$\sin1^\circ+\sin2^\circ+\sin3^\circ+\cdots+\sin180^\circ=\tan89.5^\circ.$

Grégoire Nicollier
University of Applied Sciences of Western Switzerland
Switzerland
February 3, 2016

telescoping sine

Proof 1

$\delta=\pi/180\;$ and $e^{\pm i\cdot180\delta}=-1\;$ give

$\displaystyle\begin{align*} \text{LHS}&=\sin0^\circ+\sin1^\circ+\sin2^\circ+\cdots+\sin179^\circ\\ &=\frac1{2i}\sum_{k=0}^{179}\left(e^{ik\delta}-e^{-ik\delta}\right)\\ &=\frac1{2i}\left(\frac{1-e^{ i\cdot180\delta}}{1-e^{i\delta}}-\frac{1-e^{- i\cdot180\delta}}{1-e^{-i\delta}}\right)\\ &=-i\left(\frac1{1-e^{i\delta}}+\frac{e^{i\delta}}{1-e^{i\delta}}\right)\\ &=-i\frac{e^{-i\delta/2}+e^{i\delta/2}}{e^{-i\delta/2}-e^{i\delta/2}}\\ &=\cot0.5^\circ\\ &=\text{RHS}. \end{align*}$

Proof 2

For a proof without complex numbers use

$2\sin\alpha\sin\beta=\cos(\alpha-\beta)-\cos(\alpha+\beta)$

to obtain the telescoping sum

$2\,\text{LHS}\cdot\sin0.5^\circ=\cos0.5^\circ-\cos180.5^\circ=2\cos0.5^\circ=2\,\text{RHS}\cdot\sin0.5^\circ.$

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
  • $x_n=\sin 1+\sin 3+\sin 5+\cdots+\sin (2n-1)$
  • |Contact| |Front page| |Contents| |Algebra| |Store|

    Copyright © 1996-2017 Alexander Bogomolny

     62015530

    Search by google: