Functional equation involving sine function: $ \sin x + f ( x ) = \sqrt 2 f \left( x - \frac \pi 4 \right) $ 
Let $ f : \mathbb R \to \mathbb R $ be a continuous function such that
$$ \sin x + f ( x ) = \sqrt 2 f \left( x - \frac \pi 4 \right) \text . $$
Find $ f $.

I noticed that a solution for $ f $ is the cosine function. I don't know how to continue. Is there a way I could link it to d'Alembert functional equation?
 A: Let $g(x) = f(x) - \cos x$. We have
$$\sin x + \cos x + g(x) = \sqrt 2 \cos \left( x - \frac \pi 4 \right)
+ \sqrt 2 g \left( x - \frac \pi 4 \right)$$
which yields
$$g(x) = \sqrt 2 g \left( x - \frac \pi 4 \right) \tag {*} \label { * }$$
Now note that every function $g$ satisfying \eqref{ * } gives a solution for the original equation by adding $\cos x$. There are a lot of such functions since every function $g_0 : [ 0 , \frac \pi 4 ) \to \mathbb R$ can be extended uniquely to a $g : \mathbb R \to \mathbb R$ satisfying \eqref{ * }. Further assumptions like continuity or differentiability give you some conditions on $g_0$, especially some conditions concerning the value $g_0 (x)$ near $x = 0$ and $x = \frac \pi 4$. But these assumptions are not strong enough to force $g (x) = 0$.
A: Can you assume that $f''$ exist? In this case note that is is suffices to prove that there exist a function $g$ such that 
$g(x)=\sqrt{2}g(x-\frac{\pi}{4})$ since 
$$
-\sin x+f''(x)=\sqrt{2}f''\left(x-\frac{\pi}{4}\right)
$$
Hence 
$$
(f+f'')(x)=\sqrt{2}(f+f'')\left(x-\frac{\pi}{4}\right)
$$
A: Write the eqn like this.
$${1\over \sqrt 2}\sin x+{1\over \sqrt 2}f(x)=f(x-{\pi\over 4})$$
$$\sin {\pi\over 4}\sin x+\cos {\pi\over 4}f(x)=f(x-{\pi\over 4})$$
Does it look familiar? The procedure in solving such problems is to modify values so as to make them look familiar..
