Question

In ΔABC, prove that `sin(("B" − "C")/2) = (("b" − "c")/"a")cos  "A"/(2)`.

Answer 1

By the sine rule,

`"a"/"sin A" = "b"/"sin B" = "c"/"sin C"` = k

∴ a = k sin A, b = k sin B, c = k sin C

RHS = `(("b" − "c")/"a")cos  "A"/(2)`

RHS = `(("k sin B" − "k sin C")/("k sin A"))cos  "A"/(2)`

RHS = `(("sin B" − "sin C")/("sin A"))cos  "A"/(2)`

RHS = `(2cos(("B" +  "C")/2).sin(("B" − "C")/2))/(2sin  "A"/(2) . cos  "A"/(2)).cos  "A"/(2)`

RHS = `(cos(("B" + "C")/(2)).sin(("B" − "C")/(2)))/(sin  "A"/(2)`

RHS = `(cos(π/2 − "A"/2).sin(("B" − "C")/(2)))/(sin  "A"/(2))`  ...[ ∵ A + B + C = π ]

RHS = `(sin  "A"/(2). sin(("B" − "C")/(2)))/(sin  "A"/(2)`

RHS = `sin(("B" − "C")/(2))`

LHS = `sin(("B" − "C")/(2))`

RHS = LHS

`sin(("B" − "C")/2) = (("b" − "c")/"a")cos  "A"/(2)`

Hence proved.