三角信中,1/COSa+1/COSc＝-跟号2/COSB,求COSA_C/2
三角信中,1/COSa+1/COSc＝-跟号2/COSB,求COSA_C/2
日期：2021-05-01 05:05:14 人气：1
原题:
"已知三角形ABC,三内角满足A+B=2C,1/COSA+1/COSC=-根号2/COSB,求COS（A-C)/2"
因A+B+C=π,
又A+C=2B
得B=π/3
1/cosA+1/cosC=-√2cosB
=-√2/2
(cosA+cosC)=-2√2cosAcosC
2cos(A-C)&
"已知三角形ABC,三内角满足A+B=2C,1/COSA+1/COSC=-根号2/COSB,求COS（A-C)/2"
因A+B+C=π,
又A+C=2B
得B=π/3
1/cosA+1/cosC=-√2cosB
=-√2/2
(cosA+cosC)=-2√2cosAcosC
2cos(A-C)&