Question on Cardano's Method of Solving Cubic Polynomial Equations I'm having trouble with part of a question on Cardano's method for solving cubic polynomial equations. This is a multi-part question, and I have been able to answer most of it. But I am having trouble with the last part.  I think I'll just post here the part of the question that I'm having trouble with.
We have the depressed cubic equation :
\begin{equation}
f(t) = t^{3} + pt + q = 0
\end{equation}
We also have what I believe is the negative of the discriminant :
\begin{equation}
D = 27 q^{2} + 4p^{3}
\end{equation}
We assume $p$ and $q$ are both real and $D < 0$. We also have the following polynomial in two variables ($u$ and $v$) that results from a variable transformation $t = u+v$ :
\begin{equation}
u^{3} + v^{3} + (3uv + p)(u+v) + q = 0
\end{equation}
You also have the quadratic polynomial equation :
\begin{equation}
x^{2} + qx - \frac{p^{3}}{27} = 0
\end{equation}
The solutions to the 2-variable polynomial equation satisfy the following constraints :
\begin{equation}
u^{3} + v^{3}  = -q 
\end{equation}
\begin{equation}
uv  = -\frac{p}{3}
\end{equation}
The first section of this part of the larger question asks to prove that the solutions of the quadratic equation are non-real complex conjugates. Here the solutions to the quadratic are equal to $u^{3}$ and $v^{3}$ (this relationship between the quadratic polynomial and the polynomial in two variables was proven in an earlier part of the question). I was able to do this part. The second part of this sub-question is what I'm having trouble with.
The question says, let :
\begin{equation}
u  = r\cos(\theta) + ir\sin(\theta) 
\end{equation}
\begin{equation}
v  = r\cos(\theta) - ir\sin(\theta)
\end{equation}
The question then asks the reader to prove that the depressed cubic equation has three real roots :
\begin{equation}
2r\cos(\theta) \text{ , } 2r\cos\left( \theta + \frac{2\pi}{3} \right) \text{ , } 2r\cos\left( \theta + \frac{4\pi}{3} \right)
\end{equation}
In an earlier part of the question they had the reader prove that given :
\begin{equation}
\omega = \frac{-1 + i\sqrt{3}}{2}
\end{equation}
s.t. :
\begin{equation}
\omega^{2} = \frac{-1 - i\sqrt{3}}{2}
\end{equation}
and :
\begin{equation}
\omega^{3} = 1
\end{equation}
that if $(u,v)$ is a root of the polynomial in two variables then so are :
$(u\omega,v\omega^{2})$ and $(u\omega^{2},v\omega)$. I think that the part of the question I'm having trouble with is similar. I suspect that :
\begin{equation} 
2r \cos\left( \theta + \frac{2\pi}{3} \right) = u\omega + v\omega^{2} \text{ or } u\omega^{2} + v\omega \tag{1}
\end{equation}
and :
\begin{equation} 
2r \cos\left( \theta + \frac{4\pi}{3} \right) = u\omega + v\omega^{2} \text{ or } u\omega^{2} + v\omega \tag{2}
\end{equation}
I have derived that :
\begin{equation}
\omega = \cos(\phi) + i\sin(\phi)
\end{equation}
where $\phi = \frac{2\pi}{3}$. Also :
\begin{equation}
\omega^{2} = \cos(2\phi) + i \sin(2\phi)
\end{equation}
So that the goal of the question may be to prove equations $(1)$ and $(2)$. I have tried to do this but haven't been able to.
Am I approaching this question in the correct way ? If I am approaching it the right way can someone show me how to use trigonometric identities to prove equations #1 and #2 ?

Let $w(\alpha) = \cos \alpha + i\sin \alpha$. Then
$$w(\alpha) w(\beta) = (\cos\alpha + i \sin \alpha)(\cos \beta + i\sin \beta) \\ =\cos\alpha \cos \beta - \sin \alpha \sin \beta +i(\cos\alpha \sin \beta + \sin \alpha \cos \beta) = \cos(\alpha + \beta) + i \sin(\alpha + \beta) \\= w(\alpha + \beta) .$$
An easier way to see this is to write $w(\alpha) = e^{i\alpha}$. Then
$$w(\alpha) w(\beta) = e^{i\alpha}e^{i\beta} = e^{i(\alpha + \beta)} = w(\alpha + \beta) .$$
We have
$$u\omega = rw(\theta)w(\phi) = rw(\theta+\phi) ,$$
$$u\omega^2 = rw(\theta)w(2\phi) = rw(\theta+2\phi) .$$
Moreover, since $v = \overline u$ and $\omega^2 = \overline \omega$, we get
$$v\omega^2 = \overline u \cdot \overline \omega = \overline{u\omega} ,$$
thus
$$u\omega + v\omega^2 = 2\Re (u\omega) = 2r\cos(\theta + \phi) = 2r\cos(\theta + 2\pi/3) .$$
Similarly
$$v\omega = \overline u \cdot \overline {\omega^2} = \overline{u\omega^2},$$
thus
$$u\omega^2 +  v\omega = 2\Re (u\omega^2) = 2r\cos(\theta + 2\phi) = 2r\cos(\theta + 4\pi/3) .$$
Edited:
In my opinion it is an odd aproach to apply Cardano's formula and then translate the result into a trigonometric form. A direct approach is via  angle trisection. By Moivre's formula we have
$$\cos\phi + i\sin\phi = (\cos(\phi/3) + i\sin(\phi/3))^3$$
which gives
$$\cos \phi = \cos^3(\phi/3) -3\cos(\phi/3)\sin^2(\phi/3)\\  = \cos^3(\phi/3) -3\cos(\phi/3)(1- \cos^2(\phi/3)) = 4 \cos^3(\phi/3) - 3 \cos(\phi/3) .$$
Writing $\theta = \phi/3$ and $x = 2\cos \theta$ gives us the cubic angle trisection equation
$$x^3 - 3x =  2\cos \phi \tag{1}.$$
By construction it has the obvious solution $x_0 = 2\cos \theta$. But since $\cos \phi = \cos (\phi + 2\pi) = \cos (\phi + 4 \pi)$, it also has the solutions $x_1 = 2 \cos((\phi + 2\pi)/3) = 2\cos (\theta + 2\pi/3)$, $x_2 = 2 \cos((\phi + 4\pi)/3) = 2\cos (\theta + 4\pi/3)$.
Under the assumption that $p, q$ are real and $D = 27q^2 + 4 p^3<0$ it is possible to reduce the general equation
$$t^3 + pt + q = 0 \tag{2}$$
to the angle trisection equation (1). Since $D < 0$, we must have $p < 0$. Note that therefore $D < 0$ is equivalent to  $27q^2/(-4p^3) < 1$.
Let us write $t = cx$. Then
$$x^3 + (p/c^2)x = -q/c^3 .$$
With $c = \sqrt{-p/3} > 0$ we get
$$x^3 -3x = 2(-q/2c^3) .$$
But
$$(-q/2c^3)^2 = q^2 /4(-p/3)^3 = 27q^2/(-4p^3) < 1$$
which means that
$$-q/2c^3 \in (-1,1) .$$
Therefore $\phi = \arccos(-q/2c^3)$ is a well-defined number in $(0,2\pi)$ and we get the cubic equation (1) with solutions $x_k$ as above. Therefore the solutions of (2) are
$$t_k = 2\sqrt{-p/3}\cos(\phi/3 + 2k\pi/3) , k = 0,1,2 .$$
