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