Find all continuous functions $f$ which satisfy the functional equation $$ f(x)\,f(y)-f(x+y)=\sin x\,\sin y, $$ for all $x,y\in\mathbb R$.

I can prove that $f(n\pi)=\cos\left(n\pi\right)$ for all $n\in\mathbb Z$.

First attempt. I have tried to prove that: $$ f\left(\frac{\pi}n\right)=\cos\left(\frac{\pi}n\right),\quad \text{for all}\,\,\, n\in\mathbb Z\smallsetminus\{0\} \tag{1}, $$ but I have failed.

If I prove $(1)$, then the functional equation will be solved completely using the continuity of $f$.

So how do we solve this functional equation?

  • $\begingroup$ Proving the claim would not prove all of it. How do you figure? $\endgroup$ Dec 20, 2013 at 6:32
  • $\begingroup$ Note that $f(x)f(kx)+f((k+1)x)=\sin x\sin kx$ then $x=\frac{\pi}n$ we deduce $f(\frac kn\pi)=\cos(\frac kn\pi)$ then using continuity. $\endgroup$
    – Hai Minh
    Dec 22, 2013 at 14:24

2 Answers 2


Setting $x=y=0$, we obtain that $\,\,f(0)f(0)-f(0)=0$, and thus $\,\,f(0)=0$ or $1$.

If $f(0)=0$, then setting $y=0$, we get $-f(x)=0$, which is a contradiction. Therefore $f(0)=1$.

Set $y=-x$, and get $$ f(x)f(-x)=1-\sin^2 x=\cos^2x. $$ Letting above $x=\pi/2$ we get that $f(\pi/2)=0$ or $f(-\pi/2)=0$, while letting $y=\pi/2$ or $-\pi/2$ in the original relation we get, respectively: $$ -f(x+\pi/2)=\sin x, $$ or $$ -f(x-\pi/2)=-\sin x, $$ which in both cases imply that $$ f(x)=\sin(x+\pi/2)=\cos x. $$

Note. The continuity of the function $f$ has not been used in the proof.


If you want a much more heavy-handed solution which uses continuity: first show that the function is periodic (easy, using growth at infinity). Then, a periodic function can be expanded (uniquely) in a Fourier series, from which the result follows by equating left and right sides.

  • $\begingroup$ Can you elaborate on the last part? How does the result follow by equating the left and right sides? $\endgroup$ Dec 23, 2013 at 2:09
  • $\begingroup$ Well, the sines/cosines are a basis for $L^2,$ which means there is uniqueness of representation. $\endgroup$
    – Igor Rivin
    Dec 23, 2013 at 15:32
  • 2
    $\begingroup$ Yes I understand that fact. But then what? We can prove that the function is periodic and has a period $\pi$, so we have a unique representation $f(x) = a_0 + \sum_{n} (a_n\cos(2nx) + b_n\sin(2nx))$. From this and the functional equation, how do you conclude that $f(x) = \cos x$? $\endgroup$ Dec 23, 2013 at 21:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.