Is there a function satisfying the following equation: $f(\sin x)+f(\cos x) = \frac{\tan x}{2}$?

More general than gimusi's answer, notice that

$$\sin(x)=\cos(\pi/2-x)\\\cos(x)=\sin(\pi/2-x)$$

And so

$$f(\sin(x))+f(\cos(x))=f(\sin(\pi/2-x))+f(\cos(\pi/2-x))$$

But

$$\tan(x)\ne\tan(\pi/2-x)$$

It is not possible, infact:

for $x=0: f(0)+f(1)=0$

for $x=\frac{\pi}{2}$: $f(1)+f(0)$ = RHS is not defined