= Solution
For $n\geq1$ put $q_n(t)=2nK_{n-1}(t)\sin(nt)$. Using $\sin(nt)=(e^{int}-e^{-int})/(2i)$ and the <Fourier coefficients> of $K_{n-1}$ gives
$$
\widehat q_n(k)=-ik\quad(|k|\leq n).
$$
Indeed, when $1\leq k\leq n$ only the $e^{int}$ shift contributes, with coefficient $(-in)(1-|k-n|/n)=-ik$. For $-n\leq k\leq-1$ only the opposite shift contributes and gives $in(1-|k+n|/n)=-ik$; at $k=0$ both coefficients vanish.
If $P=\sum_{|k|\leq n}a_ke^{ikt}$, <convolution> multiplies its <Fourier coefficients> by those of $q_n$. Thus $-P*q_n=\sum ika_ke^{ikt}=P'$, proving the exact identity. Since $K_{n-1}\geq0$ and has integral one,
$$
\boxed{\|P'\|_\infty\leq\|P\|_\infty\|q_n\|_1\leq2n\|P\|_\infty.}
$$
For $n=0$ the polynomial is constant and the bound is immediate. This elementary Fejér-kernel argument proves the requested estimate; the sharp <Bernstein inequality for trigonometric polynomials> has constant $n$.
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