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I have been trying to derive this approximation but have been unsuccessful in doing so. Any help would be greatly appreciated.

The image at the above mentioned link.

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Please avoid forcing people to click through-take the effort to bring the problem here. – Ross Millikan Feb 4 '11 at 6:19
I think he doesnt know how to include an image in his post. – Arjang Feb 4 '11 at 6:40

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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I have only the first term in your approximation.
Here's how: First, the angles opposite a side are denoted by the capital letter. Next, notice that the angle $B$ is equal to $\frac{\pi}{2} - (\theta + \gamma)$. Now, the Law of Cosines states $ b^2 = a^2 + c^2 - 2ac \cos B $ and $ a^2 = b^2 + c^2 - 2bc \cos A $ so if you take the difference of the two and simplify you will get $b^2 - a^2 = -c(a \cos B - b \cos A)$.

So far we have made no use of the fact that $c << a$. Let's now use this to note that $C \approx 0$ and so $A \approx \pi - B$ which implies that $\cos A \approx \cos (\pi - B) = - \cos B$. Substitute this into the difference of the two cosine laws we derived earlier. Thus, $b^2 - a^2 \approx -c(a \cos B - b (- \cos B)) = -c \cos B (a + b)$. Now divide both sides by $(a + b)$ to get $b - a \approx -c \cos B$. Finally, since $B = \frac{\pi}{2} - (\theta + \gamma)$, $\cos B = sin (\theta + \gamma)$, and so we've shown that $b - a \approx -c \sin (\theta + \gamma)$.

If I have some time later, I'll return to this.

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From the law of cosines we know that $b^2 - a^2 = c^2 - 2ac \cos B$, so $ b - a = { {-2ac \cos B} \over {b+a} } + { {c^2} \over {b+a} }$. Asuuming that $ c << a \Rightarrow b \approx a $, gets you close but the second term does not match.

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