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在三角形ABC中,已知2sin^2(A)=3sin^2(B)+3sin^2(C),cos(2A)+3cosA+3cos(B-C)=1,求A、B、C
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在三角形ABC中,已知2sin^2(A)=3sin^2(B)+3sin^2(C),cos(2A)+3cosA+3cos(B-C)=1,求A、B、C
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答案和解析
把公式cos(2A) = 1 -- 2sin²A
代入已知等式cos(2A)+3cosA+3cos(B-C)=1得:
1 -- 2sin²A + 3cosA + 3cos(B-C) = 1
∴ 2sin²A = 3cosA + 3cos(B-C)
而已知2sin²A = 3sin²B + 3sin²C
由以上两式的右边相等得:
3sin²B + 3sin²C = 3cosA + 3cosBcosC + 3sinBsinC
∴3(sinB -- sinC)² = 3cosA+3cos(B+C) = 3cosA -- 3cosA = 0
∴(sinB -- sinC)² = 0
∴sinB = sinC
∴ B = C
把B = C 代入已知等式cos(2A)+3cosA+3cos(B-C)=1得:
cos(2A)+3cosA + 3 = 1 而cos(2A) = 2cos²A -- 1
∴ (2cos²A -- 1)+ 3cosA + 3 = 1
∴2cos²A + 3cosA + 1 = 0
∴ (cosA + 1)( 2cosA + 1)= 0
∴ cosA = -- 1 或 cosA = -- 1/2
∴ A = π (舍去)或 A = 2π/3 = 120°
而 B = C
∴ B = C = π/6 = 30°
∴ A = 2π/3 = 120° B = C = π/6 = 30°
祝您学习顺利!
代入已知等式cos(2A)+3cosA+3cos(B-C)=1得:
1 -- 2sin²A + 3cosA + 3cos(B-C) = 1
∴ 2sin²A = 3cosA + 3cos(B-C)
而已知2sin²A = 3sin²B + 3sin²C
由以上两式的右边相等得:
3sin²B + 3sin²C = 3cosA + 3cosBcosC + 3sinBsinC
∴3(sinB -- sinC)² = 3cosA+3cos(B+C) = 3cosA -- 3cosA = 0
∴(sinB -- sinC)² = 0
∴sinB = sinC
∴ B = C
把B = C 代入已知等式cos(2A)+3cosA+3cos(B-C)=1得:
cos(2A)+3cosA + 3 = 1 而cos(2A) = 2cos²A -- 1
∴ (2cos²A -- 1)+ 3cosA + 3 = 1
∴2cos²A + 3cosA + 1 = 0
∴ (cosA + 1)( 2cosA + 1)= 0
∴ cosA = -- 1 或 cosA = -- 1/2
∴ A = π (舍去)或 A = 2π/3 = 120°
而 B = C
∴ B = C = π/6 = 30°
∴ A = 2π/3 = 120° B = C = π/6 = 30°
祝您学习顺利!
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