1. **Problem Statement:** Find the sum of the series $\cos 1^\circ + \cos 2^\circ + \cos 3^\circ + \cdots + \cos 180^\circ$.\n\n2. **Formula Used:** The sum of cosines of an arithmetic progression is given by:\n$$\sum_{k=1}^n \cos(k\theta) = \frac{\sin\left(\frac{n\theta}{2}\right) \cos\left(\frac{(n+1)\theta}{2}\right)}{\sin\left(\frac{\theta}{2}\right)}$$\nwhere $n$ is the number of terms and $\theta$ is the common difference in degrees.\n\n3. **Apply the formula:** Here, $n=180$ and $\theta=1^\circ$. Substitute these values:\n$$\sum_{k=1}^{180} \cos(k^\circ) = \frac{\sin\left(\frac{180 \times 1^\circ}{2}\right) \cos\left(\frac{(180+1) \times 1^\circ}{2}\right)}{\sin\left(\frac{1^\circ}{2}\right)} = \frac{\sin(90^\circ) \cos(90.5^\circ)}{\sin(0.5^\circ)}$$\n\n4. **Evaluate the trigonometric values:**\n- $\sin(90^\circ) = 1$\n- $\cos(90.5^\circ) = \cos(90^\circ + 0.5^\circ) = -\sin(0.5^\circ)$ (since $\cos(90^\circ + x) = -\sin x$)\n- $\sin(0.5^\circ)$ remains as is.\n\n5. **Simplify the expression:**\n$$\sum_{k=1}^{180} \cos(k^\circ) = \frac{1 \times (-\sin(0.5^\circ))}{\sin(0.5^\circ)} = -1$$\n\n**Final answer:**\n$$\boxed{-1}$$