$$ \theta \mapsto \frac{\sin \theta}{a + \cos \theta} \tag 1 $$

Everybody knows the right side of $(1)$ is equal to $\tan(\theta/2)$ if $a=1$ and to $\tan\theta$ if $a=0$. Obviously it's unbounded if $|a| \le 1$ and bounded otherwise. In the course of thinking about map projections I stumbled across a proof that this has maximum value $1$ and minimum $-1$ precisely if $|a|=\sqrt 2$. A consequence is that a function $\theta\mapsto\eta$ is implicitly defined by $$ \frac{\sin\theta}{a + \cos\theta} = \sin\eta \tag 2 $$ precisely if $|a|=\sqrt 2$.

My question is: Is this a standard result found in the literature somewhere --- something everyone except me knows (for appropriate values of "everybody", thus excluding those whose interests would not bring this to their attention if it's widely known)? Or to be more precise: Is this something standard that is in the literature somewhere? Are other things of interest known about this?

Appendix: One way of showing that The maximum and minimum are equal to $1$ and $-1$ respectively when $a=\sqrt 2$ is to observe that $$ \left( \frac{\sin\theta}{\sqrt2+\cos\theta} \right)^2 + \left( \frac{\sqrt 2\cos\theta + 1}{\sqrt 2 + \cos\theta} \right)^2 = 1 $$ and the value $1$ is attained when $\sin\theta = -\cos\theta>0$.


2 Answers 2


Think of the polar plot of

$$r(\theta)=\dfrac{\sin\theta}{a + \cos\theta} $$

you want to find the values of $a$ that keep this inside (or on) the unit circle.

setting $r'(\theta)=0$ gives you that $r$ has a maximum value at $\theta _{\max}$ where

$$ \cos(\theta _{\max})=-\dfrac 1a $$

So ...

$$ |r(\theta _{\max})|= \dfrac{ \sqrt{1-\left( -\dfrac 1a\right) ^2 } }{ \left | 1-\dfrac 1a \right | } = \dfrac 1{ \sqrt{a^2-1}} $$

So to keep the plot inside (or on) the unit circle you must have $a^2 \ge2$


  • $\begingroup$ This does not address the question, which was a reference request. $\qquad$ $\endgroup$ Oct 5, 2016 at 16:13

$$|\sin \eta| \le 1\\ \left|\frac {\sin\theta}{\alpha + \cos \theta}\right| \le 1$$

suppose $\alpha + \cos \theta\ge 0$

$$|\sin \theta| \le \alpha + \cos \theta\\ \sin \theta - \cos\theta \le \alpha $$


$-\sin \theta - \cos\theta \le \alpha$

in either case, how big does $\alpha$ need to be?

What is maximum of $\sin \theta - \cos\theta$

You could take a derivative and set it to 0. You can make a guess based on what you know about the symmetry of the functions.

I like this approach:

$$\sin \theta - \cos\theta = \sqrt 2 (\sin \frac {\pi}{4} \sin \theta - \cos \frac {\pi}{4} \cos \theta = -\sqrt 2 \cos (\theta + \frac {\pi}{4})$$

and the maximum is $\sqrt 2$

same deal for $-\sin \theta - \cos\theta$

$|\alpha| \ge \sqrt 2$

  • $\begingroup$ This does not address the question, which was a reference request. $\qquad$ $\endgroup$ Oct 5, 2016 at 16:13
  • $\begingroup$ I hadn't seen it before you posed the question. So, no, it is not something everyone knows, nor is it something that I have seen in the literature. $\endgroup$
    – Doug M
    Oct 5, 2016 at 16:23

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