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Question
In any Δ ABC, prove the following:
a sin A - b sin B = c sin (A - B)
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Solution
By sine rule,
`"a"/("sin A") = "b"/("sin B") = "c"/("sin C") = "k"`
∴ a = k sin A, b = k sin B, c = k sin C
LHS = a sin A - b sin B
= k sin A. sin A - k sin B. sin B
= k (sin2 A - sin2 B)
= k (sin A + sin B)(sin A - sin B)
`= "k" xx 2 sin (("A + B")/2). cos (("A - B")/2) xx 2 cos (("A + B")/2). sin (("A - B")/2)`
`= "k" xx 2 sin (("A + B")/2). cos (("A + B")/2) xx 2 sin (("A - B")/2). cos (("A - B")/2)`
= k × sin (A + B) × sin (A - B)
= k sin (π - C). sin (A - B) … [∴ A + B + C = π]
= k sin C. sin (A - B)
= c sin (A - B)
= RHS.
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