Relation between cosines and lenghts of triangle

Jul 2011
19
0
Prove that in any triangle we have the following relation \(\displaystyle \frac{a+b+c}{R}=8\cos{\frac{A}{2}}\cos{\frac{B}{2}}\cos{\frac{C}{2}}\)
 
Jul 2011
1,254
389
Belgium
What's \(\displaystyle R\)? ...

It can be useful to know that in a triangle:
A+B+C=180°
 
May 2011
99
32
Islamabad
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
 
May 2011
99
32
Islamabad
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]