Question

已知函数f（x）＝cos平方x－√3sinxcosx＋1

f(a)=5/6. π/3≤a≤2π/3 求sin2a的值

Answer 1

f(x)=cos²x -√3sinxcosx +1
=(1+cos2x)/2-(√3/2)sin2x +1
=sin(π/6 - 2x)＋3/2
因为，f(a)=5/6
所以，sin(π/6 -2a)＋3/2=5/6
即，sin(2a-π/6)＝2/3
因为，π/3≤a≤2π/3
所以， π/2≤2a-π/6≤7π/6
所以，cos(2a-π/6)＝-(√5)/3
sin2a＝sin[(2a-π/6)+(π/6)]
＝sin(2a-π/6)cos(π/6)+cos(2a-π/6)sin(π/6)
＝(2/3)×(√3/2)+(-√5/3)×(1/2)
＝(√3/3)-(√5/6)
＝(2√3 -√5)/6

Answer 2

解：
f(x)=cos²x-√3sinxcosx＋1

=(1+cos2x)/2-(√3/2)sin2x+1
=1/2+(1/2)cos2x-(√3/2)sin2x+1
=3/2+sin(π/6)cos2x-cos(π/6)sin2x
=3/2+sin(π/6-2x)
=3/2-sin(2x-π/6)
因此：
f(a)=3/2-sin(2a-π/6)=5/6
∴sin(2a-π/6)=2/3
又∵π/3≤a≤2π/3

∴π/2≤2a-π/6≤7π/6

因此：
2a-π/6=π-arcsin(2/3)
∴2a=7π/6-arcsin(2/3)
sin2a=sin[7π/6-arcsin(2/3)]
=-sin[π/6-arcsin(2/3)]
=-sin(π/6)cos[arcsin(2/3)]+cos(π/6)sin[arcsin(2/3)]
=-(1/2)*(√5/3)+(√3/2)*(2/3)
=(2√3-√5)/6