Sum 2.

$\begin{aligned} S & = \sum_{k=1}^{359} k \cos k^\circ & \small \color{#3D99F6}{\text{Using }\sum_{k=a}^b f(k) = \sum_{k=a}^b f(a+b-k)} \\ & = \color{#3D99F6}{\frac 12} \sum_{k=1}^{359} \left(k \cos k^\circ + \color{#3D99F6}{(360-k)} \cos \color{#3D99F6}{(360-k)} ^\circ \right) & \small \color{#3D99F6}{\because \cos (360-k)^\circ = \cos (-k^\circ) = \cos k^\circ} \\ & = \frac 12 \sum_{k=1}^{359} \left(k \cos k^\circ + (360-k) \color{#3D99F6}{\cos k^\circ} \right) \\ & = \frac 12 \sum_{k=1}^{359} 360 \cos k^\circ \\ & = 180 \left(\color{#3D99F6}{\sum_{k=1}^{179} \cos k^\circ} + \cos 180^\circ + \color{#3D99F6}{\sum_{k=181}^{359} \cos k^\circ} \right) & \small \color{#3D99F6}{\because \sum_{k=a}^{180-a} \cos k^\circ = 0 \text{, see Note.}} \\ & = 180 \left(\color{#3D99F6}{0} -1 + \color{#3D99F6}{0} \right) \\ & = \boxed{-180} \end{aligned}$

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$\color{#3D99F6}{\text{Note:}}$

We know that $\cos k^\circ = - \cos (180-k)^\circ$ and $\cos 90^\circ = 0$ , $\implies \displaystyle \sum_{k=a}^{180-a} \cos k^\circ = 0$ , $\implies \displaystyle \sum_{k=1}^{179} \cos k^\circ = 0$ , similarly $\displaystyle \sum_{k=181}^{359} \cos k^\circ = 0$ .

We have, $\sum_{k=1}^{359} k \cos(k^\circ) = \sum_{k=1}^{359}(360-k)\cos(k^\circ)\stackrel{(a)}{=} \frac{360}{2} \sum_{k=1}^{359} \cos(k^\circ)\stackrel{(b)}{=}-180$ where (a) follows by taking average of the previous two equal terms and (b) follows from the fact that $\sum_{k=1}^{360} \cos(k^\circ)=0$ .

We use the fact $\cos{(360 -x)}= \cos x$

Let $S =\Large \sum_{k=1}^{359} k\cos k^{\circ} = \ ?$

$S = 1\cos1 +2\cos2 +3\cos3 ........+359\cos359 +180\cos180$

$S = 360(\cos1 +\cos2 +\cos3 ......+\cos179) + 180\cos180$

We can prove

$\cos1 +\cos2 +\cos3 ......+\cos179 = 0$ Using $\cos{180 -\theta} = -\cos{\theta}$

So $S = 180\cos180 = -180$