Derivation of velocities in the Coriolis force In Fitzpatrick's Newtonian Dynamics book on the Coriolis force, he states

\begin{align}
v_{x'}&\simeq V_0\cos\theta-2\Omega t V_0\sin\lambda~\sin\theta   \tag{433}\\
v_{y'}&\simeq-V_0\sin\theta-2\Omega t V_0\sin\lambda~\cos\theta   \tag{434}
\end{align}
  To lowest order in $\Omega$, the above equations are equivalent to
  \begin{align}
v_{x'}&\simeq V_0\cos\left(\theta+2\Omega t\sin\lambda  \right)\tag{435}\\
v_{y'}&\simeq -V_0\sin\left(\theta+2\Omega t\sin\lambda  \right)\tag{436}
\end{align}

I simply don't get how the above equations are equivalent. In fact, if I take a magnitude (of the total vector) they are different. I also don't understand what "To lowest order in $\Omega$" means here.
 A: Note that Fitzpatrick states towards the beginning,

The following solution method exploits the fact that the Coriolis force is much smaller in magnitude that the force of gravity: hence, $\Omega$ can be treated as a small parameter

Generally, when statements like that are made, powers (greater than 1) of the term in question are considered to be zero:
$$
\Omega\ll1; \quad \Omega^n\,\approx0\,\,\forall\,\,n>1
$$
Thus,
$$
a\Omega+b\Omega^2\approx a\Omega
$$
for small $\Omega$.
In the small angle approximation, we have
\begin{align}
\sin\theta&\approx\theta\\
\cos\theta&\approx1-\frac{\theta^2}2\\
\tan\theta&\approx\theta
\end{align}
We can apply this to your first $v_{x'}$ equation:
\begin{align}
v_{x'}&\approx V_0\left(\cos\theta-2\Omega\sin\lambda \sin\theta t\right)\\
&\approx V_0\left(1-\frac{\theta^2}2-2\Omega\sin\lambda\theta t\right)
\end{align}
You can then complete the square to finish the solution, ignore the $\Omega^2$ term and end up with Equation (435) that the author gives.
A: Ahh, Richard Fitzpatrick.  Great guy.
Ok, If you start with the second set of expressions, use the appropriate double-angle-formula and then assume the "angle" $2\Omega \sin\lambda t$ is small  (note that the $t$ is not within the sin function!), you get the first expressions, e.g.
$$\cos(\theta+\phi)  = \cos\theta\cos\phi - \sin\theta\sin\phi,$$
and then for small $\phi$, use the first-order terms in the Taylor expansion for the trig functions, i.e. $\cos\phi \simeq 1$, and $\sin\phi\simeq \phi$.
In your case, $\phi = 2\Omega\sin\lambda t$, where again the $t$ is intended as a (linear) multiplicative factor and not inside the $\sin$ function.
Similarly, $$\sin(\theta+\phi) = \sin\theta\cos\phi + \cos\theta\sin\phi,$$
and you can take if from there...;-)
Woops, looks like Kyle beat me to the punch while I was typing this in! 
