# A functional relation which is satisfied by $\cos x$ and $\sin x$

Assume that the functions $f,g : \mathbb R\to \mathbb R$ satisfy the relations \begin{align} \left\{ \begin{array}{ll} f(x+y) &=& f(x)f(y)-g(x)g(y), \\ g(x+y) &=& f(x)g(y)+f(y)g(x), \end{array} \right. \tag{$\star$} \end{align} for all $x,y\in \mathbb R$.

1. Find $f,g$, if they are continuous.

2. Is there a pair functions $f,g$ satisfying $(\star)$ which are not continuous?

3. Characterization of all such pairs.

Update. This problem is clearly harder than the characterization of functions satisfying $$f(x+y)=f(x)+f(y),$$ as it is a system of non-linear functional equations. I imagine that, in order to attack this problem, one has to make it linear.

• As these relations are satisfied by cos and sin, as you say, I am sure that they can be written in the form $A(x+y)=A(x)A(y)$ where $A(x)=\begin{pmatrix}f(x)&-g(x)\\g(x)&f(x)\end{pmatrix}$. – Harald Hanche-Olsen Dec 20 '13 at 17:21
• Something to do with the Wronskian of $f,g$ – Torsten Hĕrculĕ Cärlemän Dec 20 '13 at 17:22
• The equations are also satisfied by $f(x)=g(x)=0$ – Mark Bennet Dec 20 '13 at 17:26
• Harald already gave you the idea to turn your problem into pretty much the problem $f(x+y)=f(x)+f(y)$. Call $h(x):=f(x)+ig(x)$ (this is the essentially the same as Harald is saying). Then the equations $(\star)$ are equivalent to $h(x+y)=h(x)h(y)$. Now, you can find that all continuous solutions of this are $h(x)=e^{Cx}$, or $h(x)=0$. – OR. Dec 20 '13 at 17:30
• For 2. you can use a discontinuous solution of $r(x+y)=r(x)+r(y)$ and put $h(x):=e^{ir(x)}$, i.e. $f(x)=\sin(r(x))$ and $g(x)=\cos(r(x))$. – OR. Dec 20 '13 at 17:35

Let me put together all in an answer.

Define $h(x):=g(x)+if(x)$. Then $(\star)$ is equivalent to $$h(x+y)=h(x)h(y).$$

If $h(x)=0$ for some $x$ then $h$ is identically zero. Assume that $h$ never vanishes.

Let $\{a_i\}_{i\in I}$ be a Hamel basis of $\mathbb{R}$ as $\mathbb{Q}$-vector space.

Put $b_i:=h(a_i)$ and define $\hat{h}(x):=e^{r(x)}$, where $r(x)$ is the unique function determined by $r(x+y)=r(x)+r(y)$, and $r(a_i)=c_i$, where $c_i$ is any number such that $e^{c_i}=b_i$, and in the sequel I will not use anything beyond this).

Then $u(x):=h(x)/\hat{h}(x)$ satisfies $u(x+y)=u(x)u(y)$, and $u(0)=1$.

If $x=\sum\frac{p_i}{q_i}a_i$, with $(p_i,q_i)=1$, then $u(x)^{\prod q_i}=\prod u(\frac{p_i}{q_i}a_i)^{\prod q_i}=1$. The last equality is because $u(\frac{p_i}{q_i}a_i)^{q_i}=u(p_ia_i)=u(a_i)^{p_i}=1$

Then $u$ is a character (group homomorphism from $\mathbb{R}$ to the unit circle).

Converselly, for every $r$ such that $r(x+y)=r(x)+r(y)$ and a character $u$ of $\mathbb{R}$ we get $h(x):=e^{r(x)}u(x)$, which satisfies $h(x+y)=h(x)h(y)$. This gives us $$f(x):=\text{Im}(h(x))$$ and $$g(x):=\text{Re}(h(x)).$$

Now, to finish, we need to 'compute' all characters of the additive group $\mathbb{R}$.

Notice that each $\frac{p}{q}\mapsto u(\frac{p}{q}a_i)$ is a character of the additive group $\mathbb{Q}$. And conversely, if $u_i$ are characters of $\mathbb{Q}$ then $$u(x)=u\left(\sum\frac{p_i}{q_i}a_i\right):=\prod u_i\left(\frac{p_i}{q_i}\right)$$ satisfies $u(x+y)=u(x)u(y)$, and $u(0)=1$.

So, what we need is to 'compute' all characters of $\mathbb{Q}$. This can be read here, for example.

Notice the above gives us a way to produce all the solutions $h$, but it is not a unique writing of them. If we get two $\hat{h}$'s which quotient is a character of $\mathbb{R}$ then we can obtain the same $h$ from them by dividing by this character.

• I agree that these pairs are indeed solutions of the functional equation. But, are you sure that: All solutions of this equation can be obtained by defining it arbitrarily on a Hamel basis of $\mathbb{R}$ as a $\mathbb{Q}$-vector space. For each of these solutions we get $$f(x):=\text{Im}(e^{r(x)})$$ and $$g(x):=\text{Re}(e^{r(x)}).$$ – Yiorgos S. Smyrlis Dec 20 '13 at 18:37
• Two criticisms: Your definition of the logarithm seems weird. Why does the interval for the angles only have length $\pi$? It certainly isn't inverse to the exponential function. But even if the definition were right, there is no way you can define a complex logarithm in such a way that it always satisfies $log(x y) = log(x) + log(y)$ for all $x,y \in \mathbb{C}$. – Brusko651 Dec 20 '13 at 19:16
• @YiorgosS.Smyrlis Ok, I think I finished writing it. – OR. Dec 20 '13 at 21:31