Help with Trigonometry Applications! Please!

For any triangle show that

cos C/2= Square Root of s(s-c)/ab

where s=1/2(a+b+c)

I know I can use the half angle formula and the laws of cosines but I dont know how!
PLEASE HELP!!!
Thanks!!

Cos^2 (C/2) = (1+cos C)/2 = [1+(a^2+b^2-c^2)/2ab]/2 = [2ab+a^2+b^2-c^2]/4ab =[(a+b)^2-c^2]/4ab
=(a+b+c)(a+b-c)/4ab = (2s)(2s-2c)/4ab = s(s-c)/ab.
Hence cos (C/2) = sqrt[s(s-c)/ab]