Finding the family of functions satisfying $f(x + y)=f(x)f(y) - f(x-y)$
$f(x)=2 \cos x$ satisfies all the conditions given here. $$f(x+y)+f(x-y)= 4 \cos x \cos y= 2 f(x) f(y) ~~~(1)$$ $$f(0)=2, f(-x)=f(x), (4 \cos^2 a-4)(4\cos^2 b-4)=16 \sin^2 a \\sin^2 b \ge 0$$ and $f'(x) =\sin x, f'(0)= 0.$ Also $f(x)=2\cos x= e^{ix}+ e^{-ix}=a^x+a^{-x} \implies a=e^{i}$.