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:

telescoping tangents

Solution

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Copyright © 1996-2017 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:

$\displaystyle\begin{align}S(n)&=\sum_{k=1}^n\frac{\sin 1}{\cos (k-1)\cos k}\\ &=\sum_{k=1}^n\frac{\sin (k-(k-1))}{\cos (k-1)\cos k}\\ &=\sum_{k=1}^n\frac{\sin k\cos(k-1)-\sin(k-1)\cos k}{\cos (k-1)\cos k}\\ &=\sum_{k=1}^n\left(\tan k - \tan(k-1)\right)\\ &=\tan n - \tan 0\\ &=\tan n. \end{align}$

Related material
Read more...

Telescoping situations

  • Leibniz and Pascal Triangles
  • Infinite Sums and Products
  • Sum of an infinite series
  • Harmonic Series And Its Parts
  • A Telescoping Series
  • An Inequality With an Infinite Series
  • That Divergent Harmonic Series
  • An Elementary Proof for Euler's Series
  • $\sin 1^{\circ}+\sin {2^\circ}+\sin 3^{\circ}+\cdots+\sin 180^{\circ}=\tan 89.5^{\circ}$
  • Problem 3824 from Crux Mathematicorum
  • $x_n=\sin 1+\sin 3+\sin 5+\cdots+\sin (2n-1)$
  • Dan Sitaru's Sum of a Series
  • A Property of Product of Special Matrices, Probably Folklore
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    Copyright © 1996-2017 Alexander Bogomolny

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