Prove by induction that the $n$-th derivative of $\cos(x)$ is given by $\cos\left(x+n\frac{\pi }{2}\right)$

Question

Given the function $f\left(x\right)=\cos(x)$. Prove by induction that for $\forall n\in \mathbb{N}$ $f^{(n)}=\cos\bigl(x+n\cdot \frac{\pi }{2}\bigr)$. Being that $f^{\left(1\right)}$ is the first derivative and so forth.

Answer 1

Well as requested, i will detail a little more the sketch of the proof. Induction proof is like domino's : you need to check that the first domino falls ( true for $n=0$ here ) and then you have to check that the $n-$th domino makes the $n+1-$th fall. Here : check the case $n=0$. You have $$ \cos\left(x+0.\frac{\pi}{2}\right)=\cos\left(x\right)=f^{\left(0\right)}\left(x\right) $$ So the given relation is true. Now suppose that it is true until a certain rank $n$, then you need to consider $f^{\left(n+1\right)}(x)$. You have $$ f^{\left(n+1\right)}(x)=\left(f^{\left(n\right)}\left(x\right)\right)'=\left(\cos\left(x+n\frac{\pi}{2}\right)\right)'=-\sin\left(x+n\frac{\pi}{2}\right)=\cos\left(x+\frac{\pi}{2}+n\frac{\pi}{2}\right) $$ The last equality comes from $\displaystyle \cos\left(x+\frac{\pi}{2}\right)=-\sin\left(x\right)$ ( check a trigonometric circle to see this ). Hence we've shown

$$ f^{\left(n+1\right)}(x)=\cos\left(x+\left(n+1\right)\frac{\pi}{2}\right) $$

The relation is hence TRUE at rank $n+1$. By induction the relation is true for ALL $n$.

Answer 2

Hint:

You just have to know that the first derivative, $f'(x)=-\sin x$, can be written as $$f'(x)=\sin\bigl(x+\tfrac\pi 2\bigr)$$ from a well-known trigonometry formula, then an easy induction will complete the proof.

Answer 3

Here is a proof that doesn't use a recurrence.

I give it because it can be rather appealing under the condition to have some practice with complex numbers.

Let us start from the famous De Moivre formula :

$$\cos(x)+i \sin(x)=e^{ix} \tag{1}$$

$n$ times differentiation of both sides gives :

$$\cos^{(n)}(x)+i \sin^{(n)}(x)=i^n e^{ix}\tag{2}$$

The RHS of (2) can be transformed

$$\cos^{(n)}(x)+i \sin^{(n)}(x)=e^{i\tfrac{\pi}{2}n}e^{ix}=e^{i(x+\tfrac{\pi}{2}n)}\tag{3}$$

Using anew De Moivre formula, we get :

$$\cos^{(n)}(x)+i \sin^{(n)}(x)=\cos(x+\tfrac{\pi}{2}n)+i\sin(x+\tfrac{\pi}{2}n)\tag{4}$$

Taking the real part of both sides terminates the proof.