sinA+sinB+sinC=cosA+cosB+cosC=0,
sin^2C=(sinA+sinB)^2
cos^2C=(cosA+cosB)^2
0<A,B,C<2π及A>B,求A-B可能值
0<A,B,C<2π及A>B,求A-B可能值
2013-12-19 6:12 am
回答 (1)
2013-12-19 7:07 am
✔ 最佳答案
sin²C + cos²C = (sinA + sinB)² + (cosA + cosB)²1 = sin²A + sin²B + 2sinA sinB + cos²A + cos²B + 2cosA cosB
1 = (sin²A + cos²A) + (sin²B + cos²B) + 2(sinA sinB + cosA cosB)
1 = 1 + 1 + 2cos(A - B)
cos(A - B) = - 1/2
由題意得 0 < A - B < 2π ,
∴ A - B = 2π/3 或 4π/3。
收錄日期: 2021-04-11 20:22:06
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20131218000051KK00164