In ` Delta ABC ` , prove that :
` sin ((B - C)/(3)) = ((b -c)/(a)) cos ((A)/(2))`

Explanation

In `Delta ABC ` ,
` a/(sinA) = b/(sinB)= c/(sin C)" "` (by sine rule)
By equal ratio theorem we can write .
` (a) /(sin A) = (b - c)/(sin B - sinC )`
` rArr (a)/(b-c) = (sin A)/(sin B - sin C)`
` rArr (b - c)/(a) = (sin B sin C)/(sin A)`
`rArr (b-c)/(a) = (2 sin ((B - C)/(2)). cos ((B + C)/(2)))/(sin A)`
` = (2 sin ((B - C)/(2)). cos ((pi - A)/(2)))/(sinA)`
(` because A + B + C = pi ` , sum of all angles of triangle)
` = (2 sin ((B-C)/(2)) . sin ((A)/(2)))/(2 sin ((A)/(2)) . cos ((A)/(2)))`
` rArr (b-c)/(a) = (sin ((B - C)/(2)))/(cos((A)/2))`
` rArr ((b - c)/(a)) . cos ((A)/(2)) = sin ((B - C)/(2)) `