Can someone confirm my method and answer for this trig problem? Find the exact value of:
$$\cos 1^{\circ}+\cos 2^{\circ}+\cos 3^{\circ}+ \ldots +\cos 358^{\circ}+\cos 359^{\circ}$$
I got $0$ as I did this by assigning either a positive or negative $x$ variable for each quadrant. Is this method valid and my answer right?
 A: The answer is $-1$.
My solution uses vector addition (equivalently, complex number addition).
Think of $\cos n^\circ$ as the $x$-coordinate of the point $(\cos n^\circ,\sin n^\circ)$, the point $n$ degrees along the circle. The sum:
$$\sum_{n=0}^{359}(\cos n^\circ,\sin n^\circ)$$
is, therefore, the sum of the vertices of a regular $360$-gon centered at the origin. (Notice that it begins at $n=0$, not $n=1$.)
This sum doesn't change when you rotate it by $1^\circ$ around the origin, since rotating a $360$-gon by $1^\circ$ doesn't change it, and since rotation around the origin is a linear map (the rotation of the sum is the sum of the rotations). The only vector that doesn't change when you rotate it around the origin is the origin itself, $(0,0)$. Therefore, the above sum equals $(0,0)$.
Looking at the $x$-coordinate, we get:
\begin{align}
0&=\cos0^\circ+\cos1^\circ+\dotsb+\cos359^\circ\\
-\cos0^\circ&= \phantom{\cos0^\circ+{}\!}\cos1^\circ+\dotsb+\cos359^\circ\\
-1&= \phantom{\cos0^\circ+{}\!}\cos1^\circ+\dotsb+\cos359^\circ\\
\end{align}
A: You can use Lagrange's trigonometric identity $$\sum_{n=1}^{N}\cos(n\theta)=-\frac{1}{2}+\frac{\sin((N+\frac{1}{2})\theta)}{2\sin(\frac{\theta}{2})}$$
Substituting $N=359$ and $\theta=\frac{\pi}{180}$, you get $$-\frac{1}{2}+\frac{\sin(360-\frac{1}{2})\frac{\pi}{180}}{2\sin(\frac{1}{2}\frac{\pi}{180})}=-\frac{1}{2}+\frac{\sin(2\pi-\frac{\pi}{360})}{2\sin(\frac{\pi}{360})}=-1$$
since $\sin(2\pi-\theta)=-\sin(\theta)$.
A: Almost. Observe that every term except for $\cos(180^{\circ})$ will cancel out.
A: Almost. Just write
$$\sum_{k=1}^{359}\cos\left ( \frac{2k\pi}{360}\right)=\cos (\pi)+\sum_{k=1}^{179}\cos\left ( \frac{2k\pi}{360}\right)+\sum_{k=181}^{359}\cos\left ( \frac{2k\pi}{360}\right) $$
Grouping in Gauss Sums yields
$$\sum_{k=1}^{179}\cos\left ( \frac{2k\pi}{360}\right)=\sum_{k=1}^{89}\cos \left (\frac{2k\pi}{360}\right)+ \cos\left (\pi-\frac{2k\pi}{360}\right)$$
$$\sum_{k=181}^{359}\cos\left ( \frac{2k\pi}{360}\right)=\sum_{k=1}^{89}\cos \left (\pi+\frac{2k\pi}{360}\right)+ \cos\left (2\pi-\frac{2k\pi}{360}\right)$$
Each term cancels out because 
$$\cos(\pi-\alpha)=-\cos (\alpha)$$
$$\cos(\pi+\alpha)=-\cos(\alpha)=-\cos(2\pi-\alpha)$$
A proof without words would go like this:

A: $$\cos 1^{\circ}+\cos 2^{\circ}+\cos 3^{\circ}+ \ldots +\cos 358^{\circ}+\cos 359^{\circ}=$$
$$\tfrac{2\sin0.5^{\circ}\cos 1^{\circ}+2\sin0.5^{\circ}\cos 2^{\circ}+2\sin0.5^{\circ}\cos 3^{\circ}+ \ldots +2\sin0.5^{\circ}\cos 358^{\circ}+2\sin0.5^{\circ}\cos 359^{\circ}}{2\sin0.5^{\circ}}$$
and you'll get it
