Question 1089830
There's almost certainly a more elegant way than this, but here goes:

WLOG, assume a = 1-x, b = x, and c = 1+x. Let I be the incenter, and r be the inradius. We have semiperimeter 3/2. Additionally, if we let D, E, and F be the points of tangency with the incircle and sides BC, CA, AB respectively, we have that *[tex \large BD = BF = \frac{1}{2}], *[tex \large CD = CE = \frac{1}{2} - x], and *[tex \large AE = AF = \frac{1}{2} + x]. Finding the area of triangle ABC using A = rs and Heron's formula we have:


*[tex \large [ABC] = \frac{3}{2}r = \sqrt{\frac{3}{2} \cdot \frac{1}{2} \cdot (\frac{1}{2} - x)(\frac{1}{2} + x)}]

Squaring both sides and simplifying gives *[tex \large 3r^2 = \frac{1}{4} - x^2]. (*)


Now we are trying to show *[tex \large \cos \frac{A-C}{2} = 2 \cos \frac{A+C}{2}]. Using sum and difference formulas we have


*[tex \large \cos \frac{A-C}{2} = \cos \frac{A}{2} \cos \frac{C}{2} + \sin \frac{A}{2} \sin \frac{C}{2}]

*[tex \large = \frac{\frac{1}{4} - x^2}{AI \cdot IC} + \frac{r^2}{AI \cdot IC}]

and also:

*[tex \large \cos \frac{A+C}{2} = \cos \frac{A}{2} \cos \frac{C}{2} - \sin \frac{A}{2} \sin \frac{C}{2}]


*[tex \large = \frac{\frac{1}{4} - x^2}{AI \cdot IC} - \frac{r^2}{AI \cdot IC}]

It follows that *[tex \large \cos \frac{A-C}{2} = 2 \cos \frac{A+C}{2}] if and only if *[tex \large \frac{1}{4} - x^2 + r^2 = 2(\frac{1}{4} - x^2 - r^2)]. This simplifies to *[tex \large 3r^2 = \frac{1}{4} - x^2] which we showed is true in (*). Therefore *[tex \large \frac{\cos \frac{A-C}{2}}{\cos \frac{A+C}{2}} = 2].