I need a proof for a scalar product - no numbers allowed

Let $$\mathbf{a}=|a|\cos\theta \mathbf{i}+|a|\sin\theta \mathbf{j}$$ and $$\mathbf{b}=|b|\cos\phi \mathbf{i}+|b|\sin\phi \mathbf{j}$$

Now, we have $$ \mathbf{a\cdot b} = |a||b|\cos\theta\cos\phi + |a||b|\sin\theta\sin\phi $$ Now use a trigonometric identity to simplify, with $\alpha=\theta-\phi$.

At 71GA's request, I'm elaborating on the relationship between my expression of $\mathbf{a}$ and the cartesian expression.

Consider the vector in the form $$ \mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} $$ We now define $$|\mathbf{a}|^2 = a_1^2+a_2^2$$ Note that this forms a right triangle relationship. For vectors with positive $a_1$, we may use the formed right triangle to determine the angle of the vector from the positive $x$-axis, $\theta$. Namely, $$ \tan(\theta) = \frac{a_2}{a_1} $$ This relationship also applies if $a_1$ is negative or zero, but the angle itself must be adapted to suit. We may then solve for $a_1$ and $a_2$, by noting that \begin{align} \frac{|\mathbf{a}|^2}{a_1^2}&=1+\frac{a_2^2}{a_1^2}\\ &=1+\tan^2(\theta)\\ &=\sec^2(\theta) \end{align} and so, we have $$ a_1^2 = |\mathbf{a}|^2\cos^2(\theta) $$ and, by appropriate choice of angle, you have $$ a_1 = |\mathbf{a}|\cos(\theta) $$ Substituting back into our equation for $\tan(\theta)$, we have $$ a_2 = |\mathbf{a}|\sin(\theta) $$ As such, we have $$ \mathbf{a} = |\mathbf{a}|\cos\theta\mathbf{i}+|\mathbf{a}|\sin\theta\mathbf{j} $$ where $\theta$ is the angle of the vector from the positive $x$-axis. And we can repeat this for vector $\mathbf{b}$, with $\phi$ as the angle. Thus, the angle between vectors $\mathbf{a}$ and $\mathbf{b}$ is $\alpha=\theta-\phi$, or is different from this by $2\pi$ (in which case the relation $\mathbf{a\cdot b}=|\mathbf{a}||\mathbf{b}|\cos(\alpha)$ is unchanged by the addition or subtraction of $2\pi$).