# Question #3ebdb

Jan 9, 2017

${d}^{n} / \left({\mathrm{dx}}^{n}\right) \cos x = \cos \left(x + \frac{n \pi}{2}\right)$

#### Explanation:

We have that:

$\frac{d}{\mathrm{dx}} \cos x = - \sin x$

${d}^{2} / \left({\mathrm{dx}}^{2}\right) \cos x = \frac{d}{\mathrm{dx}} \left(- \sin x\right) = - \cos x$

${d}^{3} / \left({\mathrm{dx}}^{3}\right) \cos x = \frac{d}{\mathrm{dx}} \left(- \cos x\right) = \sin x$

${d}^{4} / \left({\mathrm{dx}}^{4}\right) \cos x = \frac{d}{\mathrm{dx}} \left(\sin x\right) = \cos x$

And obviously after the fifth order they start repeating.

We can however find a synthetic expression valid for all orders noting that:

$\cos \left(x + \frac{\pi}{2}\right) = \cos x \cos \left(\frac{\pi}{2}\right) - \sin x \sin \left(\frac{\pi}{2}\right) = - \sin x$

$\cos \left(x + \pi\right) = \cos x \cos \left(\pi\right) - \sin x \sin \left(\pi\right) = - \cos x$

$\cos \left(x + \frac{3}{2} \pi\right) = \cos x \cos \left(\frac{3}{2} \pi\right) - \sin x \sin \left(\frac{3}{2} \pi\right) = \sin x$

$\cos \left(x + 2 \pi\right) = \cos x$

So that:

${d}^{n} / \left({\mathrm{dx}}^{n}\right) \cos x = \cos \left(x + \frac{n \pi}{2}\right)$