# Thread: Minimum value of N

1. ## Minimum value of N

$\frac{1}{sin45sin46}$ $+\frac{1}{sin47sin48}+$.................... $\frac{1}{sin133sin134}=$ $\frac{1}{sinN}$

2. Hello banku12
Originally Posted by banku12
$\frac{1}{sin45sin46}$ $+\frac{1}{sin47sin48}+$.................... $\frac{1}{sin133sin134}=$ $\frac{1}{sinN}$
This is a toughie. I found that $N = 1$ quite early (using a spreadsheet) but proving it was tricky.

First, note that:
$\sin A = \cos(90^o-A)$
$=\cos(A-90^o)$ ...(1)
Then, note that it's quite easy to prove that:
$\tan A - \tan B = \frac{\sin(A-B)}{\cos A \cos B}$ ...(2)
Then (angles in degrees):
$\frac{1}{\sin 45\sin 46} + \frac{1}{\sin 47\sin 48}+$ ... $+\frac{1}{\sin 89\sin 90}+\frac{1}{\sin 91\sin 92}+\frac{1}{\sin 93\sin 94}+...+\frac{1}{\sin 133\sin 134}$
$=\frac{1}{\cos 45 \cos 44}+\frac{1}{\cos 43 \cos 42}+$ ... $+\frac{1}{\cos 1 \cos 0}+\frac{1}{\cos 1 \cos 2}+\frac{1}{\cos 3 \cos 4}+...+\frac{1}{\cos 43 \cos 44}$, using both parts of (1)

$=\frac{1}{\cos 0 \cos 1}+\frac{1}{\cos 1 \cos 2}+...+\frac{1}{\cos 44 \cos 45}$, by re-arrangement

$= \frac{\tan1-\tan0}{\sin(1-0)}+\frac{\tan 2 - \tan 1}{\sin(2-1)}+...+\frac{\tan45 - \tan44}{\sin(45-44)}$, using (2) (That's the clever bit!)

$=\frac{\tan1-\tan0+\tan2-\tan1+...+\tan45-\tan44}{\sin 1}$

$=\frac{\tan45-\tan0}{\sin 1}$

$=\frac{1}{\sin1}$

$\Rightarrow N = 1$