Solve functional equation $f(x)+f(y)=f(x\sqrt{1-y^2}+y\sqrt{1-x^2})$

Question

Abstract

I have viewed this questions here, in which the author asked to prove that $f(4x^3-3x)+f(x)=0$ for all $x \in \mathbb{R}$. Scanning through the comments, I saw one trying to figure out the general solution, which suggests that \begin{align} f(x)=C\times \arcsin(x) \end{align} But no answer was proposed, so I suggest the following problem:

Problem

Find the continuous function $f: [-1,1] \to \mathbb{R}$, such that \begin{align} f(x)+f(y)=f(x\sqrt{1-y^2}+y\sqrt{1-x^2}) \end{align}

Approach

My initial idea was to use the Cauchy functional equation schemes to expand the condition suggested in the abstract. But no solution has been yielded.

Any help is appreciated

Answer 1

$x=y=0\;\Longrightarrow\;f(0) = 0$

$x=y=1\;\Longrightarrow\;2f(1) = f(0) = 0\;\Longrightarrow\;f(1)=0$

$x=\sin t,\,y=\cos t\;\Longrightarrow\;f(\sin t)+f(\cos t) = f(1) = 0$

$x=y=\cos t\;\Longrightarrow\;2f(\cos t) = f(2\sin t\cos t)$

$x=y=\sin t\;\Longrightarrow\;2f(\sin t) = f(2\sin t\cos t)$

Hence, $f(\sin t) = f(\cos t) = -f(\sin t)$ and so the only solution is $f=0$.