设函数f(x)=cos(2x+pai/3)+sin^2x 1,求函数的最大值和最小正周期2,设A,B,C为三角形ABC的三个内角,若cosB=1/3,f(c/2)=-1/4,且C为锐角,求sinA

9999 字1.展开后：f(x)=-(√3/2)sin2x + (1/2)
f(x)max=√3/2 -1/2
T=π
2.∵f(C/2)=-1/4
∴-(√3/2)sin2(C/2) + (1/2) = -1/4
sinC=√3/2
∵C为锐角
∴C=60°
∴cosC=1/2
A,B,C为三角形ABC的三个内角且cosB=1/3
sinB=√[1-(cosB)^2]=(2√2)/3
sinA=sin[π-(B+C)]=sin(B+C)=sinBcosC + sinCcosB = √2/3 + √3/6

1.展开后：f(x)=-(√3/2)sin2x + (1/2)
f(x)max=√3/2 + 1/2
T=π
2.∵f(C/2)=-1/4
∴-(√3/2)sin2(C/2) + (1/2) = -1/4
sinC=√3/2
∵C为锐角
∴C=60°
∴cosC=1/2
A,B,C为三角形ABC的三个内角且cosB=1/3
sinB=√[1-(cosB)^2]=(2√2)/3
sinA=sin[π-(B+C)]=sin(B+C)=sinBcosC + sinCcosB = √2/3 + √3/6

1.展开后：f(x)=-(√3/2)sin2x + (1/2)f(x)max=√3/2 -1/2T=π2.∵f(C/2)=-1/4∴-(√3/2)sin2(C/2) + (1/2) = -1/4sinC=√3/2∵C为锐角∴C=60°∴cosC=1/2A,B,C为三角形ABC的三个内角且cosB=1/3sinB=√[1-(cosB)^2]=...