Split $\cos(x-y)$ into $\phi(x)\phi(y)$

Question

As written in the topic I try to find a function $\phi$ that satisfies split $\cos(x-y) = \phi(x)\phi(y) $ $x,y \in \mathbb{R} $.

So far I'm stuck with: $\phi(x) = \cos(x)+ \sin(x)$

$(\cos(x)+ \sin(x)) \cdot (\cos(x)+ \sin(x)) = \cos(x-y) + \sin(x+y)$

Answer 1

No such function exists. We'd have that for each $x\in \Bbb R$

$$1 = \cos(0) = \cos(x-x) = \phi(x)^2 \implies \phi(x) = \pm 1$$

But of course, $\cos(x-y)$ can assume other values.

Answer 2

Suppose that there is such function.

Then $\phi(x)^2 = \cos 0 = 1 \implies |\phi(x)| = 1$. So pick any solution of the equation $x-y = \frac \pi 4$. Then $|\cos \frac \pi 4| = |\phi(x)\phi(y)| = 1$ and finally $\frac{\sqrt 2}{2} = 1$.

Therefore, there is no such $\phi$.

Answer 3

Any function $g(x,y)$ of the form $g(x,y) = \phi(x)\phi(y)$ has the property that $$ \left|\begin{aligned} g(a,a)\quad&g(a,b)\\ g(b,a)\quad&g(b,b) \end{aligned}\right| = \left|\begin{aligned} \phi(a)\phi(a)\quad&\phi(a)\phi(b)\\ \phi(b)\phi(a)\quad&\phi(b)\phi(b) \end{aligned}\right| = \phi(a)^2\phi(b)^2 - \phi(a)^2\phi(b)^2 =0 $$ for any values $a,b$. Now try $a=0,b=\pi/2$ using $g(x,y) = \cos(x-y)$: $$ \left|\begin{aligned} \cos(0-0)\quad&\cos(0-\pi/2)\\ \cos(\pi/2-0)\quad&\cos(\pi/2-\pi/2) \end{aligned}\right| = \left|\begin{aligned} 1\quad&0\\ 0\quad&1 \end{aligned}\right| = 1 $$ Therefore, $\cos(x-y)$ cannot be written in that form.

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Remark
The standard product to sum formula is $$ \cos(x-y) + \cos(x+y) = 2 \cos x \cos y $$