Two inertial frames (different angles) 
The motion of the earth relative to a star changes the angle at which the star is perceived. Consider to inertial frames of reference I and I'. I' moves with velocity w relative to I along the x-axis. Show that
a) by looking at it in a non-relativistic way (classic) one can show the following correlation
$\tan(\Delta\theta)=\frac{w\sin(\theta)}{c-w\cos(\theta)}$
where $\theta$ is the angle for I and $\theta'$ the angle for I'. $\Delta\theta=\theta'-\theta$ is the perceived angle-difference.
b) by looking at it in a relativistic way that
$\cos(\theta)=\frac{\beta+\cos(\theta')}{1+\beta\cos(\theta')}$ and $\sin(\Delta\theta)=\sin(\theta)\frac{w}{c}$
with $\beta=\frac{w}{c}$.

I've tried getting a visual aid for the problem by drawing all the information I got, but I still can't seem to find an approach to get to the desired equation. I've tried using the law of cosine, but it got me nowhere. Anyone got any ideas?

 A: a) Assume that light has to travel 1 m to reach the origin of the reference frame I, so it'll take $t=1/c $ seconds to reach.
Now, using trigonometry (the sine rule in the first equation), we get:
$$ \frac{\sin{\Delta\theta}}{\omega t} = \frac{\sin{\theta}}{ct} $$ 
$$ct\cos{\Delta\theta} = 1-{\omega t}\cos{\theta}$$
Dividing the first equation with the second,
$$ \tan{\Delta\theta} = \frac{\omega\sin{\theta}}{c(1-\frac{\omega}{c}\cos{\theta})} = \frac{\omega\sin{\theta}}{c-\omega\cos{\theta}} $$
A: Hint for (a) :  

Use the Figure above to prove that :  
\begin{equation}
\tan\left(\phi^{\prime}-\phi\right)=\dfrac{w\sin\phi}{c-w\cos\phi}=\dfrac{w\sin\theta}{c+w\cos\theta}\;=\;-\tan\left(\theta^{\prime}-\theta\right)
\tag{a-01}
\end{equation}
This explains why I find the last equality instead of the first as I post in one of my comments : we refer to different angles.  
To give some further help, the vectors in the Figure are defined as follows using the $\:\phi=\pi-\theta\:$ angles
\begin{align}
\mathbf{u} & \equiv c\cdot\left(\cos\phi,\sin\phi\right)
\tag{a-02}\\
\mathbf{u}^{\prime} & \equiv u^{\prime}\cdot\left(\cos\phi^{\prime},\sin\phi^{\prime}\right) , \quad u^{\prime}=\|\mathbf{u}^{\prime}\| 
\tag{a-03}\\
\mathbf{w} & \equiv w\cdot\left(1,0\right) , \quad w=\|\mathbf{w}\| 
\tag{a-04}
\end{align}
and of course
\begin{equation}
\mathbf{u}^{\prime} =\mathbf{u} -\mathbf{w} 
\tag{a-05}
\end{equation} 

Hint for (b) :  

Use the Figure above to prove that :  
\begin{align}
\cos\phi & =\dfrac{\beta+\cos\phi^{\prime}}{1+\beta\cos\phi^{\prime}}=\dfrac{\beta-\cos\theta^{\prime}}{1-\beta\cos\theta^{\prime}}=-\cos\theta \qquad \text{or}
\tag{b-01}\\
\cos\phi^{\prime} & =\dfrac{-\beta+\cos\phi}{1-\beta\cos\phi}=-\dfrac{\beta+\cos\theta}{1+\beta\cos\theta}=-\cos\theta^{\prime}
\tag{b-02}
\end{align}
To give some further help, the vectors in the Figure are defined as follows using the $\:\phi=\pi-\theta\:$ angles
\begin{align}
\mathbf{u} & \equiv c\cdot\left(\cos\phi,\sin\phi\right)
\tag{b-03}\\
\mathbf{u}^{\prime} & \equiv c\cdot\left(\cos\phi^{\prime},\sin\phi^{\prime}\right)
\tag{b-04}\\
\mathbf{w} & \equiv w\cdot\left(1,0\right) , \quad w=\|\mathbf{w}\| 
\tag{b-05}
\end{align}
and of course the velocity vector $\mathbf{u}^{\prime}$ is the relativistic sum of velocities $\mathbf{u}$ and $-\mathbf{w}$ :
\begin{align}
u_{x}^{\prime} & =\dfrac{u_{x}-w}{1-\dfrac{u_{x}w}{c^{2}}}
\tag{b-06x}\\
u_{y}^{\prime} & =\dfrac{u_{y}}{\gamma\left(1-\dfrac{u_{x}w}{c^{2}}\right)}
\tag{b-06y}\\
\gamma &=\left(1-\dfrac{w^{2}}{c^{2}}\right)^{-1/2}
\tag{b-06$\gamma$}
\end{align}
