Telescoping Tangents
Here's a problem I have borrowed from Imad Zak facebook group. The problem is by Mikolaj Hajduk and the solution is by Marian Dinca. The problem is simple but I am glad to add another example to the telescoping collection:
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Copyright © 1996-2018 Alexander Bogomolny
For $n\gt 1,\,$ find the value $S(n),$
$\displaystyle S(n) =\sum_{k=1}^n\frac{\sin 1}{\cos (k-1)\cos k}.$
We invoke the formula for the sine of a sum:
$\displaystyle\begin{align}S(n)&=\sum_{k=1}^n\frac{\sin 1}{\cos (k-1)\cos k}\\ &=\sum_{k=1}^n\frac{\sin (k-(k-1))}{\cos (k-1)\cos k}\\ &=\sum_{k=1}^n\frac{\sin k\cos(k-1)-\sin(k-1)\cos k}{\cos (k-1)\cos k}\\ &=\sum_{k=1}^n\left(\tan k - \tan(k-1)\right)\\ &=\tan n - \tan 0\\ &=\tan n. \end{align}$
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Copyright © 1996-2018 Alexander Bogomolny
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