If A, B and C are interior angles of a triangle ABC, then show that `"sin"((B+C)/2)=cos(A/2)` .

Explanation

(i) `2 sin.(A)/(2) sin.(C)/(2) = 2 sqrt(((s-b) (s-c))/(bc)). sqrt(((s -a) (s-b))/(ab))`
`= (2(s-b))/(b) sqrt(((s-a) (s-c))/(ac))`
`= (2s - 2b)/(b). sin.(B)/(2)`
`=(a + c - b)/(b). sin.(B)/(2)`
`= (2b -b)/(b). sin.(B)/(2)` (as a, b, c are in A.P.)
`= sin.(B)/(2)`
(ii) `a cos^(2).(C)/(2) + c cos^(2).(A)/(2) = (a s (s - c))/(ab) + (c s(s -a))/(bc)`
`= (s)/(b) (s - c + s -a)`
`= (a + b+ c)/(2b) b`
`= (2b + b)/(2)`
`= (3b)/(2)`