Is the $n^{th}$ derivative of $\sin(x)$ just a translation of $\sin(x)$? I noticed that
$$\frac d{dx}\sin x=\cos x=\sin\left(x+\frac\pi2\right)$$
$$\frac{d^n}{dx^n}\sin x=\sin\left(x+\frac{\pi n}2\right)$$
Does this hold for any positive real value of $n$? 
If so, does anybody have any reasoning behind why it's just a simple translation?
 A: Yes, you're right about the derivatives (although, as pointed out in the comments and other answers, only for natural derivatives, i.e. differentiating a natural number of times (I guess you can generalise to integers if you keep choosing the right constant term for each antidifferentiation)).
As for the reason why, I like to think of $\sin$ as one part of a whole, namely as one coordinate of circular motion at unit speed along the unit circle in the plane. The velocity is always perpendicular to the position vector, so it rotates at the same rate, but stays $90^\circ$ ahead. The velocity vector rotates in a circular motion, which means that the acceleration does so too, only it's $90^\circ$ ahead of the velocity vector. The jerk vector is $90^\circ$ ahead of the acceleration vector, and so on.
A: No, it is not.  For example, the fractional derivative of order $1/2$ is, according to Maple, 
$$ \sqrt {2}\cos \left( x \right) {\it FresnelC} \left( {\frac {\sqrt {2x}
}{\sqrt {\pi}}} \right) +\sqrt {2}\sin \left( x \right) {\it 
FresnelS} \left( {\frac {\sqrt {2x}}{\sqrt {\pi}}} \right) 
$$
EDIT: There are indeed several definitions.  In the one being used here,  for $0 < n < 1$, 
$$D^n f(x) = \frac{1}{\Gamma(1-n)} \int_0^x (x-t)^{-n} f'(t)\; dt $$
A: Indeed, we know that
$$
\frac{d}{dx} \sin{x} = \cos{x} = \sin{\left( x + \frac{\pi}{2} \right) }
$$
and then
$$
\frac{d^2}{dx^2} \sin{x} = \cos{\left(x + \frac{\pi}{2} \right)}= \sin{\left(x + \frac{2\pi}{2} \right)}
$$
Proof by induction:
Let's assume that 
$$
\frac{d^k}{dx^k} \sin{x} = \sin{\left( x+ \frac{k\pi}{2} \right)}
$$
for some integer $k$. We have already seen that this is true for $k=0,1,2$.
Then we take $k$ to $k+1$:
$$
\frac{d^{k+1}}{dx^{k+1}} \sin{\left(x \right)}= \frac{d}{dx} \sin{\left( x+ \frac{k\pi}{2} \right)} = \cos{\left(x + \frac{k\pi}{2} \right)} 
$$
$$
= \sin{\left(x + \frac{\pi}{2} + \frac{k\pi}{2} \right)}= \sin{\left( x + \frac{(k+1)\pi}{2}\right)}
$$
which shows that the assumption is correct for whole numbers.
