Approximation to $\sqrt{\cos(\theta)}$? I have this formula, (it is just the law of cosines angle formula):
$$
d = \sqrt{a^2 + b^2 - 2ab \ cos(\theta)}
$$
Here is my issue. I am wondering if there is a way to 'extract' the $cos$ term. My motivation here is because I want to show that $d$ is in fact sinusoidal - or perhaps very close to sinusoidal - as $\theta$ is varied, while $a$ and $b$ are fixed. 
So, might there be a way to simplify, or 'extract' the $cos$ term here? What is the approximation to $\sqrt{cos(\theta)}$? Thank you.

Well, it's rather simple 
$$d = \sqrt{a^2 + b^2 - 2ab \ cos(\theta)}$$
$$d^2 = a^2 + b^2 - 2ab \ cos(\theta)$$
$$ d^2 - a^2 - b^2=2ab \ cos(\theta)$$
$$ \frac{d^2 - a^2 - b^2}{2ab} = \ cos(\theta)$$
$$ \frac{d^2}{2ab} - \frac{a^2 + b^2}{2ab} = \ cos(\theta)$$
and since $a$ and $b$ are fixed, $k = \frac{a^2 + b^2}{2ab}$
$$ \ cos(\theta) = \frac{d^2}{2ab} -k$$
with $d\leq a+b$ for the triangle inequality.
Finally $\sqrt{cos(\theta)} = \sqrt{\frac{d^2}{2ab} - \frac{a^2 + b^2}{2ab}}$
Anyway, assuming $a=b$ for simplicity, $$d=a\sqrt{2-2\cos(\theta)}=2|sin(\frac{\theta}{2})|$$ and this shows why $d$ is sinusoidal.
Furthermore, let $a=b+k$, then $$d = \sqrt{a^2 + a^2 +2ak+k^2 - 2a(a+k) \ cos(\theta)}$$
$$ d = \sqrt{2a^2-2a^2\cos\theta+2ak-2ak\cos\theta+k^2} $$
$$ d = \sqrt{2a^2(1-\cos\theta)+2ak(1-\cos\theta)+k^2} $$
$$ d = \sqrt{2a^2(2\sin^2(\frac{\theta}{2})+2ak(2\sin^2(\frac{\theta}{2})+k^2} $$
$$ d = \sqrt{4a^2\sin^2(\frac{\theta}{2})+4ak\sin^2(\frac{\theta}{2})+k^2} $$
which does have a sinusoidal pattern, but it's more difficult to unveil when $k \ne 0$
