Show that $d(x,y) = \cos(x-y) + 3$ for $x \ne y$ is a metric. 
Show that $d(x,x) = 0$ for $x\in\mathbb{R}$ and $d(x,y) = \cos(x-y) + 3$ for $x,y\in\mathbb{R}$ is a metric.

I have a problem with showing that $d(x,y) \le d(x,z) + d(y,z)$.
I showed that
$$d(x,z) + d(y,z) = \cos x\cos z+\sin x\sin z + \cos y\cos z + \sin y\sin z + 6$$
What should I do next?
 A: There's a bit of a red herring in this question. All you need from the metric is that, for two different points $x$ and $y$, the distance between them satisfies $2 \leq d(x,y) \leq 4$. You don't need anything about $cos$ other than the fact that it's between $-1$ and $1$.
Given this property, you should be able to find an upper bound to $d(x,y)$, the left side of the triangle inequality, and a lower bound for $d(x,z) + d(y,z)$, the right hand side. With these two bounds you'll be able to show that the triangle inequality holds for all different $x$, $y$, and $z$.
A: \begin{align}&d(x,z)+d(z,y)-d(x,y)=\\&=\begin{cases} \cos(x-z)+\cos(y-z)-\cos(x-y)+3&\text{if }x\ne y\land y\ne z\land z\ne x\\ 2\cos(x-z)+6&\text{if }x=y\ne z\\ 0&\text{if }x=z\ne y\\ 0&\text{if }z=y\ne x\\ 0&\text{if }x=y=z\end{cases}\end{align}
It's apparent by $-1\le \cos t\le 1$ that the quantity is $\ge0$.
A: You have: $d(x,y) \le d(x,z)+d(y,z) \iff \cos x\cos y+\cos y\cos z+\cos z\cos x+\sin x\sin y+\sin y\sin z+\sin z\sin x \le 3\iff \cos(x-y)-\cos(y-z)-\cos(z-x) \le 3$. This is clearly true since $-1\le\sin t\le 1$ for all real $t$.
A: Notice that $-1 ≤ \cos(x-y)≤1$ and hence $2≤d(x,y)≤4$ for every $x,y\in \mathbb R$.
Thus we have
$$
d(x,y)≤4=2+2≤d(x,z)+d(z,y).
$$
