Can someone confirm my method and answer for this trig problem?

Almost. Observe that every term except for $\cos(180^{\circ})$ will cancel out.

The answer is $-1$.

My solution uses vector addition (equivalently, complex number addition).

Think of $\cos n^\circ$ as the $x$-coordinate 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}

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 $\sin(2\pi-\theta)=-\sin(\theta)$.

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