# Leo Giugiuc's Trigonometric Lemma

Leo Giugiuc has posted the following statement and its proof at the CutTheKnotMath facebook page:

Let $x$ and $y$ be real numbers such that $0\lt x\lt y\lt\pi.$

- If $x+y\lt\pi$ then $\sin x\lt\sin y.$
- If $x+y\gt\pi$ then $\sin x\gt\sin y.$

### Proof 1 of 1.

In this case necessarily $\displaystyle x\lt\frac{\pi}{2}$ because, otherwise, $\displaystyle y\gt\frac{\pi}{2},$ contradicting the premise. By the hypothesis, $x\lt y$ and also $x\lt\pi -y.$ But, one of the two quantities - $y$ and $\pi -y$ - is bound to be in $\displaystyle \left[0,\frac{\pi}{2}\right]$ and, since *sine* is strictly increasing on $\displaystyle \left[0,\frac{\pi}{2}\right],$ either $\sin x\lt\sin y$ or $\sin x\lt\sin (\pi -y),$ which proves #1 because $\sin y=\sin (\pi -y).$

### Proof 2 of 1.

Consider $\Delta ABC,$ with angles $x$ at $A,$ $y$ at $B,$ and $\pi -x-y$ at $C.$ Note that $BC\lt AC,$ for $x\lt y.$ By the Law of Sines, $\displaystyle\frac{AC}{BC}=\frac{\sin x}{\sin y},$ implying $\sin x\gt\sin y.$

### Proof of 2.

In this case, $0\lt\pi y\lt\pi -x$ and $(\pi -y)+(\pi -x)\lt 2\pi -\pi =\pi.$ Hence, by the first part, with $\pi -y$ for $x$ and $\pi -x$ for $y,$ we conclude that $\sin (\pi -y)\lt\sin (\pi -x),$ or, equivalently, $\sin y\lt\sin x.$

|Contact| |Front page| |Contents| |Algebra|

Copyright © 1996-2018 Alexander Bogomolny