If the function f satisfies the relation f(x+y)+f(x−y) =2f(x)f(y)∀x, y∈R  and  f(0)≠0, then

f(x+y)+f(x−y)=2f(x)⋅f(y)               (1)Put x = 0. Then f(y)+f(−y)=2f(0)f(y)           (2)Put x = y = 0. Then f(0)+f(0)=2f(0)f(0)∴ f(0)=1[as⁡f(0)≠0]∴ f(−y)=f(y)                        [From (2)]Hence, the function is even. Then f(−2)=f(2)=a.