= Solution
For a periodic function use the <modulus of continuity>
$$
\omega(g,\delta)=\sup_{\theta\in\mathbb R,\ |h|\le\delta}
|g(\theta+h)-g(\theta)|.
$$
On the interval, take the supremum over pairs of points whose distance is at most $\delta$. The <mean value theorem> applied to cosine, whose derivative has absolute value at most one, gives
$$
|\cos(\theta+h)-\cos\theta|\le|h|.
$$
Both cosine values lie in $[-1,1]$, so for $|h|\le\delta$,
$$
|\widetilde f(\theta+h)-\widetilde f(\theta)|
=|f(\cos(\theta+h))-f(\cos\theta)|
\le\omega(f,\delta).
$$
Taking the supremum proves
$$
\boxed{\omega(\widetilde f,\delta)\le\omega(f,\delta)}.
$$
The <cosine substitution for polynomial approximation> is also a linear isometry in the <supremum norm>, since cosine maps a full period onto $[-1,1]$; its image consists of even continuous periodic functions.
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