I don't think it is a very good idea to try an find a geometric picture for every trig identity because that would take exceedingly long. I will show you that your formula comes from the idea that a rotation by $n\theta$ is equal to $n$ rotations by $\theta$.
Denote $\langle a,b\rangle$ as the vector to the ordered pair $(a,b)$. We will discuss rotations of these vectors. First, consider that the rotation of the vector $\langle 1,0\rangle$ by an angle of $\theta$ is given by $\langle \cos\theta,\sin\theta\rangle$. Similarly, rotation of the vector $\langle 1,0\rangle$ by $3\theta$ is given by $\langle \cos(3\theta),\sin(3\theta)\rangle$.
I will now define a new type of multiplication between vectors. Let $$
\langle a,b\rangle\ltimes\langle c,d \rangle = \langle ac-bd,ac+bc\rangle$$
It so happens that if the vector $\langle a,b\rangle$ makes an angle $\phi_1$ with the horizontal and $\langle c,d \rangle$ makes an angle $\phi_2$ then $\langle a,b\rangle\ltimes\langle c,d \rangle$ makes an angle $\phi_1 + \phi_2\!$. Using this, there is another way to find the rotation of a vector by $3\theta$. We can rotate it three times by $\theta$ rather than rotating it once by $3\theta$. In other words, we will find $\left(\langle \cos\theta,\sin\theta \rangle \ltimes \cos\theta,\sin\theta \rangle\right) \ltimes \cos\theta,\sin\theta \rangle.$ Using this, one could suspect that
$$\begin{align}\langle \cos(3\theta),\sin(3\theta)\rangle &= \left(\langle \cos\theta,\sin\theta \rangle \ltimes \cos\theta,\sin\theta \rangle\right) \ltimes \cos\theta,\sin\theta \rangle \\
&= \langle \cos^2\!\theta-\sin^2\!\theta,2\sin\theta\cos\theta\rangle \ltimes \langle \cos\theta, \sin\theta \rangle \\
&= \langle \cos^{3}\!\theta-\sin^2\!\theta\cos\theta-2\!\sin^2\!\theta\!\cos\theta, 2\!\sin\theta\!\cos^2\!\theta+\sin\theta\cos^2\!\theta-\sin^3\!\theta\rangle \\
&=\langle 4\cos^3\!\theta-3\cos\theta,3\sin\theta-4\sin^3\!\theta\rangle\end{align}$$
Where the last step follows from the Pythagorean identities. If we then equate the first elements and the second elements we find the familiar results
$$\begin{align} \cos(3\theta) &= 4\cos^3\!\theta-3\cos\theta \\ \sin(3\theta) &= 3\sin\theta-4\sin^3\!\theta\end{align}$$
This is essentially a result of complex numbers and is called De moivre's formula. You can see it being used to prove your identity here. Our type of multiplication mirrors complex multiplication which is inherently rotational in nature. You can see why here. De moivre's formula says that a rotation by $n\theta$ is equal to $n$ rotations by $\theta$.
The best geometric proof of your formula comes from this basic idea which is so simple that no picture is needed to represent it.
This is advantageous because it can be used to show other identites such as $$\cos(5\theta) = 16 \cos^5\! \theta - 20 \cos^3 \!\theta + 5 \cos \theta$$