Prove that cosine is norm decreasing Prove that:
$|\cos(y) - \cos(x)| <|x-y| ,\forall (x,y)\in(\mathbb{R} \times \mathbb{R}) /\triangle(\mathbb{R})$
My attempt
It looks like we need to prove that the cosine function is norm decreasing.So the goal is to show that $$d(f(x),f(y))<d(x,y),\forall x\neq y$$
By the mean value theorem, Let $f(x)=\cos(x)$ then $f'(x)=-\sin(x)$. So $$-\sin(c)= \frac{\cos(y)-\cos(x)}{y-x}$$
$$\Rightarrow (y-x)(- \sin(c))=\cos(y)-\cos(x) $$
I am not sure, but can we conclude anything about cosine being bounded here? Why or why not?
By some trig identities we know that:
$$i) \sin(A+B)=\sin(A)cos(B) +\cos(A) \sin (B)$$
$$ii) cos(A+B)= \cos(A) \cos(B) - \sin(A) \sin(B) $$
$$iii) \cos(x)-\cos(y)= -2\sin(\frac{x-y}{2})(\sin\frac{x+y}{2})$$
Plugging $iii$) in $d(f(x)-f(y))<d(x,y)$:
$$|-2\sin(\frac{x-y}{2})(\sin\frac{x+y}{2})|<|x-y|$$
$$= 2|\sin(\frac{x-y}{2})(\sin\frac{x+y}{2})|<|x-y|$$
Don't really know where to go from here
Oh and another thing... why is the domain $\forall (x,y)\in(\mathbb{R} \times \mathbb{R}) /\triangle(\mathbb{R})$, what does this even mean?
 A: The set on which this is true should be written as
$$ \mathbb{R} \times \mathbb{R} \setminus \Delta(\mathbb{R}), $$
I believe: this is the set of pairs of real numbers without the diagonal: elements of the form $(x,x)$. $\Delta(\mathbb{R}) = \{(x,y):x=y \} $. This is obviously necessary since both sides are zero if $x=y$, so not unequal.

Pursuing your trigonometric identity idea, you have that
$$ \cos{(A+B)} = \cos{A}\cos{B}-\sin{A}\sin{B} \\
\cos{(A-B)} = \cos{A}\cos{B}+\sin{A}\sin{B}, $$
so subtracting gives
$$ \cos{(A+B)}-\cos{(A-B)} = - 2\sin{A}\sin{B} . $$
Replacing $A$ and $B$ by $(x\pm y)/2$ then gives
$$ \cos{x}-\cos{y} = -2\sin{\left( \frac{x+y}{2} \right)}\sin{\left( \frac{x-y}{2} \right)}. $$
Now, we want to show that the right-hand side is less in absolute value than $\lvert x-y \rvert$. Obviously $\left|\sin{\left( \frac{x+y}{2} \right)}\right|\leq 1$, so it suffices to show that $2\left|\sin{\left( \frac{x-y}{2} \right)}\right|<\lvert x-y \rvert$ for $\lvert x-y \rvert \neq 0$, which can be rewritten as
$$ \lvert\sin{u}\rvert<\lvert u \rvert $$
for $u \in \mathbb{R} \setminus \{0\}$. This is obviously true for $\lvert u\rvert>1$ since $\sin{u}$ is bounded between $1$ and $-1$. Moreover, it suffices to check for $u>0$ since $\sin{u}$ is odd. A simple geometrical argument can now be used: draw a circle with centre $O$, and draw two lines through $O$ that contain an angle $u$, meeting the circle at $A$ and $B$ respectively. Then $\triangle AOB$ has area $\frac{1}{2}R^2\sin{u}$, while the sector $OAB$ has area $\frac{1}{2}R^2u$, and is obviously larger than the triangle. The result follows.
A: Hint: Suppose $y>x$. By fundamental theorem of calculus, $\cos(y)-\cos(x)=\int_x^y -\sin(t)\, \textrm{d}t$. It is enough to notice that $ \int_x^y \lvert\sin(t)\rvert \,\textrm{d}t<\int_x^y 1\,\textrm{d}t$.
