若x,y為實數,求所有滿足恆等式sinx+cosy=f(x)+f(y)+g(x)-g(y)的函數f(x)和g(x)?

設 sin(x)+cos(y)=f(x)+f(y)+g(x)-g(y) for all x, y in R.

即 for all x, y in R,
    (f(x)+g(x)) + (f(y)-g(y)) = sin(x)+cos(y)

故 for all x in R
    f(x) + g(x) + (f(0)-g(0)) = sin(x) + 1,
for all y in R,
    f(y) - g(y) + (f(0)+g(0) = cos(y)

第2式以 x 代  y, 而後兩式相加減, 得
    2f(x) + 2f(0) = sin(x) + cos(x) + 1
    2g(x) - 2g(0) = sin(x) - cos(x) + 1

x = 0 代入可得 f(0) = 1/2.

所以,
   f(x) = (sin(x) + cos(x))/2
   g(x) = (sin(x) - cos(x))/2 + C
式中 C 是任意常數.

sin(x) + cos(y) ≡ f(x) + f(y) + g(x) - g(y)

代 y = x 得 sin(x) + cos(x) ≡ f(x) + f(x) + g(x) - g(x) = 2 f(x)
即 f(x) = [ sin(x) + cos(x) ] / 2

故 sin(x) + cos(y) ≡ [ sin(x) + cos(x) ] / 2 + [ sin(y) + cos(y) ] / 2 + g(x) - g(y)
2 sin(x) + 2 cos(y) ≡ sin(x) + cos(x) + sin(y) + cos(y) + 2 g(x) - 2 g(y)
sin(x) + cos(y) ≡ cos(x) + sin(y) + 2 g(x) - 2 g(y)
2 g(x) - sin(x) + cos(x) ≡ 2 g(y) - sin(y) + cos(y)
即 2 g(x) - sin(x) + cos(x) 是一常數 k。
2 g(x) - sin(x) + cos(x) = k
g(x) = [sin(x) - cos(x)]/2 + k/2