Consider the cardioid $\rho=2a\left( 1 - \cos \phi \right) $. Show that the angle between the tangent vector and an arbitrary point (different from the origin) of the curve is half the polar angle.
Expressed in polar coordinates, I have the curve
$$\mathbf x \left( \phi \right) = \left( \rho, \phi \right) = \left( 2a\left( 1 - \cos \phi \right), \phi \right) $$
I tried to do the problem using cartesian coordinates, as I am used to do, but things were not nice. So I have looked up the formula for the tangent vector in polar coordinates (and I understood where it comes from).
$$\mathbf t _ \mathbf x \left( \phi \right) = \left( \frac{ \mathrm d \rho}{\mathrm d \phi}, \rho \right) = \left( 2 a \sin \phi, \rho \right)=\left( 2 a \sin \phi, 2a \left( 1 - \cos \phi \right) \right)$$
Then I know that the angle $\theta$ between these two vectors is given by
$$\cos \theta = \frac{\left\langle \mathbf x , \mathbf t _\mathbf x \right\rangle }{\left\Vert \mathbf{x}\right\Vert \left\Vert \mathbf{t}_\mathbf x\right\Vert }$$
And the inner product in polar coordinates is the product of the first coordinates times the cosine of the difference of the second coordinates, so
$$\left\langle \mathbf x , \mathbf t _\mathbf x \right\rangle = 4 a^2 \sin \phi \left( 1 - \cos \phi \right) \cos \left( \phi - \rho \right)$$
The norms of each vector are the first coordinate as they are in polar form, so
$$\cos \theta = \frac{\left\langle \mathbf x , \mathbf t _\mathbf x \right\rangle }{\left\Vert \mathbf{x}\right\Vert \left\Vert \mathbf{t}_\mathbf x\right\Vert }=\frac{4a^2\sin \phi \left( 1-\cos \phi \right) \cos \left( \phi - \rho \right)}{4a^2 \sin\phi \left( 1 - \cos \phi \right) }=\cos \left(\phi - \rho \right)$$
So $\theta = \phi - \rho + 2 \pi n$, where $\rho=2a\left( 1 - \cos \phi \right) $, but I think I have to prove that $\theta = \frac{\phi}{2}$. Am I missing something?