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

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

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

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

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