- #1

etotheipi

Gold Member

2019 Award

- 309

- 66

## Main Question or Discussion Point

I've discussed this with a few people and I've found it pretty interesting (ok, maybe not

For instance, I'll tend to think of an equation in the form of two fractions (ignoring the 1's on the denominators, though I've put them in for clarity), and will just "slide" the quantities across either of the diagonals, since I've thought this is a pretty quick method:

##(c^{2}) = (a^{2} + b^{2} - 2ab\cos{\theta})##

##\frac{(c^{2} - a^{2} - b^{2})}{1} = \frac{-2ab\cos{\theta}}{1} \implies \frac{1}{-2ab} = \frac{\cos{\theta}}{(c^{2} - a^{2} - b^{2})}## etc.

I was wondering if there were any methods that anyone else tends to use that are more efficient? Sorry for the rather broad question!

**that**interesting, but still...) how there's quite a bit of difference in internal thought process for performing the bog-standard algebraic manipulations/rearrangements that come up all of the time.For instance, I'll tend to think of an equation in the form of two fractions (ignoring the 1's on the denominators, though I've put them in for clarity), and will just "slide" the quantities across either of the diagonals, since I've thought this is a pretty quick method:

##(c^{2}) = (a^{2} + b^{2} - 2ab\cos{\theta})##

##\frac{(c^{2} - a^{2} - b^{2})}{1} = \frac{-2ab\cos{\theta}}{1} \implies \frac{1}{-2ab} = \frac{\cos{\theta}}{(c^{2} - a^{2} - b^{2})}## etc.

I was wondering if there were any methods that anyone else tends to use that are more efficient? Sorry for the rather broad question!

Last edited: