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c^2cos^2B+b^2cos^2C+bc cos(B-C) =1/2 (a^2+b^2+c^2)? |
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Answer» c2 cos2B + b2 cos2 C + bc cos (B-C) = (cos B + b cos C)2 - 2 bc cos B cos C + bc cos (B-C) (∵ a2+b2 = (a+b)2 - 2ab) = a2 - bc (cos(B+C) + cos(B-C)) + bc cos (B-C) (∵ 2 cosB cosC = cos(B+C) + cos(B-C)) And b cos C + c cos B = a = a2 - bc cos (B+C) = a2 - bc cos(180° - A) (∵ A+B+C = 180°, therefore B+C = 180° - A) = a2 + bc cos A = a2 + bc x \(\frac{b^2+c^2-a^2}{2bc}\) (∵ cos A \(=\frac{b^2+c^2-a^2}{2bc})\) = a2 + 1/2 (b2+c2-a2) = 1/2 (2a2+b2+c2-a2) = 1/2 (a2+b2+c2) Hence proved. |
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