Use of small approximations to get the following formula

Question

I cannot see how the following formula has been found

$\displaystyle \cos(\theta + \frac{v\epsilon^2 \Omega}{L}+O(\epsilon^4))-\cos(\theta) = -\epsilon^2\frac{\Omega v}{L}\sin(\theta)+O(\epsilon^4)$

I tried to use the formula $\cos \theta - \cos \varphi = -2\sin\left( {\theta + \varphi \over 2}\right) \sin\left({\theta - \varphi \over 2}\right)$ but to no avail

Answer 1

$$ \cos\big( \theta + \varepsilon^2 \frac{v\Omega}{L} + O(\varepsilon^{4}) \big) = \cos(\theta)\cos\big( \varepsilon^{2} \frac{v\Omega}{L} + O(\varepsilon^4) \big) - \sin(\theta)\sin\big( \varepsilon^2 \frac{v\Omega}{L} + O(\varepsilon^4) \big) $$

For $\varepsilon$ small :

$$ \cos \big( \varepsilon^{2} \frac{v\Omega}{L} + O(\varepsilon^4) \big) = 1 + O(\varepsilon^4) $$

and

$$ \sin \big( \varepsilon^{2} \frac{v\Omega}{L} + O(\varepsilon^4) \big) = \varepsilon^{2} \frac{v\Omega}{L} + O(\varepsilon^4). $$

As a consequence :

$$ \begin{align*} \cos\big( \theta + \varepsilon^2 \frac{v\Omega}{L} + O(\varepsilon^{4}) \big) - \cos(\theta) &= {} \cos(\theta) \big( 1 + O(\varepsilon^{4} \big) - \sin(\theta) \big( \varepsilon^{2} \frac{v\Omega}{L} + O(\varepsilon^4) \big) - \cos(\theta) \\[2mm] &= - \varepsilon^{2} \frac{v\Omega}{L} \sin(\theta) + O(\varepsilon^4). \end{align*} $$

Answer 2

It just the taylor expansion at $\theta$ of $cos(\theta+x) \sim cos(\theta)-sin(\theta)x$, where $x=\epsilon^2\frac{\Omega v}{L}+O(\epsilon^4)$