Prove that $\cos (\theta_1 + \theta_2) = \cos \theta_1 \cos \theta_2 - \sin \theta_1 \sin \theta_2$ by considering the dot product of the unit vectors $\mathbf{v_1}$ and $\mathbf{v_2}$. These vectors are at the angles $\theta_1$ above the $x$-axis and $\theta_2$ below the $x$-axis, respectively.
Solution
As directed, we try to relate $\cos (\theta_1 + \theta_2)$ to the dot product $\mathbf{v_1} \cdot \mathbf{v_2}$.
To apply this formula, we note $|\bfv_1|=1$ and $|\bfv_2|=1$ because the vectors have unit length. The angle between the vectors is $\theta_1 + \theta_2$.
To compute $\bfv_1 \cdot \bfv_2$, we need to find the components of each vector.
The unit vector at angle $\theta_1$ above the $x$-axis is given by $$\mathbf{v_1} = \cos\theta_1 \ \mathbf{i} + \sin\theta_1 \ \mathbf{j}$$
The unit vector at angle $\theta_2$ below the $x$-axis is the same as the unit vector at angle $-\theta_2$ above the $x$-axis: $$\mathbf{v_2} = \cos( -\theta_2) \ \mathbf{i} + \sin( - \theta_2) \ \mathbf{j} = \cos\theta_2 \ \mathbf{i} - \sin \theta_2 \ \mathbf{j}$$
We now compute $\mathbf{v_1} \cdot \mathbf{v_2} = \cos \theta_1 \cos \theta_2 - \sin \theta_1 \sin \theta_2$.
