Given the triangle ABC, what is cos(A/2)*cos(B/2)*cos(C/2)?

Since we'll have to evaluate the product of cosines, we'll
apply the law of cosines:

cos A/2= sqrt
(p(p-a)/b*c),

Half perimeter p: p=(a+b+c)/2 and a,b,c, are
the length of the opposite sides to the A,B,C, angles.

cos
B/2= sqrt (p(p-b)/a*c)

cos C/2= sqrt
(p(p-c)/b*a)

We'll evaluate the
product:

cos A/2*cos B/2*cos C/2= sqrt
[p(p-a)*p(p-b)*p(p-c)/a^2*b^2*c^2]

cos A/2*cos B/2*cos
C/2=p*sqrt[p(p-a)*(p-b)*(p-c)]/a*b*c

We recognize the
Heron's formula for evaluating the area:

A =
sqrt[p(p-a)*(p-b)*(p-c)]

cos A/2*cos B/2*cos
C/2=p*A/a*b*c

The requested product is: cos
A/2*cos B/2*cos C/2=p*A/a*b*c