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Question
In ∆ABC, prove that `sin ((A - B)/2) = ((a - b)/c) cos C/2`
Solution
`sin((A - B)/2) = ((a - b)/c) cos C/2`
Consider `(a - b)/c`
= `(k sin A - k sin B)/(k sin C)` ....(Sine rule)
= `(sin A - sin B)/sin C`
= `(2cos((A + B)/2)sin((A - B)/2))/(sin C)`
= `(2sin C/2 * sin((A - B)/2))/(2sin C/2 cos C/2)` ....(∵ A + B + C = π)
= `sin((A - B)/2)/(cos C/2)`
∴ `(a - b)/2 = sin((A - B)/2)/(cos C/2)`
∴ `sin ((A - B)/2) = ((a - b)/c) cos C/2`
Hence proved.
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