For what range in $x$ and $y$ is $(\frac{\pi}{y} \cos x \sin y - \cos y \sin x)\sin y > \sin x$ true?

Question

I am trying to determine a relation between $x$ and $y$ such that

$$(\frac{\pi}{y} \cos x \sin y - \cos y \sin x)\sin y > \sin x$$

is true, and I cannot see a way to do this. Any help would be appreciated.

Answer 1

Note that the equation is periodic in $x$, in the sense that if $[x,y] \in S$ (with $S$ as your solution set), then $[x+2\pi, y] \in S$.

Hence it stands to reason that the separation line between $S$ and $\neg S$ can be given as $x(y)$ and is periodic.

Solving $$ \left(\frac{\pi}{y} \cos (x) \sin (y)− \cos (y) \sin (x)\right)\sin (y) - \sin (x) = 0 $$ (the separation line) for $x$ should give (I didn't compute this by hand) $$ x_n(y) = \cot^{-1}\left(\frac{y \csc(y) \left(\cos (y)+\csc (y)\right)}{\pi }\right) + n\cdot \pi \quad \text{for } n \in \mathbb Z. $$

Brilliant mathematical insights, or in my case, looking at the actual values, would yield that (now $n \in \mathbb R$):

• for $n \in [0,1]$: $[x_n(y),y] \notin S$
• for $n \in (1,2)$: $[x_n(y),y] \in S$
• for $n \in [3,4]$: $[x_n(y),y] \notin S$
• and so forth, and thus, unless I'm mistaken:

The solution set $S$ is given by $$ [x_n(y),y], \quad n \in (2m-1,2m), \quad m \in \mathbb Z. $$

A final note, the distinction between the open interval $(\cdot,\cdot)$ and the closed one $[\cdot,\cdot]$ stems from your strict $>$ relation (meaning that the separation line is not in $S$). If you want the solution for $\geqslant$, swap the $(\cdot,\cdot)$s and $[\cdot,\cdot]$s.