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If $$f(x) + f(y) =f\left(x\sqrt{1-y^2 }+y\sqrt{1-x^2 }\right)\text,$$ prove that $$f(4x^3 -3x) + 3f(x) =0\text.$$

I started by substituting $x = y$, in the expression and I get $$2f(x)=f\left(2x\sqrt{1-x^2}\right)\text.$$ Then, I also substitute $y = \sqrt{(1-x^2)}$ and I get $$f(x) +f\left(x\sqrt{1-x^2}\right) = 0\text.$$ How shall I proceed further and solve this problem?

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  • $\begingroup$ Do you want $f(x)+f(y) =f(x\sqrt{1-y^2}+y\sqrt{1-x^2})$? $\endgroup$ Jun 7 '18 at 18:47
  • $\begingroup$ @LordSharktheUnknown Sorry, I wrote it wrong. I have edited it now, $\endgroup$
    – ShiS
    Jun 7 '18 at 18:49
  • $\begingroup$ So $f$ should be $C\arcsin$? $\endgroup$ Jun 7 '18 at 18:50
  • $\begingroup$ Yes, that satisfies the equation. But how shall I prove this in general.Or how will I get this solution by solving the equation forward $\endgroup$
    – ShiS
    Jun 7 '18 at 18:51
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    $\begingroup$ @LordSharktheUnknown Ineterestingly, $\arcsin$ does not work as $f(0)+f(0)=f(0)$ implies $f(0)=0$ and then $f(1)+f(1)=f(0)$ implies $f(1)=0$. In fact, transporting the equation back to angles, one gets a (local form of) Cuachy's functional equation. Therefore the only continuous solution is $f(x)=0$. $\endgroup$ Jun 7 '18 at 20:12
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I'll assume the arguments of $f$ are always between $-1$ and $1$. Then $$2f(x)=f(2x\sqrt{1-x^2})$$ and $$3f(x)=f(x)+f(2x\sqrt{1-x^2})=f(y)$$ where $$ \begin{split} y &= x\sqrt{1-4x^2(1-x^2)}+\sqrt{1-x^2}2x\sqrt{1-x^2}\\ &= x\sqrt{1-4x^2+4x^4}+2x\left(1-x^2\right)\\ &= x\left(1-2x^2\right)+2x\left(1-x^2\right)=3x-4x^3. \end{split} $$ But $f(y)+f(-y)=f(0)=f(0)+f(0)$ and so $3f(x)=f(-y)$.

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Easy way to prove that for every real number $ x $ with $ - 1 \le x \le 1 $, $ f ( x ) = 0 $ (using some steps already mentioned in the comments, but repeated here to be complete):

Let $ x = y = 0 $ in $$ f ( x ) + f ( y ) = f \Big( x \sqrt { 1 - y ^ 2 } + y \sqrt { 1 - x ^ 2 } \Big) \tag { * } \label { * } $$ and you get $ f ( 0 ) = 0 $. Again, let $ x = y = 1 $ in \eqref{ * } and you get $ f ( 1 ) = 0 $. Thus, letting $ y = 1 $ in \eqref{ * } you'll have $ f ( x ) = f \big( \sqrt { 1 - x ^ 2 } \big) $. This also shows that $ f ( | x | ) = f ( x ) $. Now, substituting $ | x | $ for $ x $ and $ \sqrt { 1 - x ^ 2 } $ for $ y $ in \eqref{ * }, you get $ f ( x ) = 0 $.

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