已知函数f（x）=2cos2x+sinx的平方-4cosx,求f(π/3)的值

f(x)
=2cos2x+sin²x-4cosx
=2(2cos²x-1)+(1-cos²x)-4cosx
=4cos²x-2+1-cos²x-4cosx
=3cos²x-4cosx-1
f(π/3)
=3cos²(π/3)-4cos(π/3)-1
=3/4-2-1
=-9/4

=2*2*（-0.5）+sin60的平方-4*0.5
剩下自己求下