# Chaitanya's Random Pages

## May 31, 2013

### ATP World Tour players of the year

Filed under: sport — ckrao @ 12:32 pm

Below is a list of ATP World Tour tennis players of the year dating back to 1989. In most cases their ranking had suffered the previous year due to injury. Information here is from Wikipedia and atpworldtour.com. There are several big names featuring here. I was led to this list after thinking about how well Tommy Haas has been doing in recent times after all the injuries he has been through.

 Year Player Ranking at end of year Ranking at end of previous year Notes 1989 Goran Prpić 28 226 1990 Thomas Muster 7 21 struck by a car in 1989 1991 Jimmy Connors 49 936 reached SF of US Open at age 39 1992 Henri Leconte 61 159 reached SF in French Open 1993 Mikael Pernfors 32 237 1994 Guy Forget 40 158 1995 Derrick Rostagno 110 no ranking 1996 Stephane Simian 92 333 1997 Sergi Bruguera 8 82 reached final of French Open 1998 Younes El Aynaoui 45 237 1999 Chris Woodruff 51 1342 2000 Sergi Bruguera (2) 85 378 2001 Guillermo Cañas 15 231 2002 Richard Krajicek 112 no ranking 2003 Mark Philippoussis 9 80 reached Wimbledon final 2004 Tommy Haas 17 no ranking 2005 James Blake 23 97 had broken neck, shingles 2006 Mardy Fish 47 225 2007 Igor Andreev 33 91 2008 Rainer Schüttler 33 99 2009 Marco Chiudinelli 56 779 2010 Robin Haase 65 451 2011 Juan Martín del Potro 11 258 2012 Tommy Haas (2) 21 205

## May 25, 2013

### Sine of half angles

Filed under: mathematics — ckrao @ 8:34 am

I found the following cool-looking formulas in [1].

$\displaystyle \sin \frac{45^{\circ}}{2^n} = \frac{\sqrt{2 - \sqrt{2 + \sqrt{2 + \ldots + \sqrt{2}}}}}{2}$
(n + 1 nested roots)

$\displaystyle \sin \frac{15^{\circ}}{2^n} = \frac{\sqrt{2 - \sqrt{2 + \sqrt{2 + \ldots + \sqrt{3}}}}}{2}$
(n + 2 nested roots)

$\displaystyle \sin \frac{18^{\circ}}{2^n} = \frac{\sqrt{8 - 2\sqrt{8 + 2\sqrt{8 + \ldots + 2\sqrt{10 + 2\sqrt{5}}}}}}{4}$
(n + 2 nested roots)

To prove them, we use the following form double angle formula for cosine.

$\displaystyle 4 \sin^2 x = 2(1 - \cos 2x) = 2 - \sqrt{4 - 4 \sin^2 2x }\quad \quad (1)$

Now we evaluate this for different values of $x$. Firstly $\sin^2 45^{\circ} = 1/2$ implying from (1) that

$4 \sin^2 22.5^{\circ} = 2 - \sqrt{2}.\quad \quad \quad \quad (2)$

Secondly,

\begin{aligned}4 \sin^2 15^{\circ} &= 2(1 - \cos 30^{\circ})\\ &= 2 - \sqrt{3},\end{aligned}

so that from (1),

$\displaystyle 4 \sin^2 7.5^{\circ} = 2 - \sqrt{4 - (2 - \sqrt{3}} = 2 - \sqrt{2 + \sqrt{3}}. \quad \quad (3)$

To find $\sin 18^{\circ}$ one could apply Ptolemy’s theorem to four points of a regular pentagon or alternatively set $y = 36^{\circ}$ and write

\begin{aligned} \cos 3y &= -\cos (180^{\circ} - 3y)\\&= - \cos 2y.\end{aligned}

Using the double and triple angle formulae for cosine this becomes

\begin{aligned} 4 \cos^3 y - 3\cos y = 1 - 2\cos^2 y \\ \hbox{i.e.}\ 4\cos^3 y + 2\cos^2 y - 3\cos y - 1 &= 0\\ \hbox{i.e.} \ (\cos y + 1) (4 \cos^2 y - 2 \cos y - 1) &= 0.\end{aligned}

From this the only valid (positive) solution is $\cos y = (2 + 2\sqrt{5})/8$, so from (1)

$16 \sin^2 18^{\circ} = 8(1 - \cos 36^{\circ}) = 6 - 2\sqrt{5}.$

Hence applying (1) again,

$\displaystyle 16 \sin^2 9^{\circ} = 2(4 - \sqrt{16 - 16 \sin^2 18^{\circ} }) = 8 - 2\sqrt{10 + 2\sqrt{5}}.\quad \quad (4)$

Now we set up a hypothesis for mathematical induction. Assume that

$\displaystyle 4 \sin^2 \frac{45^{\circ}}{2^k} = 2 - \underbrace{\sqrt{2 + \sqrt{2 + \ldots + \sqrt{2}}}}_{\hbox{k nested roots}}$
$\displaystyle 4 \sin^2 \frac{15^{\circ}}{2^k} = 2 - \underbrace{\sqrt{2 + \sqrt{2 + \ldots + \sqrt{3}}}}_{\hbox{k+1 nested roots}}$
$\displaystyle 16 \sin^2 \frac{18^{\circ}}{2^k} = 8 - 2\underbrace{\sqrt{8 + 2\sqrt{8 + \ldots + 2\sqrt{10 + 2\sqrt{5}}}}}_{\hbox{k+1 nested roots}}$

Note that these are equivalent to our initial three formulas. In (2)-(4) we have seen that these results are true for $k = 1$. Then using (1) and the inductive hypothesis,

\begin{aligned} 4 \sin^2 \frac{45^{\circ}}{2^{k+1}} &= 2 - \sqrt{4 - \sin^2 \frac{45^{\circ}}{2^{k}}}\\&= 2 - \sqrt{4 - \left(2 - \underbrace{ \sqrt{2 + \sqrt{2 + \ldots + \sqrt{2}}}}_{\hbox{k nested roots}}\right)} \\ &= 2 - \underbrace{\sqrt{2 + \sqrt{2 + \ldots + \sqrt{2}}}}_{\hbox{k+1 nested roots}}\end{aligned}

Similarly,

\begin{aligned} 4 \sin^2 \frac{15^{\circ}}{2^{k+1}} &= 2 - \sqrt{4 - \sin^2 \frac{15^{\circ}}{2^{k}}}\\&= 2 - \sqrt{4 - \left(2 - \underbrace{ \sqrt{2 + \sqrt{2 + \ldots + \sqrt{3}}}}_{\hbox{k+1 nested roots}}\right)} \\ &= 2 - \underbrace{\sqrt{2 + \sqrt{2 + \ldots + \sqrt{3}}}}_{\hbox{k+2 nested roots}}\end{aligned}

and

\begin{aligned} 16 \sin^2 \frac{18^{\circ}}{2^{k+1}} &= 8 - 2\sqrt{16 - 16 \sin^2 \frac{18^{\circ}}{2^{k}}}\\&= 8 - 2\sqrt{16 - \left(8 - 2\underbrace{ \sqrt{8 + 2\sqrt{8 + \ldots + 2\sqrt{10 + 2\sqrt{5}}}}}_{\hbox{k+1 nested roots}}\right)} \\ &= 8 - 2\underbrace{\sqrt{8 + 2\sqrt{8 + \ldots + 2\sqrt{10 + 2\sqrt{5}}}}}_{\hbox{k+2 nested roots}}.\end{aligned}

In each case we have shown that if the result is true for $n = k$, it is true for $n = k+1$. By the principle of mathematical induction, the result is true for $n = 1,2, \ldots$ and we are done.

#### Reference

[1] E. Maor, Trigonometric Delights, Princeton University Press, 1998.

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