已知三角形ABC的三个内角,满足A+C=2B,设x=cos((A-C)/2),f(x)=cosB(1/cosA+1/co
已知三角形ABC的三个内角,满足A+C=2B,设x=cos((A-C)/2),f(x)=cosB(1/cosA+1/co
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
已知△ABC的三内角A、B、C满足A+C=2B,设x=cos(A-C)/2,f(x)=cosB(1/cosA+1/cos
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的三个内角满足:A+C=2B,(1/cosA)+(1/cosC)=-(根号2/cosB) 求cos(A-
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=负的根号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
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
设函数f(x)=cos(2x+派/3)+sin平方x,设A,B,C为△ABC的三个内角,若cosB=1/3,f(C/2)
设函数f(x)=cos(2x+π/3)+sinx^2,设A,B,C为三角形的三个内角,若cosB=1/3,f(c/2)=
设函数f(X)=cos(2x+π/3)+sin方x.设A、B、C为△ABC的三个内角,若cosB=1/3,f(C/3)=