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Good Day, I was trying to solve the below problem:

Simplify $$\sum_{m=1}^{6} \frac{1}{(\sin k)(\sin k - \cos k)}$$ where $k = \theta + \frac{m \pi}{4} \text{and } 0 < \theta < \frac{\pi}{2}. $

I was thinking of decomposing the fraction somehow and getting a telescoping sum or something, but was unable to do so. I am absolutely clueless and there is nothing that I know that simplifies or progresses on the problem. The only thing that I think works is $$\frac{1}{(\sin k)(\sin k - \cos k)} = \frac{1}{\sin^2k-\sin k \cos k} = \frac{2}{1 - \cos 2k-\sin2k}$$ but again I've no idea how to proceed.

Any help would be appreciated. Thanks.

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  • $\begingroup$ $\sin(\theta+\pi/2)=\cos\theta$ and $\cos(\theta+\pi/2)=-\sin\theta$ so this becomes a case where you can easily write this in terms of just $\sin2\theta$ and $\cos2\theta,$ and the terms have a frequency of $4,$ so when $k=1$ and $k=5$ the terms are equal. When $k=2$ and $k=6$ the terms are equal. $\endgroup$ Aug 28, 2021 at 14:23
  • $\begingroup$ Given there are just $6$ terms and you have already written it as a function of $ \sin 2k$ and $\cos 2k$, instead of trying to find a telescoping sum, you should just convert all terms as a function of $\sin 2\theta$ and $\cos 2\theta$. Do it for odd $m$ and then for even $m$ as Thomas Andrews suggested. $\endgroup$
    – Math Lover
    Aug 28, 2021 at 14:38
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    $\begingroup$ Related math.stackexchange.com/questions/464031/… and math.stackexchange.com/questions/425966/… $\endgroup$ Aug 28, 2021 at 15:14

1 Answer 1

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$\displaystyle\sum_{m=1}^{6} \frac{1}{(\sin k)(\sin k - \cos k)}$

$\displaystyle\implies\frac{1}{\sqrt2}\sum_{m=1}^{6} \frac{1}{(\sin k)(\sin (k -\frac{\pi}{4})}$

$\displaystyle\implies\sum_{m=1}^{6} \frac{\sin\frac{\pi}{4}}{\sin k\sin (k -\frac{\pi}{4})}$

$\displaystyle\implies\sum_{m=1}^{6} \frac{\sin\bigg(k-(k-\frac{\pi}{4})\bigg)}{(\sin k)(\sin (k -\frac{\pi}{4})}$

$\displaystyle\implies\sum_{m=1}^{6} \frac{\sin\bigg(A-B\bigg)}{(\sin A)(\sin B)}$

$\displaystyle\implies\displaystyle\sum_{m=1}^{6}(\cot B-\cot A)$

$\displaystyle\implies\sum_{m=1}^{6}(\cot(k -\frac{\pi}{4}) -\cot k)$

$\displaystyle\implies\sum_{m=1}^{6}\bigg(\cot(\theta+(m-1)\frac{\pi}{4}) -\cot (\theta +m\frac{\pi}{4})\bigg)$

$\displaystyle\implies\sum_{m=1}^{6}g(m-1)-g(m)$ where $g(m)=\cot (\theta+\frac{m\pi}{4})$

Now it can be telescoped to $g(0)-g(6)$

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