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Thread: Minimum value of N

  1. #1
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    Minimum value of N

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

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

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


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


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


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


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

    $\displaystyle \Rightarrow N = 1$
    Grandad
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