1. Prove result for functions?

Let $f(x) + f(y) = f\left(x\sqrt{1 - y^2} + y\sqrt{1 - x^2}\right)$. Prove that $f(4x^3 - 3x) + 3f(x) = 0$.

2. My 2 cents on this question:

We can write 3f(x) = 4f(x) - f(x) and move the -f(x) to the right hand side.

Let's simplify 4f(x) first, and ultimately the entire left hand side.

$2f(x)= f(x) + f(x) = f\left(x\sqrt{1 - x^2} + x\sqrt{1 - x^2}\right)$
so
$2f(x) = f(2x\sqrt{1-x^2})$
$4f(x) = 2f(x)+2f(x) = f(2x\sqrt{1-x^2}) + f(2x \sqrt{1-x^2}$
$4f(x)=f (2x\sqrt{1-x^2}\sqrt{1-(2x\sqrt{1-x^2})^2}+2x\sqrt{1-x^2}\sqrt{1-(2x\sqrt{1-x^2})^2})$

$4f(x) = f(4x\sqrt{1-x^2}\sqrt{4x^4-4x^2+1})=f(4x\sqrt{(1-x^2)(4x^4-4x^2+1)})$
$= f(4x\sqrt{1-5x^2+8x^4-4x^6}) =f(4x(2x^2-1)\sqrt{1-x^2}) = f((8x^3-4x)\sqrt{1-x^2})$

Now we need to combine this with f(4x^3 -3x):

$f(4x^3-3x) + 4f(x) = f(4x^3-3x) +f((8x^3-4x)\sqrt{1-x^2})$

Let $A=4x^3-3x$
$B = (8x^3-4x)\sqrt{1-x^2}$

We need:

$1-A^2 = (1-x^2) (2x-1)^2 (2x+1)^2$

$1-B^2 = (8x^4 -8x^2+1)^2$

$\sqrt{1-A^2} = \sqrt{(1-x^2)} (2x-1) (2x+1)$
$\sqrt{1-B^2} = (8x^4 -8x^2+1)$

Remember that

$f(A) + f(B) =f(A \sqrt{1- B^2}+ B \sqrt{1- A^2}$

So we have:

$f(A) + f(B) = f((4x^3-3x)(8x^4-8x^2+1)+(8x^4-8x^2+1)^2(2x-1)(2x+1)\sqrt{1-x^2})$

The left hand side argument, assuming 1-1, should equal the argument of the right hand side, x.

3. Originally Posted by fardeen_gen
Let $f(x) + f(y) = f\left(x\sqrt{1 - y^2} + y\sqrt{1 - x^2}\right)$. Prove that $f(4x^3 - 3x) + 3f(x) = 0$.
Put $x=y=0 , f(0)+f(0)=f(0),i.e. f(0)=0$

Put $y=-x , f(x)+f(-x)=f(0)=0,i.e. f(-x)=-f(x)$

Put $y=x , 2f(x)=f\left(2x\sqrt{1-x^2}\right)$

$3f(x)=f(x)+f\left(2x\sqrt{1-x^2}\right)$

$3f(x)=f\left(x\sqrt{1-4x^2(1-x^2)}+2x\sqrt{1-x^2}.\sqrt{1-x^2}\right)$

$3f(x)=f\left(x(1-2x^2)+2x(1-x^2)\right)$

$3f(x)=f(3x-4x^3)$

$3f(x)=-f(4x^3-3x)$

$f(4x^3-3x)+3f(x)=0$