Thread: Relation between cosines and lenghts of triangle

Prove that in any triangle we have the following relation

What's ? ...

It can be useful to know that in a triangle:
A+B+C=180°

R is radius of the circumscribed circle.

RHS = 8cos(A/2) cos(B/2) cos(C/2) = 8 sqrt {[s^3(s-a)(s-b)(s-c)]/(abc)^2 }

= 8s(area of triangle)/abc

now using s = (a+b+c)/2 and 4(area of triangle)/abc = 1/R you get required result

could you be more speciffic?

about what, if you point out the problem

Originally Posted by waqarhaider

RHS = 8cos(A/2) cos(B/2) cos(C/2) = 8 sqrt {[s^3(s-a)(s-b)(s-c)]/(abc)^2 }

about this

cos(A/2) = sqrt[ s(s-a)/bc] , cos(B/2) = sqrt[ s(s-b)/ac] and cos(C/2) = sqrt[s(s-c)/ab]