# 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?

• Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking.
– Community Bot
Commented Feb 25, 2023 at 17:59
• @BrunoB yes it does by definition Commented Feb 25, 2023 at 18:07
• @SineoftheTime Oh I see, it's a piecewise definition, gotcha. Commented Feb 25, 2023 at 18:09
• The formula of $d$ isn't correct. Take $x\neq y$ instead of $x, y\in\Bbb{R}$ Commented Feb 25, 2023 at 18:12

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 $$z$$.
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 $$t$$.
Notice that $$-1 ≤ \cos(x-y)≤1$$ and hence $$2≤d(x,y)≤4$$ for every $$x,y\in \mathbb R$$.
Thus we have $$d(x,y)≤4=2+2≤d(x,z)+d(z,y).$$