21th USA Mathematical Olympiad 1992 Problem

In the (previous blog post), we were asked to prove a 21th USA Mathematical Olympiad 1992 Problem:

Let $k=1^{\circ}$, show that $\displaystyle \sum_{n=0}^{88}\dfrac{1}{\cos (nk) \cos(n+1)k}=\dfrac{\cos k}{\sin^2 k}$

But we have already worked out an trigonometric identity where:

$\dfrac{\sin 1^{\circ}}{\sin x^{\circ} \sin (x^{\circ}+1^{\circ})}=\cot x^{\circ}-\cot (x^{\circ}+1^{\circ})$

Since

$\sin  x^{\circ}=\cos (90^{\circ}-x^{\circ})$ and $\sin  (x^{\circ}+1^{\circ})=\cos (90^{\circ}-(x^{\circ}+1^{\circ}))$

We can alter the way we have it at the LHS of the identity above in such a way that we obtain:

$\dfrac{\sin 1^{\circ}}{\cos (90^{\circ}-x^{\circ}) \cos (90^{\circ}-(x^{\circ}+1^{\circ}))}=\cot x^{\circ}-\cot (x^{\circ}+1^{\circ})$

Therefore, if we have $x=1^{\circ}$ all the way up to $x=89^{\circ}$, adding them all up we see that:

$\dfrac{\sin 1^{\circ}}{\cos (89^{\circ}) \cos (88^{\circ})}=\cot 1^{\circ}-\cot (2^{\circ})$

$\dfrac{\sin 1^{\circ}}{\cos (88^{\circ}) \cos (87^{\circ})}=\cot 2^{\circ}-\cot (3^{\circ})$

$\dfrac{\sin 1^{\circ}}{\cos (87^{\circ}) \cos (86^{\circ})}=\cot 3^{\circ}-\cot (4^{\circ})$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \vdots$

$\dfrac{\sin 1^{\circ}}{\cos (3^{\circ}) \cos (2^{\circ})}=\cot 87^{\circ}-\cot (88^{\circ})$

$\dfrac{\sin 1^{\circ}}{\cos (2^{\circ}) \cos (1^{\circ})}=\cot 88^{\circ}-\cot (89^{\circ})$

$\dfrac{\sin 1^{\circ}}{\cos (1^{\circ}) \cos (0^{\circ})}=\cot 89^{\circ}-\cot (90^{\circ})$

Therefore we get the addend as follows:

$\dfrac{\sin 1^{\circ}}{\cos (89^{\circ}) \cos (88^{\circ})}+\dfrac{\sin 1^{\circ}}{\cos (88^{\circ}) \cos (87^{\circ})}+\cdots+\dfrac{\sin 1^{\circ}}{\cos (1^{\circ}) \cos (0^{\circ})}$

$=\cot 1^{\circ}-\cot (2^{\circ})+\cot 2^{\circ}-\cot (3^{\circ})+\cdots+\cot 88^{\circ}-\cot (89^{\circ})+\cot 89^{\circ}-\cot (90^{\circ})$

$=\cot 1^{\circ}-\cot (90^{\circ})$

$=\cot 1^{\circ}$

$k=1^{\circ}$ yields:

$\displaystyle \therefore \sum_{n=0}^{88}\dfrac{\sin 1^{\circ}}{\cos (n^{\circ}) \cos(n^{\circ}+1^{\circ})}=\cot 1^{\circ}=\dfrac{\cos 1^{\circ}}{\sin 1^{\circ}}$

i.e.

$\displaystyle \sum_{n=0}^{88}\dfrac{1}{\cos (n^{\circ}) \cos(n^{\circ}+1^{\circ})}=\dfrac{\cos 1^{\circ} }{\sin^2 1^{\circ}}$

we're hence done.