Functional equation $f(r \cos \varphi)+f(r \sin \varphi)=f(r)$ 
Find all such monotonous function $f:[0,+\infty)\to \mathbb R$, such that for any real $r\ge0$ and $\varphi \in \left[\frac{\pi}6, \frac{\pi}4\right]$,
$$f(r \cos \varphi)+f(r \sin \varphi)=f(r)\text.$$

My work so far:

*

*If $\varphi \in \left[\frac{\pi}6, \frac{\pi}4\right]$ then $$\frac{\sqrt2}2\le\cos \varphi\le\frac{\sqrt3}2$$
$$\frac12\le\sin \varphi\le\frac{\sqrt2}2$$


*If $r=0$ then $f(0)=0$


*If $\varphi= \frac{\pi}{4}$ then $$f(r)=4f\left(\frac r2\right)$$


*$f(x)=cx^2$ for $x\ge0$ and $c\in\mathbb R$
 A: Consider a monotonous function $ f : [ 0 , + \infty ) \to \mathbb R $ satisfying the functional equation
$$ f ( r \cos \phi ) + f ( r \sin \phi ) = f ( r ) \tag 0 \label 0 $$
for every $ r \in [ 0 , + \infty ) $ and every $ \phi \in \big[ \frac \pi 6 , \frac \pi 4 \big] $. You can show that there is a $ c \in \mathbb R $ such that
$ f ( x ) = c x ^ 2 $ for all $ x \in [ 0 , + \infty ) $. As you've mentioned, functions of this form satisfy the above criterions, and thus they form the class of all solutions.
The trick is to observe that when you consider $ ( r , \phi ) $ as polar coordinates of a point (noting that we have $ r \ge 0 $ and that's valid), $ ( x , y ) = ( r \cos \phi , r \sin \phi ) $ will be the cartesian coordinates of the same point. In that case, having $ \frac \pi 6 \le \phi \le \frac \pi 4 $ is equivalent to having $ y \le x \le 2 y $.
So we start by considering $ x $ and $ y $ so that $ y \le x \le 2 y $. Then letting $ r = \sqrt { x ^ 2 + y ^ 2 } $, we would have $ r \in [ 0 , + \infty ) $ and there would be a $ \phi \in \big[ \frac \pi 6 , \frac \pi 4 \big] $ so that
$ x = r \cos \phi $ and $ y = r \sin \phi $. Therefore by \eqref{0} we have
$$ f ( x ) + f ( y ) = f \left( \sqrt { x ^ 2 + y ^ 2 } \right) \tag 1 \label 1 $$
for every $ x $ and $ y $ with $ y \le x \le 2 y $. Define $ g : [ 0 , + \infty ) \to \mathbb R $ by $ g ( z ) = f \left( \sqrt z \right) $. Then by \eqref{1} we have
$$ g ( x ) + g ( y ) = g ( x + y ) \tag 2 \label 2 $$
for every $ x $ and $ y $ with $ 0 \le y \le x \le 4 y $. As $ 0 \le 0 \le 0 \le 4 \cdot 0 $, we can let $ x = y = 0 $ in \eqref{2} and get $ g ( 0 ) = 0 $. We can now inductively show that
$$ g ( n x ) = n g ( x ) \tag 3 \label 3 $$
for every nonnegative integer $ n $. This holds for $ n = 0 $ and $ n = 1 $. For $ n \ge 2 $, we either have $ n = 2 m $ or $ n = 2 m + 1 $ for some nonnegative integer $ m $ less than $ n $. If $ n = 2 m $, then since $ 0 \le m x \le m x \le 4 m x $, substituting $ m x $ for both $ x $ and $ y $ in \eqref{2} we get $ 2 g ( m x ) = g ( 2 m x ) $. As $ m < n $, by induction hypothesis we have $ g ( m x ) = m g ( x ) $, which leads to $ g ( n x ) = n g ( x ) $. If $ n = 2 m + 1 $, since $ n \ge 2 $ we have $ 0 \le m x \le ( m + 1 ) x \le 4 m x $, and substituting $ ( m + 1 ) x $ for $ x $ and $ m x $ for $ y $ in \eqref{2} we get $ g \big( ( m + 1 ) x \big) + g ( m x ) = g \big( ( 2 m + 1 ) x \big) $. As $ m + 1 < n $, by induction hypothesis we have $ g \big( ( m + 1 ) x \big) = ( m + 1 ) g ( x ) $ and $ g ( m x ) = m g ( x ) $, which again leads to $ g ( n x ) = n g ( x ) $.
Now, for $ n > 0 $, we can substitute $ \frac x n $ for $ x $ in \eqref{3} and get
$$ g \left( \frac x n \right) = \frac 1 n g ( x ) \text . \tag 4 \label 4 $$
Substituting $ m x $ for $ x $ in \eqref{4} for some nonnegative integer $ m $ and using \eqref{3}, we find out that $ g ( q x ) = q g ( x ) $ for every nonnegative rational number $ q $. In particular, defining $ c = g ( 1 ) $, we have $ g ( q ) = c q $ for every nonnegative rational number $ q $. Since $ f $ is monotonous, $ g $ is monotonous, too, and hence for every nonnegative real number $ x $ between two nonnegative rational numbers $ p $ and $ q $, $ g ( x ) $ is between $ c p $ and $ c q $. As nonnegative rational numbers are dense in nonnegative real numbers, this forces $ g ( x ) $ to be equal to $ c x $, and thus $ f ( x ) = c x ^ 2 $, as desired.
