Apologies, I can't include images - not enough rep.
An example exercise dealing with the Law of Cosines to prove Heron's formula for the area of a triangle has me befuddled.
After substitution using the Law of Cosines into a formula for area involving cosine :
$$ A^2 = \frac{1}{4} a^2b^2 \frac {(a+b+c)(a+b-c)}{2ab}\frac {(c+a-b)(c-a+b)}{2ab} $$
The handout simplifies this directly to:
$$ A^2 = \frac{(a+b+c)}{2} \frac{(a+b-c)}{2} \frac {(c+a-b)}{2}\frac {(c-a+b)}{2} $$
I see how the $a^2b^2$ with the two factors of $ab$. Then, I see how the 2 rationals on the right are factored into four.
But what happened to the factor of $\frac14$??
Direct-Link to the image of the above problem cut from professor's precalculus handout: http://s13.postimg.org/5os0orttj/Precalc_Lawof_Cosines_NYU_PDF3.png
Original source -- Problem On page 3 of this pdf:
As Ahmed Hussein noticed, it gets split up between the $2$'s in the denominators:
$$\frac14=\frac12\cdot\frac12$$
This accounts for all of the $2$'s in the denominators.