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)=-1/4,c为锐角求sinA

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)=-1/4,c为锐角求sinA
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)=-1/4,c为锐角
求sinA

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)=-1/4,c为锐角求sinA
f(C/2) = cos(C+pi/3) + (1-cosC)/2 = 1/2 - sinC * sqrt(3)/2 = -1/4
sin C = 3/4 * 2 / sqrt(3) = sqrt(3)/2
sinA = sin(B+C) = sinB cos C + cosB sinC = sqrt(8)/3 * 1/2 + 1/3 * sqrt(3)/2 =
= [sqrt(8) + sqrt(3)]/6