SOLUTION: Let a, b and c be sides of a triangle in arithmetic progression in this order. Show that cos((A-C)/2)/cos((A+C)/2)=2

SOLUTION: Let a, b and c be sides of a triangle in arithmetic progression in this order. Show that cos((A-C)/2)/cos((A+C)/2)=2

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Question 1089830: Let a, b and c be sides of a triangle in arithmetic progression in this order. Show that cos((A-C)/2)/cos((A+C)/2)=2
Answer by richard1234(7193) (Show Source): You can put this solution on YOUR website!
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

,

, and

. Finding the area of triangle ABC using A = rs and Heron's formula we have:

Squaring both sides and simplifying gives

. (*)

Now we are trying to show

. Using sum and difference formulas we have

and also:

It follows that

if and only if

. This simplifies to

which we showed is true in (*). Therefore

.
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