The integer part of $\sum_{k=0}^{44}\frac{1}{\cos(k^\circ)\cos((k+1)^\circ)}$

Question

What is the integer part of the number $$\sum_{k=0}^{44}\frac{1}{\cos (k^\circ)\cos((k+1)^\circ)}$$

I tried to solve it using partial fractions but could not get a result. Please help me out.

Answer 1

Notice for any $a < b$, we have

$$\tan b - \tan a = \frac{\sin b\cos a - \sin a\cos b}{\cos b\cos a} = \frac{\sin(b-a)}{\cos b\cos a}$$ This implies $$\begin{align}\sum_{k=0}^{44} \frac{1}{\cos(k^\circ)\cos((k+1)^\circ)} &= \frac{1}{\sin 1^\circ}\sum_{k=0}^{44} \left(\tan((k+1)^\circ) - \tan(k^\circ)\right)\\ &= \frac{1}{\sin 1^\circ}\left(\tan(45^\circ) - \tan(0^\circ)\right)\\ &= \frac{1}{\sin\frac{\pi}{180}} \approx 57.29868849855018 \end{align}$$ The integer we seek is $57$.

Answer 2

Playing around.

Let $s(n) =\sum_{k=0}^{n}\frac{1}{\cos(k)\cos(k+1)} $.

Looking at the sum in Wolfy, it peaks at about 12 and has a period of about 22 with 7 peaks. The extreme values are about $\pm 10$. This is probably due to $\pi \approx \frac{22}{7}$.

In particular, $s(44) \approx 1.9$ so the integer part is 1.

Also,

$\begin{array}\\ \frac{1}{\cos(k)}-\frac1{\cos(k+1)} &=\frac{\cos(k+1)-\cos(k)}{\cos(k)\cos(k+1)}\\ &=-2\frac{\sin(k+\frac12)\sin(\frac12)}{\cos(k)\cos(k+1)}\\ \end{array} $

but this doesn't seem to help much.

Can't think of anything else.