= Solution
The <first Jackson theorem for periodic approximation> states that an absolute constant $C_J$ exists such that, for every continuous $2\pi$-periodic <function> $g$ and integer $n\ge1$, there is a <trigonometric polynomial> $q_n$ of degree at most $n$ satisfying
$$
\|g-q_n\|_\infty\le C_J\omega(g,1/n).
$$
Apply it to the continuous periodic <function> $g(\theta)=f(\cos\theta)$, which is even. Replace the approximant by its even part
$$
q_n^{\rm ev}(\theta)=\frac12\bigl(q_n(\theta)+q_n(-\theta)\bigr).
$$
Since $g(-\theta)=g(\theta)$, the <triangle inequality> gives $\|g-q_n^{\rm ev}\|_\infty\le\|g-q_n\|_\infty$. This symmetrization cancels all sine terms, so $q_n^{\rm ev}=\sum_{j=0}^n\alpha_j\cos(j\theta)$. The <Chebyshev polynomials> satisfy $T_j(\cos\theta)=\cos(j\theta)$ and have degree $j$; the latter follows inductively from $T_0=1$, $T_1=x$ and $T_{j+1}=2xT_j-T_{j-1}$. Therefore
$$
p_n(x)=\sum_{j=0}^n\alpha_jT_j(x)
$$
is an algebraic <polynomial> of degree at most $n$, and $p_n(\cos\theta)=q_n^{\rm ev}(\theta)$. Because cosine maps onto $[-1,1]$, the <supremum norms> agree exactly:
$$
\|f-p_n\|_{[-1,1]}=\|g-q_n^{\rm ev}\|_{\mathbb T}.
$$
Finally, the definition of the best error, periodic Jackson approximation, and part (a) give the successive bounds
$$
E_n(f)\le\|f-p_n\|_\infty
\le C_J\omega(g,1/n)
\le C_J\omega(f,1/n).
$$
Thus
$$
\boxed{E_n(f)\le C_J\omega(f,1/n),\qquad n\ge1.}
$$
Neither the constant nor the choice of approximation space depends on $f$. The <cosine substitution for polynomial approximation> and even symmetrization are what preserve the required degree bound.
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