Thread: triangles is tough

in a triangle ABC where a,b,c are the sides opposite to the angles A,B,C respectively
AC is double of AB, then find the value of

Originally Posted by grgrsanjay

in a triangle ABC where a,b,c are the sides opposite to the angles A,B,C respectively
AC is double of AB, then find the value of

=

=

Since (B+C)/2 = π/2 - A/2, sin[(B+C)/2] = cos(A/2) and cos[(B+C)/2] = sin(A/2).

Substtitue these values in the above equation and simplify.