StatusX
Jan25-06, 01:55 PM
Here's a strange identity I found. Nothing spectacular, and it should be clear from the form of it how I derived it, but I though it was kind of cool, and I've never seen this type of geometric identity.
Given a triangle with side lengths A, B, and C, (with B<A, and maybe some other restrictions I've missed) and angle \gamma opposite C, then the following identity holds:
\ln \left( \frac{A}{C} \right) = \frac{B}{A} \cos (\gamma) + \frac{1}{2}\left( \frac{B}{A} \right)^2 \cos (2\gamma) + ...
Anyone ever seen this before? Do you know if it can be put in a more pleasing form, or applied to anything useful? I know it quickly gives a few taylor series with certain values for the sides and angle. It was derived algebraically, but maybe there's a really clever geometric proof?
EDIT: Actually, I was thinking, wouldn't this give a method for constructing (ie, with a compas and straightedge) the logs of numbers, albeit in an infinite number of steps? I don't know much about constructing numbers, but it does seem to be a conducive form.
Given a triangle with side lengths A, B, and C, (with B<A, and maybe some other restrictions I've missed) and angle \gamma opposite C, then the following identity holds:
\ln \left( \frac{A}{C} \right) = \frac{B}{A} \cos (\gamma) + \frac{1}{2}\left( \frac{B}{A} \right)^2 \cos (2\gamma) + ...
Anyone ever seen this before? Do you know if it can be put in a more pleasing form, or applied to anything useful? I know it quickly gives a few taylor series with certain values for the sides and angle. It was derived algebraically, but maybe there's a really clever geometric proof?
EDIT: Actually, I was thinking, wouldn't this give a method for constructing (ie, with a compas and straightedge) the logs of numbers, albeit in an infinite number of steps? I don't know much about constructing numbers, but it does seem to be a conducive form.