What is the $n$th derivative of $\sin(x)$? I don't understand why for $f(x) = \sin(x)$, $\;f^{(n)}(x) = \sin(n\pi/2 + x)$ 
It does not make sense to me; can anyone help with the reasoning?
Thank you. 
 A: The 1st derivative is, as you know, $\;f'(x)=\cos x$ and it is known from basic trigonometry that $$\cos x =\sin\bigl(x+\tfrac\pi 2\bigr),$$
whence, by an easy induction,
$$f^{(n)}(x) =\sin\bigl(x+\tfrac{n\pi} 2\bigr).$$
Note:
It is proved in the same way that, if $g(x)=\cos x$,
$$g^{(n)}(x) =\cos\bigl(x+\tfrac{n\pi} 2\bigr).$$
A: You can reason by induction. For $n=0$, it is trivially true. Assume $f^{(k)}(x)=\sin\left(\frac{k\pi}{2}+x\right)$. Then differentiating, we get $$f^{(k+1)}(x)=\cos\left(\frac{k\pi}{2}+x\right)=\sin\left(\frac{k\pi}{2}+\frac{\pi}{2}+x\right)=\sin\left(\frac{(k+1)\pi}{2}+x\right),$$ which completes our induction.
A: Part I:
$\sin'(x) = \cos (x)$ and $\cos'x = -\sin x$.
If we iterate these $n$ times then for $f(x) = \sin x$ if 
$n = 4k$ for some $k$ then $f^n(x) =\sin x$.
If $n = 4k + 1$ then $f^n(x) = \cos x$
If $n = 4k + 2$ then $f^n(x) = -\sin x$
And if $n = 4k + 3$ then $f^n(x) = -\cos x$.
.....
Now lets walk away and notice.
Part II:
If we have a $\sin x$ and $\cos x$ and we rotate $x$ ninety degrees or by $\frac {\pi}2$ then the result is $\sin \mapsto \cos x$ and $\cos x \mapsto -\sin x$ so $\sin(\frac {\pi}2 + x) = \cos x$.
We rotate again and we get $\sin(\pi + x) = -\sin x$.
And again $\sin(\frac {3\pi}2 + x) = -\cos x$.
And again $\sin (2\pi + x) = \sin x$.
Iterating it:
$\sin (n\frac {\pi}2 + x) = :$
Well if  $n = 4k$ for some $k$ then $\sin(n\frac \pi 2 + x) = \sin(4k\frac\pi 2 + x)=\sin(2k\pi + x) = \sin x$.
If $n = 4k + 1$ then $\sin(n\frac \pi 2 + x) = \sin((4k+1)\frac\pi 2 + x)=\sin(2k\pi+\frac \pi 2 + x)= \sin(\frac \pi 2 + x) = \cos x$.
If $n = 4k + 2$ then $\sin(n\frac \pi 2 + x) = \sin((4k+2)\frac\pi 2 + x)=\sin(2k\pi+\pi + x)= \sin( \pi  + x) = -\sin x$
And if $n=4k + 3$ then $\sin(n\frac \pi 2 + x) = \sin((4k+3)\frac{3\pi} 2 + x)=\sin(2k\pi+\frac {3\pi 2} + x)= \sin(-\frac \pi 2 + x) = -\cos x$
....
Now we compare Part I to Part II we see
If $n = 4k$ then
$f^n(x)  = \sin x= \sin (2k\pi + x) = \sin(n\frac{\pi}2 + x)$
If $n = 4k + 1$ 
$f^n(x) = \cos x = \sin (2k\pi + \frac \pi 2 + x)  = \sin(n\frac{\pi}2 + x)$
If $n = 4k + 2$
$f^n(x) = -\sin x = \sin (2k\pi + \pi + x) = = \sin(n\frac{\pi}2 + x)$
If $n = 4k + 3$
$f^n(x) = -\cos x = \sin (2k\pi + \frac {3\pi}2 + x) = = \sin(n\frac{\pi}2 + x)$
