
|q'(\theta_x)| \leq n \|p\|_{L^\infty([-1,1])}.
\end{align*}
[guided]
We now use the trigonometric input. The standard Bernstein inequality for trigonometric polynomials says that if $r: \mathbb{R} \to \mathbb{C}$ is a trigonometric polynomial of degree at most $n$, then for every $\theta \in \mathbb{R}$,
\begin{align*}
|r'(\theta)| \leq n\|r\|_{L^\infty(\mathbb{R})}.
\end{align*}
The previous step verified the only structural hypothesis needed for this result: $q$ is a trigonometric polynomial of degree at most $n$. Applying the estimate with $r=q$ gives
\begin{align*}
|q'(\theta)| \leq n\|q\|_{L^\infty(\mathbb{R})}
\end{align*}
for every $\theta \in \mathbb{R}$.
It remains to identify the norm of $q$. Since $q(\theta)=p(\cos\theta)$ and $\cos(\mathbb{R})=[-1,1]$, the set of values attained by $q$ on $\mathbb{R}$ is exactly the set of values attained by $p$ on $[-1,1]$. Therefore
\begin{align*}
\|q\|_{L^\infty(\mathbb{R})} = \|p\|_{L^\infty([-1,1])}.
\end{align*}
Substituting this norm identity into the trigonometric Bernstein estimate and then evaluating at $\theta=\theta_x$ gives
\begin{align*}
|q'(\theta_x)| \leq n \|p\|_{L^\infty([-1,1])}.
\end{align*}
[/guided]
[/step]
[step:Convert the sine factor into the algebraic endpoint weight]
From the first step,
\begin{align*}
\sin \theta_x\, |p'(x)| = |q'(\theta_x)|.
\end{align*}
Combining this identity with the estimate from the previous step gives
\begin{align*}
\sin \theta_x\, |p'(x)| \leq n \|p\|_{L^\infty([-1,1])}.
\end{align*}
Since $\theta_x \in (0,\pi)$ and $\cos \theta_x = x$, we have
\begin{align*}
\sin \theta_x = \sqrt{1-\cos^2 \theta_x} = \sqrt{1-x^2}.
\end{align*}
Substituting this into the preceding inequality yields
\begin{align*}
\sqrt{1-x^2}\, |p'(x)| \leq n \|p\|_{L^\infty([-1,1])}.
\end{align*}
This is the desired Bernstein inequality for algebraic polynomials.
[guided]
The first step produced the exact identity
\begin{align*}
\sin \theta_x\, |p'(x)| = |q'(\theta_x)|.
\end{align*}
The trigonometric Bernstein step produced the estimate
\begin{align*}
|q'(\theta_x)| \leq n \|p\|_{L^\infty([-1,1])}.
\end{align*}
Combining these two statements gives
\begin{align*}
\sin \theta_x\, |p'(x)| \leq n \|p\|_{L^\infty([-1,1])}.
\end{align*}
Now we translate the trigonometric factor back into the algebraic variable. Since $\theta_x \in (0,\pi)$, we have $\sin \theta_x>0$. Since also $\cos\theta_x=x$, the identity $\sin^2\theta + \cos^2\theta = 1$ gives
\begin{align*}
\sin \theta_x = \sqrt{1-\cos^2\theta_x} = \sqrt{1-x^2}.
\end{align*}
Substituting this equality into the previous inequality yields
\begin{align*}
\sqrt{1-x^2}\, |p'(x)| \leq n \|p\|_{L^\infty([-1,1])}.
\end{align*}
This is exactly the Bernstein inequality for algebraic polynomials at the fixed point $x \in (-1,1)$.
[/guided]
[/step]
