If sin(theta/2)=squareroot[(x-1)/(2x)], then cos(theta)= ?
(The explanation is more important than the answer.)

searching google for "trigonometric identities" found quite a few
pages that had summaries of the topic, in particular we need the "half
angle formula", which is

sin(t/2) = +/- sqrt((1-cos(t)) / 2)

(found at http://aleph0.clarku.edu/~djoyce/java/trig/identities.html)

so in this case we need to set the insides of the two square roots
equal, and our equation is

(1-cos(t))/2 = (x-1) / 2x.

so we multiply both sides by 2....
1-cos(t) = (x-1) / x

and then subtract both sides from 1...  (alternatively, subtract 1
from each side and then multiply by -1)
cos(t) = 1 - (x - 1) / x

now we can notice that (x - 1) / x is the same as (x / x) - (1 / x), so 

cos(t) = 1 - ((x / x) - (1 / x)) = 1 - (1 - 1/x) so....

cos(t) = 1/x.   (!)

if you want more background, I again recommend Dr. Math:

http://mathforum.org/library/drmath/sets/high_trigonometry.html

good luck!

--David

Thanks for the links.