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Question
In Δ ABC, prove that `cos(("A" - "B")/2) = (("a" + "b")/"c")sin "C"/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 = `(("a" + "b")/"c") sin "C"/2`
`= (("k sin A" + "k sin B")/("k sin C")) sin "C"/2`
`= (("sin A" + "sin B")/"sin C") sin "C"/2`
`= (2sin (("A" + "B")/2). cos ("A" - "B")/2)/(2"sin" "C"/2. cos "C"/2). sin "C"/2`
`= ("sin" ("A" + "B")/2. cos ("A" - "B")/2)/(cos "C"/2)`
`= (sin (pi/2 - "C"/2).cos ("A" - "B")/2)/(cos "C"/2) ...[because "A" + "B" + "C" = pi]`
`= (cos "C"/2.cos ("A" - "B")/2)/(cos "C"/2)`
`= cos (("A" - "B")/2)`
= LHS
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