<s>[INST] 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?
 [/INST] 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.
</s>