Baby Rudin Theorem 8.14

Question

I am unable to understand why $g(t)\cos(\frac{t}{2})$ & $g(t)\sin(\frac{t}{2})$ becomes bounded on $[-\pi,\pi] $. To conclude this Rudin has used two given facts $$\\$$ 1. For some x, there are constants $\delta$>0 & M<$\infty$ such that $|f(x+t)-f(x)|\leq M|t| , \forall t\in (-\delta ,\delta)$ $$\\$$2. $ g(t)=\frac{f(x-t)-f(x)}{\sin(t/2)}$ , for $0<| t|\leq \pi$ and $g(0)=0$ .$$\\$$Thanks.

Answer 1

Had to look in the book, but $f$ is assumed to be Riemann integrable, and therefore bounded ( by definition ). The only place $g$ might not be bounded then is at $0$, but by property $1$ it is bounded, since $$ \left|\frac{f(x-t)-f(x)}{\sin(t/2)}\right| < \left|\frac{Mt}{\sin(t/2)}\right| $$

which is bounded at $0$