I'm looking at circular motion, and looking at the radial vector $r$ and seeing what happens when $r$ changes by a small $\varepsilon$ (if $r$ was initally $b$). I'm trying to form a second order ODE for $\varepsilon$ and I have a $\sin(b+\varepsilon)$ to take care of and the solutions show that: if $r \to b + \varepsilon$ then $\sin(r) = \sin(b)\cos(\varepsilon) + \cos(b)\sin(\varepsilon) \approx (-1)^n\varepsilon$.
I understand the equality part from the double angle formulae but where did the $(-1)^n\varepsilon$ come from?
Thinking about it this must be a typo and I think it should be $\sin(r)\approx (-1)^n\epsilon+\sin(b)$
since $r\to b+\epsilon$$\,$, $b$ could be larger or smaller than $r$, so equivilantley, we could have $\epsilon\gt 0$ or $\epsilon\lt 0\,$, thus:
$\cos(b)\sin(\epsilon)\approx(-1)^n\epsilon$
and $\sin(b)\cos(\epsilon)\approx\sin(b)$
thus:
$\sin(r)=\cos(b)\sin(\epsilon)+\sin(b)\cos(\epsilon)\approx(-1)^n\epsilon+\sin(b)$
Which actually does make sense since it is saying that if $r$ varies slightly, the value $\sin(r)$ changes slightly as well.