Let f (x) defined on R satisfy f (x + y) = f (X-Y) = 2F (x) f (y) and f (π / 2) = 0 for any x, y ∈ R If f (0) ≠ 0, try to find f (π), Get rid of talking in detail. If I speak well, I will understand it after reading it once or twice. I will add at least 50 points Let f (x) defined on R satisfy f (x + y) + F (X-Y) = 2F (x) f (y) and f (π / 2) = 0 for any x, y ∈ R, If f (0) ≠ 0, try to find f (π), There's a little mistake on it. It's not an equal sign, it's a plus sign. I understand it after one or two times. I'll add at least 50 points

Let f (x) defined on R satisfy f (x + y) = f (X-Y) = 2F (x) f (y) and f (π / 2) = 0 for any x, y ∈ R If f (0) ≠ 0, try to find f (π), Get rid of talking in detail. If I speak well, I will understand it after reading it once or twice. I will add at least 50 points Let f (x) defined on R satisfy f (x + y) + F (X-Y) = 2F (x) f (y) and f (π / 2) = 0 for any x, y ∈ R, If f (0) ≠ 0, try to find f (π), There's a little mistake on it. It's not an equal sign, it's a plus sign. I understand it after one or two times. I'll add at least 50 points

Because any x, y ∈ r satisfies f (x + y) + F (X-Y) = 2F (x) f (y), let x = y = 0, then 2F (0) = 2 [f (0)] ^ 2, because f (0) ≠ 0, then f (0) = 1 is obtained by deleting f (0) on both sides of the above equation, so that x = y = π / 2 has f (π / 2 + π / 2) + F (0) = 2 [f (π / 2)] ^ 2 = 0, so f (π) = - 1