Cosine of Triangle numbers

Let $T_n = \frac{n(n+1)}2.$ Then $\begin{aligned} \cos(T_{359-n}) &= \cos\left( \frac{(359-n)(360-n)}2 \right) \\ &= \cos\left( 180 \cdot 359 - \frac{719}2 n + \frac{n^2}2 \right) \\ &= \cos\left(180 \cdot 359 -360n + \frac{n^2+n}2 \right) \\ &= \cos\left(180+\frac{n^2+n}2\right) \\ &= -\cos\left(\frac{n^2+n}2 \right) = -\cos(T_n). \end{aligned}$ So all but the last two terms in the sum pair off and sum to zero, e.g. $\cos(T_1) + \cos(T_{358}) = 0, \cos(T_2) + \cos(T_{357}) = 0,$ etc.

So the sum equals $\cos(T_{359}) + \cos(T_{360}) = \cos(180 \cdot 359) + \cos(180 \cdot 361) = -1 + -1 = \fbox{-2}.$

Let $S$ be the given sum. Then

$\begin{aligned} S & = \sum_{n=1}^{360} \cos \left(\frac {n(n+1)}2^\circ\right) \quad \quad \small \blue{\text{By reflection } \sum_{k=a}^b f(k) = \sum_{k=a}^b f(a+b-k)} \\ & = \sum_{n=1}^{360} \cos \left(\frac {(360-(n-1))(360-(n-2))}2^\circ\right) \quad \quad \small \blue{\text{Since }\cos \theta = \cos (\theta \bmod 360^\circ)} \\ & = \sum_{n=1}^{360} \cos \left(-(2n-3)180^\circ+\frac {(n-1)(n-2)}2^\circ\right) \quad \quad \small \blue{\text{and }2n-3 \text{ is odd}} \\ & = \sum_{n=1}^{360} \cos \left(-180^\circ+\frac {(n-1)(n-2)}2^\circ\right) \quad \quad \small \blue{\text{and }\cos(-\theta) = \cos \theta} \\ & = \sum_{n=1}^{360} \cos \left(180^\circ-\frac {(n-1)(n-2)}2^\circ\right) \quad \quad \small \blue{\text{and }\cos(180^\circ-\theta) = -\cos \theta} \\ & = - \sum_{n=1}^{360} \cos \left(\frac {(n-1)(n-2)}2^\circ\right) \\ & = - \underbrace{\cos 0^\circ}_{\blue{n=1}} - \underbrace{\cos 0^\circ}_{\blue{n=2}} - \sum_{n=3}^{360} \cos \left(\frac {(n-1)(n-2)}2^\circ\right) \\ & = - 1 - 1 - \sum_{n=1}^{358} \cos \left(\frac {n(n+1)}2^\circ\right) \\ & = - 1 - 1 - \blue{\sum_{n=1}^{360} \cos \left(\frac {n(n+1)}2^\circ\right)} + \cos \left(\frac {359(360)}2\right) + \cos \left(\frac {360(361)}2\right) \\ & = -1 - 1 - \blue S - 1 -1 \\ \implies 2S & = - 4 \\ S & = \boxed{-2} \end{aligned}$