On the derivation of the north-south aberration angle In A.P. French's Special relativity, page $39$, the author said the north-south aberration angle $\alpha$ below, in figure (b), is equal to $v\sin(\theta_{0})/{c}$, where $\theta_{0}$ is the angle when the earth is stationary (no aberration).

How did $v\sin(\theta_{0})/c$ came about?
This value seems rather suspicious, especially because of the $\sin(\theta_{0})$. According to this Wikipedia article, the angle $\theta$ above (named $\phi$ in the article) satisfies $$\tan(\theta)=\frac { \sin(\theta_{0})}{v/c +  \cos (\theta_{0})},$$
thus,
$$\tan(\alpha)=\tan(\theta_{0}-\theta)=\frac{\frac{\sin(\theta_{0})}{\cos(\theta_{0})}-\frac { \sin(\theta_{0})}{v/c +  \cos (\theta_{0})}}{1+\frac{\sin(\theta_{0})}{\cos(\theta_{0})}\times \frac { \sin(\theta_{0})}{v/c +  \cos (\theta_{0})}}.$$
I highly doubt that taking the $\tan^{-1}$ of the above would result in $v\sin(\theta_{0})/c$.
Is there a simpler way to compute the aberration angle to check whether it conforms with $v\sin(\theta_{0})/c$?
 A: Use 4-vectors (with $c\equiv 1$):
$$ k_{\mu} = (\omega, k_x, k_y, k_z) = k(1, -\cos{\theta_0}, -\sin{\theta_0}, 0) $$
is the wave-vector in the stationary frame. The angle of arrival is given by
$$\theta = \tan^{-1}{\frac{k_y}{k_x}}=\tan^{-1}{\frac{-\sin\theta_0}{-\cos\theta_0}}=\theta_0$$
Boost by $v$:
$$ k'_{\mu} = (\gamma(\omega-vk_x), \gamma(k_x-\omega v), k_y, k_z) $$
$$ k'_{\mu} = k(\gamma(1+v\cos\theta_0), -\gamma(\cos\theta_0+v), -\sin\theta_0, 0) $$
so:
$$\theta' = \tan^{-1}{\frac{k'_y}{k'_x}}=\tan^{-1}{\frac{\sin\theta_0}{\gamma(\cos\theta_0+v)}}$$
which reduced to the wiki expression at solar system velocities ($\gamma < 1+10^{-8}$).
Note: I once did this with aerospace engineers trying to point a star tracker in deep space. They were taking a 3+1 approach and were in knots.  They down right thought it was physicists' voodoo.
So now that we have a rigorous expression, we need to consider the fact that $v \ll c$ and that $\alpha \ll 1$. That means two things (with $\gamma \approx 1$):
$$ \tan{\theta} = \frac{\sin\theta_0}{\cos\theta_0 +v}=\frac{\tan\theta_0}{1+v/\cos{\theta_0}}\approx \tan\theta_0[1-v/\cos\theta_0]$$
Meanwhile, a Taylor expansion about $\theta_0$ gives:
$$ \tan\theta \equiv \tan{(\theta_0-\alpha)} \approx \tan\theta_0 - \alpha\cdot[\tan x]'_{|x=\theta_0}=\tan\theta_0 - \alpha/\cos^2{\theta_0}$$
$$ \tan\theta = \tan\theta_0[1 - \frac{\alpha}{\cos\theta_0\sin\theta_0}]$$
Setting:
$$ \frac v{\cos\theta_0} = \frac{\alpha}{\cos\theta_0\sin\theta_0}$$
and reintroducing $c$ gives:
$$\alpha = \frac v c \sin\theta_0$$
A: 
If you construct the right-angle triangle as shown in red on fig. (b) then it is easy to see that $\sin\alpha = \text{something}/c$. And by breaking $v$ into components you can see (from the doodle on the right) that the something is equal to $v\sin\theta_0$.
Then, as mentioned in my comment, all that's left is to say that $\alpha$ is small (i.e. $v\ll c$) so that $(v\sin\theta_0) /c=\sin\alpha\approx\alpha$.
(Please excuse the shoddy MS Paint editing.)
