Showing the $n$-th derivative of $\cos x$ by induction

Question

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.

Answer 1

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?

Answer 2

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.

Answer 3

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

Answer 4

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,

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

Answer 5

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.