Is Cosine the only function satisfying $ f'(x)= f(x+\frac{\pi}{2})$? Basically most of us know that $\frac {\textrm{d}}{\textrm{d}x} \cos x = -\sin x$ . Also $ \cos (x+\frac{\pi}{2})=-\sin x$
That makes $$ \frac {\textrm{d}}{\textrm{d}x} \cos x = \cos (x+\frac{\pi}{2}) $$ So, out of curiosity I wondered what other functions could have this property. It's pretty interesting that by translating the graph of a function by some vector we get the graph of its derivative. 
Therefore, here's my attempt:  I tried to find the defining property of this function. Here we go:
$$ f'(x)= f(x+\frac{\pi}{2}) $$ That makes
$$ f''(x)= \left( f'(x)\right)'= f(x+\frac{\pi}{2})'=f'(x+\frac{\pi}{2})=f(x+\pi)$$
Generally $$f^n(x)=f(x+n\frac{\pi}{2})$$
With $f^n(x) $ being the $n$-th derivative of $f$.
 And I'm stuck at this step. What makes cosine adapt to this property is that $ \cos (x+\pi)=-\cos x$ which allows it to be the solution of the following differential equation:
$$f''(x)+f(x)=0$$
Though I don't want to admit that $f(x+\pi)=-f(x) $  So, could you please help me? Isn't there any way to determine all the functions that satisfy the previously acknowledged property? Are they infinite?

We have:
$$\sin(x+\pi/2)=-\sin(\pi/2-x)=\cos x=(\sin x)'$$
Actually, $\cos(x+a)$ for any $a$ will work.
Applying Fourier transform rules, your original equation becomes:
$$i\omega \hat f(\omega)=e^{i\omega \pi/2}\hat f(\omega)$$
So, when $i\omega \neq e^{i\omega\pi/2}$, $\hat f$ must be zero. The set of solutions to $i\omega =e^{i\omega\pi/2}$ is discrete, and contains only $-1$ and $1$ on the real line. Are there complex solution? I don't know.
The result is that $\hat f$ must be a linear combination of delta functions at the roots $\omega_1,\omega_2,...$, an thus that $f$ is a linear combination of $e^{i\omega_k x}$, The cases $\omega=1,-1$ give $\sin x$ and $\cos x$.
(This corresponds to the comment given above, where $\frac{\ln \lambda}{\lambda}=\frac{\pi}{2}$, by setting $\lambda =i\omega$.)
