How do we compute the modulus of continuity for log and sin? How do we actually compute the modulus of continuity for $\log(x)$, $x \in [1,+\infty)$ or for $\sin(x)$, $x \in \mathbb{R}$? I am trying to use the definition, i.e. 
$$
\omega(f,\delta) = \sup_{|x-y|<\delta} |f(x)-f(y)|,
$$
but I do not see the way to compute the actual supremum.
 A: Answer for $f(x)=\log(x), x \ge 1$:
W.l.o.g. we can assume that $y \ge x$, then
$$\omega(f,\delta) = \sup_{|x-y|<\delta} |\log(x)-\log(y)| = \sup_{x \le y < x+\delta} \log\left(\frac{y}{x}\right).$$
Since $\log(\frac{y}{x})$ is continuous and monotonically increasing if considered as a function of $y$ with fixed $x$, the supremum (with $x$ fixed) is actually the value for $y=x+\delta$:
$$\omega(f,\delta) = \sup_x \log\left(\frac{x+\delta}{x}\right) = \sup_x \log\left(1 +\frac{\delta}{x}\right).$$
Again, since the logarithm is increasing, the highest values of the expression under the supremum is achieved when the argument is as big as possible. Since we have $x \ge 1$ and $1+\frac{\delta}x$ is decreasing, this happens for $x=1$, so we finally have
$$\omega(f,\delta) =\log(1+\delta)$$
For $f(x)=\sin(x)$, I'll give some hints that should help you to get on the right track:
1) Can you give a constant upper bound for $\omega(\sin(x),\delta)$ that is true for all $\delta$? 
2) Using 1), can you give the exact value of $\omega(\sin(x),\delta)$ for 'big' $\delta$? How small can you make the $\delta$ such that it still works?
3) Looking at the solution above for $\log(x)$, monotony of $f$ plays an important role, as is makes evaluation of the supremum easier. Find where $\sin(x)$ has 'stretches' of monotony!
4) Look at the graph of $\sin(x)$ and try to formulate an idea where the $x,y$ should be chosen to create a 'big' $|\sin(x)-\sin(y)|$ for values of $\delta$ not covered by 2). Try to prove that idea, using sine addition theorems.
