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I was asked to show that the $n$-th derivative of $\cos x$ is $\cos(\frac{n\pi}{2} + x)$.

My progress :

By induction, I proved it was true for $n=1$. Then I assumed it was true for $n = k$ so now I had to prove it was true for $n=k+1$. $$ \cos\left(\frac{\pi(k+1)}{2} + x \right)$$ $$ = \cos\left(\frac{k\pi+\pi}{2} + x \right)$$ $$ = \cos\left(\frac{k\pi}{2} + \frac{\pi}{2} + x \right).$$

Can somebody please assist me into proving my answer? Thanks.

Asaf Karagila
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  • Can a moderator please delete the question? I have understood the answer. –  Aug 28 '14 at 15:40
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    No; someone may want to view the question for help with similar issues. The site isn't just for answering questions then removing them. – Jam Aug 28 '14 at 15:42
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    No, the question stays. Moderators have been notified of the defacing you are attempting. – Andrés E. Caicedo Aug 28 '14 at 15:43
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    @user108104 That's got to be the third time you've defaced it, right? The site isn't just about you. Thanks for getting a mod, Andres. – Jam Aug 28 '14 at 15:44
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    @user108104 why are you so determined to have your question removed from the site, even after other users have nicely explained why your question should stay up? – vociferous_rutabaga Aug 28 '14 at 15:45
  • Because I would like to ask a question similar to this one so it will be a duplicate. –  Aug 28 '14 at 15:46
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    @user108104 when a moderator arrives he will restore the question and lock it so you cannot vandalize it. You should accept this with grace, or you may end up getting formally reprimanded or worse. – Adam Hughes Aug 28 '14 at 15:47
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    @user108104 if the question is dissimilar enough to warrant a separate post, then there will be no problem with that. Just be sure to emphasize how your second question differs from this one in your question, and perhaps link this one. Your current behavior is likely to lead to a situation where you can no longer ask any questions. – vociferous_rutabaga Aug 28 '14 at 15:48
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    @user108104 If you have another question, ask it. If it's the same question, don't. Seems simple, no? Don't remove information because it's not useful to you (the site is for archiving answers, not just writing them). – Jam Aug 28 '14 at 15:48

5 Answers5

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Next use the identity for $\cos(a+b)$, where here $a = \frac{k \pi}{2} + x$ and $b = \frac{\pi}{2}$. Can you take it from there?

rogerl
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  • Well I then went off to say that it follows by $$cos(\frac{k\pi+\pi}{2}).cosx -sin(\frac{k\pi+\pi}{2}).sinx$$ . I don't know what to do next – Aspiring Mathlete Aug 28 '14 at 14:43
  • You need to read what I said again. Group the terms differently before applying the identity. – rogerl Aug 28 '14 at 14:44
  • Oh sorry. I meant to say that I got $$ -sin(\frac{k\pi}{2}+x) $$ and I do not know the value of $k$, so I wouldn't know how to reduce it. – Aspiring Mathlete Aug 28 '14 at 14:48
  • OK, so what are you trying to show? What is the $k^{\mathrm{th}}$ derivative of $\cos x$ according to the inductive hypothesis? You want to show that the expression you just got is equal to the first derivative of that, right? – rogerl Aug 28 '14 at 14:50
  • I am trying to prove that it is true for $n = k+1 $ – Aspiring Mathlete Aug 28 '14 at 14:51
  • Right. So saying that it is true for $n=k+1$ means that the expression you got is equal to the $(k+1)^{\mathrm{st}}$ derivative of $\cos x$. You can compute that derivative by taking the derivative of the $k^{\mathrm{th}}$ derivative --- and you know what the $k^{\mathrm{th}}$ derivative is by the inductive hypothesis. – rogerl Aug 28 '14 at 14:53
  • I don't know what the kth derivative is. I assumed it was true to work out what $ (k+1)st $ derivative of cos$x$ which is what I have shown above. – Aspiring Mathlete Aug 28 '14 at 14:57
  • Exactly. You are using induction, and that is your inductive hypothesis (assumption). You are assuming that the $k^{\mathrm{th}}$ derivative is $\cos\left(\frac{k \pi}{2} + x\right)$. – rogerl Aug 28 '14 at 14:59
  • So I have basically finished the proof? – Aspiring Mathlete Aug 28 '14 at 15:01
  • Why don't you post your entire proof as an edit to your question, and then we can continue if necessary. – rogerl Aug 28 '14 at 15:56
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Let the result be true for $n=k$

Therefore, the $k$-th derivative of $\cos x$ is $\cos (\frac{k\pi}{2}+x)$

So, the $(k+1)$-th derivative is \begin{align} -\sin\left(\frac{k\pi}{2}+x\right)\cdot 1 &=-\sin \left(\frac{(k+1)\pi}{2}-\frac{\pi}{2}+x\right)\\ &=\sin\left(\frac{\pi}{2}-\frac{(k+1)\pi}{2}-x\right) && \text{(Since $\sin(-x)=-\sin x$)}\\ &=\cos \left(\frac{(k+1)\pi}{2}+x\right) \end{align}

Hence the proof, by induction.

egreg
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Diya
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If you are familiar with complex numbers, here is another point of view : for $z \in \mathbb{C}$, $\Re(z)$ denotes the real part of $z$.

Note that, for all $x$ in $\mathbb{R}$, $\cos(x) = \Re \big( e^{ix} \big)$.

If $f$ denotes the function $x \, \mapsto \, e^{ix}$, for $n \in \mathbb{N}$, $\displaystyle f^{(n)}(x) = i^{n} e^{ix} $ ($f^{(n)}$ is the $n$-th derivative of $f$). Since $i = e^{i \frac{\pi}{2}}$, you get :

$$ f^{(n)}(x) = e^{ix + in\frac{\pi}{2}} $$

Taking the real part gives :

$$ \Re \big( f^{(n)} \big)(x) = \cos \big( x + n\frac{\pi}{2} \big) $$

As a consequence,

$$ \cos^{(n)}(x) = \Re \big( f^{(n)} \big)(x) = \cos \big( x + n\frac{\pi}{2} \big). $$

pitchounet
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You should also be careful about what you are actually trying to prove. The inductive hypothesis should be

$$\textrm{Assume that } \frac{d^k}{dx^k}\left(\cos x\right) = \cos\left(\frac{k\pi}{2} + x\right).$$

Differentiate both sides, then try to rewrite the resulting right-hand side as you mention in your question to get it into the desired form.

MPW
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You could set $y=x+\frac {n\pi}2$ and use the chain rule to prove the inductive step.

Mark Bennet
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