2. Determine all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ for which $f(x+y)=f(x-y)+2 f(y) \cos x$ holds for all real numbers $x$ and $y$.
IV/2. By substituting $x=0$ into the functional equation, we get $f(-y)=-f(y)$ for any real number $y$. The function $f$ is odd. We also substitute $x=\frac{\pi}{2}$ and derive $f\left(y+\frac{\pi}{2}\right)=f\left(\frac{\pi}{2}-y\right)$ for all real numbers $y$. Considering the oddness, we get $f\left(y+\frac{\pi}{2}\right)=-f\left(y-\frac{\pi}{2}\right)$. From the original equation for $y=\frac{\pi}{2}$, it follows that

$$
f\left(x+\frac{\pi}{2}\right)=f\left(x-\frac{\pi}{2}\right)+2 \cdot f\left(\frac{\pi}{2}\right) \cdot \cos x
$$

Since $f\left(x+\frac{\pi}{2}\right)=-f\left(x-\frac{\pi}{2}\right)$, we have $f\left(x+\frac{\pi}{2}\right)=f\left(\frac{\pi}{2}\right) \cdot \cos x$ or

$$
f(x)=f\left(\frac{\pi}{2}\right) \cdot \cos \left(x-\frac{\pi}{2}\right)=f\left(\frac{\pi}{2}\right) \cdot \sin x
$$

Let $a=f\left(\frac{\pi}{2}\right)$. Then $f(x)=a \cdot \sin x$. We can verify that this function satisfies the functional equation for any real number $a$.

![](https://cdn.mathpix.com/cropped/2024_06_07_c45646d9586b6df064a6g-14.jpg?height=66&width=1645&top_left_y=1800&top_left_x=208)

![](https://cdn.mathpix.com/cropped/2024_06_07_c45646d9586b6df064a6g-14.jpg?height=63&width=1633&top_left_y=1853&top_left_x=220)

![](https://cdn.mathpix.com/cropped/2024_06_07_c45646d9586b6df064a6g-14.jpg?height=58&width=1633&top_left_y=1907&top_left_x=220)

![](https://cdn.mathpix.com/cropped/2024_06_07_c45646d9586b6df064a6g-14.jpg?height=60&width=1633&top_left_y=1957&top_left_x=220)

Conclusion: $f(x)=f\left(\frac{\pi}{2}\right) \cdot \sin x$ or $f(x)=a \cdot \sin x \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$

It is verified that the functions $f(x)=a \cdot \sin x$ are solutions (this point is also recognized if the contestant guesses the solution $f(x)=a \cdot \sin x$) 2 points (If the contestant guesses and verifies at least one of the solutions $f(x)=0$ or $f(x)=\sin x$, they get 1 point).