Proving $f(x,y)=\left ( \frac{1}{4}\sin(x+y),1+\frac{2}{3}\arctan(x-y) \right )$ is a contraction

Question

$f(x,y)=\left ( \frac{1}{4}\sin(x+y),1+\frac{2}{3}\arctan(x-y) \right )$

Prove that $f$ is a contraction

$\mathbb{R}^2$ is equipped with $d((x,y),(x',y'))=|x-x'|+|y-y'|$

My problem is that I'm not able to find an upper bound of $|\sin(x+y)-\sin(x'+y')|$

Answer 1

By MVT $|\sin(A)-\sin(B)|\leq |A-B|$.