Question

With the usual notations, prove that `(sin("A" - "B"))/(sin ("A" + "B")) = ("a"^2 - "b"^2)/"c"^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 = `("a"^2 - "b"^2)/"c"^2 = ("k"^2 "sin"^2 "A" - "k"^2 "sin"^2 "B")/("k"^2 "sin"^2 "C")`

`= ("sin"^2"A" - "sin"^2"B")/("sin"^2"C")`

`= ((sin "A" + sin "B")(sin "A" - sin "B"))/[sin{pi - ("A" + "B")}]^2`    .....[∵ A + B + C = π]

`= (2 sin  (("A" + "B")/2). cos  (("A" - "B")/2) xx 2 cos  (("A" + "B")/2). sin  (("A" - "B")/2))/("sin"^2 ("A + B"))`

`= (2 sin  (("A" + "B")/2). cos  (("A" + "B")/2) xx 2 sin  (("A" - "B")/2). cos  (("A" - "B")/2))/("sin"^2 ("A + B"))`

`= ("sin" ("A" + "B"). sin("A" - "B"))/("sin"^2 ("A" + "B"))`

`= (sin ("A - B"))/(sin ("A + B"))` = LHS