[proofplan] We convert the trigonometric polynomial into a Laurent polynomial on the unit circle, then multiply by $z^n$ to obtain an ordinary algebraic polynomial of degree at most $2n$. The derivative of the trigonometric polynomial becomes exactly the polar derivative expression $zP'(z)-nP(z)$ for that algebraic polynomial. Applying the sharp Bernstein polar derivative inequality on the unit circle gives the desired factor $n$, and the real-valued case follows because real-valued trigonometric polynomials are a subclass of complex-valued ones. [/proofplan]

[step:Handle the constant case separately] If $n=0$, then $t(x)=c_0$ for every $x \in \mathbb{R}$. Hence $t'(x)=0$ for every $x \in \mathbb{R}$, and therefore $\|t'\|_{L^\infty(\mathbb{R})} = 0 = 0 \cdot \|t\|_{L^\infty(\mathbb{R})}$. Thus the theorem holds for $n=0$. For the rest of the proof, assume $n \in \mathbb{N}$. [/step]

[step:Represent the trigonometric polynomial as a Laurent polynomial on the unit circle]Define the Laurent polynomial $T: \mathbb{C} \setminus \{0\} \to \mathbb{C}$ by \begin{align*} T(z) = \sum_{k=-n}^{n} c_k z^k. \end{align*} For each $x \in \mathbb{R}$, with $z=e^{ix}$, we have $|z|=1$ and \begin{align*} t(x) = T(e^{ix}). \end{align*} Since the map $x \mapsto e^{ix}$ parametrizes the unit circle, it follows that \begin{align*} \|t\|_{L^\infty(\mathbb{R})} = \sup_{|z|=1} |T(z)|. \end{align*}[/step]

[guided]The point of this step is to replace the real variable $x$ by the complex unit-circle variable $z=e^{ix}$. Define $T: \mathbb{C} \setminus \{0\} \to \mathbb{C}$ by \begin{align*} T(z) = \sum_{k=-n}^{n} c_k z^k. \end{align*} This is a Laurent polynomial because negative powers of $z$ may occur. When $z=e^{ix}$, each term $z^k$ is exactly $e^{ikx}$, so \begin{align*} T(e^{ix}) = \sum_{k=-n}^{n} c_k e^{ikx} = t(x). \end{align*} The map $x \mapsto e^{ix}$ has image precisely the unit circle $\{z \in \mathbb{C}: |z|=1\}$. Therefore taking the supremum of $|t(x)|$ over $x \in \mathbb{R}$ is the same as taking the supremum of $|T(z)|$ over $|z|=1$: \begin{align*} \|t\|_{L^\infty(\mathbb{R})} = \sup_{x \in \mathbb{R}} |T(e^{ix})| = \sup_{|z|=1} |T(z)|. \end{align*}[/guided]

[step:Convert the Laurent polynomial into an algebraic polynomial] Define the algebraic polynomial $P: \mathbb{C} \to \mathbb{C}$ by $P(z)=z^nT(z)$. Expanding the definition of $T$, we obtain \begin{align*} P(z) = \sum_{k=-n}^{n} c_k z^{n+k}. \end{align*} Thus $P$ has degree at most $2n$. Moreover, for $|z|=1$, \begin{align*} |P(z)| = |z|^n |T(z)| = |T(z)|. \end{align*} Hence \begin{align*} \sup_{|z|=1} |P(z)| = \|t\|_{L^\infty(\mathbb{R})}. \end{align*} [/step]

[step:Relate the angular derivative to the polar derivative of $P$] For $z \in \mathbb{C} \setminus \{0\}$, differentiating $P(z)=z^nT(z)$ gives \begin{align*} P'(z) = n z^{n-1}T(z) + z^n T'(z). \end{align*} Multiplying by $z$ and subtracting $nP(z)$ gives \begin{align*} zP'(z)-nP(z) = z^{n+1}T'(z). \end{align*} For $x \in \mathbb{R}$, set $z=e^{ix}$. By the chain rule applied to $t(x)=T(e^{ix})$, \begin{align*} t'(x) = i e^{ix} T'(e^{ix}) = izT'(z). \end{align*} Since $|z|=1$, the two preceding identities imply \begin{align*} |t'(x)| = |T'(z)| = |z^{n+1}T'(z)| = |zP'(z)-nP(z)|. \end{align*} [/step]

[step:Apply the sharp polar Bernstein inequality on the unit circle] We use the following standard sharp form of Bernstein's inequality for algebraic polynomials on the unit circle: if $P:\mathbb{C}\to\mathbb{C}$ is a polynomial of degree at most $2n$, then for every $z \in \mathbb{C}$ with $|z|=1$, \begin{align*} |zP'(z)-nP(z)| \leq n \sup_{|\zeta|=1} |P(\zeta)|. \end{align*} (citing a result not yet in the wiki: Bernstein's polar derivative inequality for algebraic polynomials) The polynomial $P$ constructed above has degree at most $2n$, so the inequality applies. Therefore, for every $x \in \mathbb{R}$ and $z=e^{ix}$, \begin{align*} |t'(x)| = |zP'(z)-nP(z)| \leq n \sup_{|\zeta|=1} |P(\zeta)|. \end{align*} Using the equality of suprema already proved, \begin{align*} |t'(x)| \leq n \|t\|_{L^\infty(\mathbb{R})}. \end{align*} Taking the supremum over $x \in \mathbb{R}$ yields \begin{align*} \|t'\|_{L^\infty(\mathbb{R})} \leq n \|t\|_{L^\infty(\mathbb{R})}. \end{align*} [/step]

[step:Recover the real-valued case] If $t$ is real-valued, then it is also a complex-valued trigonometric polynomial with the same pointwise absolute value and the same derivative as a real function. The complex-valued estimate therefore applies without changing either norm: \begin{align*} \|t'\|_{L^\infty(\mathbb{R})} \leq n \|t\|_{L^\infty(\mathbb{R})}. \end{align*} This completes the proof. [/step]