<s>[INST] Approximation For Difference Of Two Sides Of A Triangle I have been trying to derive this approximation but have been unsuccessful in doing so. Any help would be greatly appreciated.

 [/INST] The vectors ${\mathbf{a}}$ and ${\mathbf{c}}$, in terms of their magnitudes and the angles $\theta$ and $\gamma$, are
$$
{\mathbf{a}} = (a \sin \theta, a \cos \theta)
$$
and
$$
{\mathbf{c}} = (c \cos \gamma, c \sin \gamma).
$$
Assuming $c \ll a$, we can expand $b-a$ in powers of the small parameter $\varepsilon = c/a$ as follows:
$$
\begin{eqnarray}
b-a &=& -a + \lVert{\mathbf{b}}\rVert \\
  &=& -a + \lVert{\mathbf{a}} - {\mathbf{c}}\rVert \\
  &=& -a + \sqrt{(\mathbf{a} - \mathbf{c})\cdot(\mathbf{a} - \mathbf{c})} \\
  &=& -a + \sqrt{a^2 - 2{\mathbf{a}}\cdot{\mathbf{c}} + c^2} \\
  &=& -a + a \sqrt{1 - 2 \varepsilon\left(\sin\theta \cos\gamma + \cos\theta \sin\gamma\right) + \varepsilon^2} \\
  &=& -a + a \sqrt{1 - 2 \varepsilon\sin(\theta+\gamma) + \varepsilon^2} \\
  &=& -a + a \left(1 + \frac{1}{2}\left(- 2 \varepsilon\sin(\theta+\gamma) + \varepsilon^2\right) - \frac{1}{8}\left(- 2 \varepsilon\sin(\theta+\gamma) + \varepsilon^2\right)^2 + O(\varepsilon^3)\right) \\
  &=& a \left(- \varepsilon\sin(\theta+\gamma) + \frac{1}{2}\varepsilon^2 - \frac{1}{2}\varepsilon^2\sin^2(\theta+\gamma) + O(\varepsilon^3)\right) \\
  &=& - c\sin(\theta+\gamma) + \frac{c^2}{2a}\cos^2(\theta+\gamma) + O\left(\frac{c^3}{a^2}\right),
\end{eqnarray}
$$
which is the desired result.  The series expansion $\sqrt{1+x} = 1 + (1/2)x - (1/8)x^2 + O(x^3)$ was used to approximate the square root.
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