You could prove this algebraically by vector algebra.
Assume there exist two vectors $\vec u$ and $\vec v$ in $\mathbb{R}^2$ which make angles $\theta$ and $\phi$ with the positive $x$-axis, such that $\theta>\phi$. Their components would be $|\vec u|\cos\theta$, $|\vec v|\cos\phi$ (along the $x$-axis) and $|\vec u|\sin\theta$, $|\vec v|\sin\phi$ (along the $y$-axis). We know they would make an angle $\theta-\phi$ with each-other. Now, if we take their dot product,
$$\vec u \cdot \vec v=|\vec u||\vec v|\cos (\theta-\phi)$$
$$\vec u \cdot \vec v=(|\vec u|\cos\theta\times|\vec v|\cos\phi)+(|\vec u|\sin\theta\times|\vec v|\sin\phi)$$
Equating both equations for the dot product,
$$|\vec u||\vec v|\cos\theta\cos\phi+|\vec u||\vec v|\sin\theta\sin\phi=|\vec u||\vec v|\cos (\theta-\phi)$$
Dividing both sides by $|\vec u||\vec v|$,
$$\cos\theta\cos\phi+\sin\theta\sin\phi=\cos (\theta-\phi)$$
If the vectors were on opposite sides of the $x$-axis, their angle would be $\theta+\phi$ and the $y$-component of the vector below the $x$-axis would be negative.
With this information, on conducting the same steps as we did before, we obtain:
$$\cos\theta\cos\phi-\sin\theta\sin\phi=\cos (\theta+\phi)$$
Now, if we took the cross product of the initial vectors $|\vec u|$ and $|\vec v|$ instead,
$$\vec u \times \vec v=\begin{bmatrix}
\hat i & \hat j & \hat k \\
|\vec u|\cos\theta & |\vec u|\sin\theta & 0\\
|\vec v|\cos\phi & |\vec v|\sin\phi & 0\\
\end{bmatrix}=|\vec u||\vec v|\cos\theta\sin\phi-|\vec u||\vec v|\sin\theta\cos\phi$$
$$|\vec u \times \vec v|=|\vec u||\vec v|\sin (\theta-\phi)$$
Remember how we said $\theta>\phi$? This indicates that the cross product will produce a vector pointing into the plane; as proven by the Right Hand Screw Rule. Hence we multiply this magnitude by $(-1)$ as it points in the negative $z$-direction.
$$|\vec u \times \vec v|=-|\vec u||\vec v|\sin (\theta-\phi)$$
Equating this with the equation we derived from the cross-matrix,
$$|\vec u||\vec v|\cos\theta\sin\phi-|\vec u||\vec v|\sin\theta\cos\phi=-|\vec u||\vec v|\sin (\theta-\phi)$$
Dividing both sides by $-|\vec u||\vec v|$,
$$\sin\theta\cos\phi-\cos\theta\sin\phi=\sin (\theta-\phi)$$
If the vectors were on opposite sides of the $x$-axis, their angle would be $\theta+\phi$ and the $y$-component of the vector below the $x$-axis would be negative.
With this information, on conducting the same steps as we did before, we obtain:
$$\sin\theta\cos\phi+\cos\theta\sin\phi=\sin (\theta+\phi)$$
And there, we proved all four equations using vector algebra:
$$\sin\theta\cos\phi+\cos\theta\sin\phi=\sin (\theta+\phi)$$
$$\sin\theta\cos\phi-\cos\theta\sin\phi=\sin (\theta-\phi)$$
$$\cos\theta\cos\phi-\sin\theta\sin\phi=\cos (\theta+\phi)$$
$$\cos\theta\cos\phi+\sin\theta\sin\phi=\cos (\theta-\phi)$$
