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Question
In any Δ ABC, prove the following:
a2 sin (B - C) = (b2 - c2) sin A.
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
RHS = (b2 - c2) sin A
= (k2 sin2B - k2 sin2C)sin A
= k2 (sin2B - sin2C) sin A
= k2 (sin B + sin C)(sin B - sin C) sin A
= `"k"^ 2 xx 2 "sin" (("B + C")/2). cos(("B - C")/2) xx 2 cos (("B + C")/2).sin (("B - C")/2) xx sin "A"`
= `"k"^ 2 xx 2 "sin" (("B + C")/2). cos(("B + C")/2) xx 2 sin (("B - C")/2). cos (("B - C")/2) xx sin "A"`
= k2 x sin(B + C) x sin (B - C) x sin A
= k2 . sin (π - A). sin (B - C). sin A ....[∵ A + B + C = π]
= k2. sin A. sin (B - C). sin A
= (k sin A)2. sin(B - C)
= a2 sin (B - C)
= LHS
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