Question

for ∆ABC,sin(B+C)/2=..............?​

Answer 1

the answer is cos A/2

Answer 2

For a ∆ABC,
A+B+C = π
So,
A+B+C/2 = π/2
So,
(B+C)/2 = π/2 - A/2
So,
Sin(B+C)/2 = sin(π/2 - A/2) = cos A/2

Now,
cos A/2 = √ {s(s-a)/bc}
where, s is the semiperimeter of the triangle, and a, b ,c are the lengths of the sides opposite to vertices A,B & C respectively.