What is the $n$th derivative of $\sin(x)$?

Question

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.

Answer 1

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 $g(x)=\cos x$, $$g^{(n)}(x) =\cos\bigl(x+\tfrac{n\pi} 2\bigr).$$

Answer 2

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.

Answer 3

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 $n = 4k + 3$

$f^n(x) = -\cos x = \sin (2k\pi + \frac {3\pi}2 + x) = = \sin(n\frac{\pi}2 + x)$