# On equivalences of trigonometric inequalities

This question is in error. (The source had a typo. See this answer.)

Let $$a$$ be a real positive number. \begin{align} \text{(I)}& & a &> \frac{\sin(y_1(a))}{y_1(a)} & &\text{where y_1(a) is the unique root of}& y &= a \cot(y) ~~\text{in}~~ (0,\frac{\pi}{2}) \\ \text{(II)}& & a &>\xi & &\text{where ~~~\xi~\,~~ is the unique root of}& \xi^2 &= \cos(\xi) \quad\text{in}~~ (0,\frac{\pi}{2}) \end{align}

I have read that the two statements are equivalent, i.e. (I)$$\iff$$(II). Does anyone have any hint how to prove this equivalence?

• Is there a typo? I think for inequality (I) the defining equation for $y_1(a)$ should be $y = a \tan y$. That way things work out trivially. – Lee David Chung Lin Apr 21 at 11:55

Let $$a$$ be a real positive number. \begin{align} \text{(I)}& & a &> \frac{\sin(y_1(a))}{y_1(a)} & &\text{where y_1(a) is the unique root of}& y &= a \cot(y) ~~\text{in}~~ (0,\frac{\pi}{2}) \\ \text{(II)}& & a &>\frac{\sin\xi}{\xi} & &\text{where ~~~\xi~\,~~ is the unique root of}& \xi^2 &= \cos(\xi) \quad\text{in}~~ (0,\frac{\pi}{2}) \end{align}