Plotting $\cos(x) >\cos (y) $

Question

I am new to the website so I may go wrong at many places but please bear with me. I came across this : $$\cos (x) - \cos(y) > 0$$ and tried to plot it using simple trigonometry. But I was unable to do so after repeated attempts. My Working: $$\cos (x) > \cos (y) $$ $$\therefore 2\sin \left(\frac{x+y}{2} \right) \times \sin \left(\frac{x-y}{2} \right) > 0 $$ Which implies both $$\sin \left(\frac{x+y}{2} \right)$$and $$\sin \left(\frac{x-y}{2} \right)$$ must have equal signs. After this I could not make any further conclusions. Any help?

Answer 1

Let $A$ be the set of all points $(x,y)$ such that $\cos(x) > \cos(y)$. $A$ has several symmetries:

• $(x,y) \in A \implies (x \pm 2\pi, y) \in A$.
• $(x,y) \in A \implies (x, y \pm 2\pi) \in A$.
• $(x,y) \in A \implies (-x, y) \in A$.
• $(x,y) \in A \implies (x, -y) \in A$.
• $(x,y) \in A \implies (y, x) \not\in A$.

Consider the triangle $T_0 = \{(x,y) \in \mathbb{R}^2 \mid x < y \text{ and } x,y \in [0, \pi] \}$. As $\cos$ is decreasing on $[0, \pi]$, it follows that $T_0 \subset A$.

Finally, we can take $T_0$ and flip it over the $y$-axis to get $T_1 \in A$. Take $T_0 \cup T_1$ and flip it over the $x$-axis to get $T_2 \in A$. Finally, take $T_3 = T_0 \cup T_1 \cup T_2$ and tile the entire plane with it to get $T^\star \in A$.

It is easy to see that reflecting $T^\star$ over the line $y=x$ gives $\mathbb{R}^2 \setminus T^\star$. Thus, $\mathbb{R}^2 \setminus T^\star \cap A = \emptyset \implies T^\star = A$.

Answer 2

Let indicate for $\cos y \neq 1$

$$z=\arccos y\neq0$$

then

$$\cos (x) - \cos(y) > 0 \iff \cos (x) > \cos(y) $$ $$\iff 0+2k\pi\le x< z+2k\pi, \quad 2\pi-z+2k\pi<x\le 2\pi+2k\pi$$

To plot the region in the $x-y$ plane, let consider at first the condition $$\cos x=\cos y\iff y=x+2k\pi \,\lor \, y=-x+2k\pi $$

which is a set of lines parallel to the bisector lines.

Then from the initial observation it is easy to find the region bounded by the lines for which the inequality is satisfied.

Answer 3

In light blue the region $\cos(x)>\cos(y)$

Answer 4

$$ \cos(x) > \cos(y) \quad\Leftrightarrow\quad x > y + 2k\pi \quad\vee\quad x > -y + 2k\pi, \qquad k \in \mathbb{Z} $$

Note: $ x > -y + 2k\pi $, the minus in $ y $ comes from the fact that $ \cos$ is an even function i.e. $\cos(a) = \cos(-a) $.

PD: I'm not allowed to write comments.