Question 601506
Since cos(x+k*2pi) = cos(x) for any integer k, this means that cos(a-2pi) = cos(a) and cos(a+4pi) = cos(a)


This means that 


f(a) + f(a-2pi) + f(a+4pi) = f(a) + f(a) + f(a) = 3*f(a) = 3*(-1/12) = -3/12 = -1/4


So f(a) + f(a-2pi) + f(a+4pi) = -1/4