Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 75 / 4 / a / Solution

For a function on an interval, the modulus of continuity is $\omega (f,t)={\sup }_{|x-y|\le t}|f(x)-f(y)|$, with the pair restricted to that interval. For a periodic function the equivalent definition is ${\sup }_{\theta ,|h|\le t}|g(\theta +h)-g(\theta )|$. The mean value theorem and $|\sin \theta |\le 1$ giveBoth cosine values lie in $[-1,1]$. Hence for $|h|\le t$,Taking the supremum provesIf the periodic modulus of continuity is defined using circular distance instead, choose lifts at that shortest distance and apply the same argument. The cosine substitution for polynomial approximation therefore does not enlarge this modulus.