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Proving that $ \frac{1}{\sin(45°)\sin(46°)}+\frac{1}{\sin(47°)\sin(48°)}+...+\frac{1}{\sin(133°)\sin(134°)}=\frac{1}{\sin(1°)}$

Find the smallest postive integer n such that :

$\dfrac{1}{\sin 45^\circ \sin 46^\circ} + \dfrac1{\sin 47^\circ \sin 48^\circ} + \dots + \dfrac1{\sin 133^\circ \sin 134^\circ} = \dfrac1{\sin n^\circ}$

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John Chang
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  • wolfram alpha gives n $\approx$ -0.408264°, which makes me wonder if there is a integer value of n. I used arcsin( 1/( sum( 1/ ( sin(2*i+1) * sin(2*i+2))) for i = 22 to 66 ))*180/pi to evaluate n. – mythealias Nov 12 '12 at 07:59
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    @mythealias: note that the 2*i+1 are in degrees, not radians... – Robert Israel Nov 12 '12 at 08:10
  • based on Robert's comment if i convert to degree, i get $n = 1$. earlier i had tried $sin(23)$ and wolfram alpha used degrees so i though it takes arguments in degree. n = arcsin( 1/( sum( 1/ ( sin(pi/180*(2*i+1)) * sin(pi/180*(2*i+2)))) for i = 22 to 66 ))*180/pi – mythealias Nov 12 '12 at 08:16
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    Is this from an old contest, a homework, etc.? – Mark S. Nov 12 '12 at 18:41
  • The same question (with the $n=1^\circ$ answer) appeared in december 2011 here. – Raymond Manzoni Nov 13 '12 at 10:01

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