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F4 A Maths - prove

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In triangle ABC, show that (a) bcosC - ccosB = (b^2 -c^2) / a (b) sin^2A + sin^2B - sin^2C = 2sinAsinBcosC

最佳解答:

(a) In the triangle ABC, by using Cosine Rule, we have cos C = (a^2 + b^2 - c^2)/2ab and cos B = (a^2 + c^2 - b^2)/2ac So we have L.H.S. = b cos C - c cos B = b[(a^2 + b^2 - c^2)/2ab] - c[(a^2 + c^2 - b^2)/2ac] = (a^2 + b^2 - c^2)/2a - (a^2 + c^2 - b^2)/2a = [(a^2 + b^2 - c^2) - (a^2 + c^2 - b^2)]/2a = (a^2 + b^2 - c^2 - a^2 - c^2 + b^2)/2a = (b^2 - c^2)/a = R.H.S. (b) L.H.S. = sin^2 A + sin^2 B - sin^2 C = sin^2 A - sin^2 C + sin^2 B = (sin A + sin C)(sin A - sinC) + sin^2 B = [2 sin 0.5(A + C) cos 0.5(A - C)][2 cos 0.5(A + C) sin 0.5(A - C)] + sin^2 B ------ (by compound angle formulae) = [2 sin 0.5(A + C) cos 0.5(A + C)][2 cos 0.5(A - C) sin 0.5(A - C)] + sin^2 {180 - (A + C)] ------ (because A + B + C = 180) = sin (A + C) sin (A - C) + sin^2 (A + C) = sin (A + C) [sin (A - C) + sin (A + C)] = sin (180 - B) [2 sin 0.5(A - C + A + C) cos 0.5(A - C - A - C)] ------ (by compound angle formulae) = sin B [2 sin A cos (-C)] = 2 sin A sin B cos C ------ (because cos (-C) = cos C) = R.H.S.

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