Question

函数f(x)=-根号3sin²x+sinx×cosx,a属于(0,π),若f(a/2)=1/4-根号3/2,求sina的

Answer 1

f(x)=-√3sin²x+sinxcosx
=-√3/2*(1-cos2x)+1/2*sin2x
=-√3/2+√3/2*cos2x+1/2*sin2x
=sin(2x+π/3)-√3/2
∴f(a/2)=sin(a+π/3)-√3/2=1/4-√3/2
∴sin(a+π/3)=1/4
∴a+π/3=2kπ+arcsin1/4，或a+π/3=2kπ+π-arcsin1/4
∴a=2kπ-π/3+arcsin1/4，或a=2kπ+2π/3-arcsin1/4 (k∈Z)

额，晕，没看到是求sina
sin(a+π/3)=1/4，∴cos(a+π/3)=±√[1-(1/4²)]=±√15/4
而sina=sin[(a+π/3)-π/3]=sin(a+π/3)*1/2-cos(a+π/3)*√3/2
当cos(a+π/3)=√15/4时，sina=1/4*1/2-√15/4*√3/2=(8-3√5)/8；
当cos(a+π/3)=-15/4时，sina=1/4*1/2+√15/4*√3/2=(8+3√5)/8

望采纳