已知三角形ABC,三内角满足A+B=2C,1/COSA+1/COSC=负根号2处以COSB,求COS(A-C)/2
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/05/21 08:59:55
已知三角形ABC,三内角满足A+B=2C,1/COSA+1/COSC=负根号2处以COSB,求COS(A-C)/2
因A+B+C=π,又A+C=2B
得B=π/3
1/cosA+1/cosC=-2√2
=>(cosA+cosC)=-2√2cosAcosC
=>2cos(A-C)/2cos(A+C)/2=-√2[cos(A+C)+cos(A-C)]
=>cos(A-C)/2=-√2[-1/2+cos(A-C)]
=>cos(A-C)/2=-√2[-1/2+2cos²(A-C)/2-1]
=>4cos²(A-C)/2+√2cos(A-C)/2-3=0(|A-C|/2
得B=π/3
1/cosA+1/cosC=-2√2
=>(cosA+cosC)=-2√2cosAcosC
=>2cos(A-C)/2cos(A+C)/2=-√2[cos(A+C)+cos(A-C)]
=>cos(A-C)/2=-√2[-1/2+cos(A-C)]
=>cos(A-C)/2=-√2[-1/2+2cos²(A-C)/2-1]
=>4cos²(A-C)/2+√2cos(A-C)/2-3=0(|A-C|/2
已知三角形ABC,三内角满足A+B=2C,1/COSA+1/COSC=负根号2处以COSB,求COS(A-C)/2
已知三角形ABC的三个内角满足:A+C=2B,(1/cosA)+(1/cosC)=-(根号2/cosB) 求cos(A-
已知三角形ABC的三个内角A,B,C满足A+C=2B,1/cosA+1/cosC=负的根号2/cosB,求cos(A-C
已知△ABC的三个内角A、B、C满足A+C=2B,且1/cosA+1/cosC=-根号2/cosB,求cos[(A-c)
已知三角形ABC的三个内角A,B,C满足:A+C=2B,1/cosA+1/cosC=-√2/cosB,求cos(A-C)
已知三角形ABC的三个内角A.B.C成等差数列,且1/cosA+1/cosC= - 根号2/cosB,求cos【(A-C
三角函数求值√表根号,已知三角形ABC满足A+C=2B.且1/cosA+1/cosC=-√2/cosB,求cos(A-C
已知三角形ABC的三个内角,满足A+B=2B,设x=cos(A-C)/2,f(x)=cosB(1/cosA+1/cosC
在三角形ABC中,内角A,B,c的对边a,b,c.已知(2c-a)/b=(cosA-2cosC)/cosB.1、求sin
已知△ABC的三个锐角A,B,C满足A+C=2B,1/cosA +1/cosC=-√2/cosB,求cos(A/2-C/
在三角形ABC中内角的对边分别为a.b.c已知(cosA-2cosC)/cosB=(2c-a)/b 1)求sinC/si
已知三角形ABC中,三内角A,B,C 满足A:B:C=1:2:2,求1-cosA+cosB-cosAcosB的值.