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Answer 1 · 2017-01-09T11:17:44

$\frac{d^{n}}{{d x}^{n}} cos x = cos (x + \frac{n π}{2})$ $d^n/(dx^n) cosx = cos(x+(npi)/2)$

We have that:

$\frac{d}{d x} cos x = - sin x$ $d/(dx) cosx = -sinx$

$\frac{d^{2}}{{d x}^{2}} cos x = \frac{d}{d x} (- sin x) = - cos x$ $d^2/(dx^2) cosx = d/(dx) (-sinx) = -cosx$

$\frac{d^{3}}{{d x}^{3}} cos x = \frac{d}{d x} (- cos x) = sin x$ $d^3/(dx^3) cosx = d/(dx) (-cosx) = sinx$

$\frac{d^{4}}{{d x}^{4}} cos x = \frac{d}{d x} (sin x) = cos x$ $d^4/(dx^4) cosx = d/(dx) (sinx) = cosx$

And obviously after the fifth order they start repeating.

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

$cos (x + \frac{π}{2}) = cos x cos (\frac{π}{2}) - sin x sin (\frac{π}{2}) = - sin x$ $cos(x+pi/2) =cosxcos(pi/2)- sinxsin(pi/2) = -sinx$

$cos (x + π) = cos x cos (π) - sin x sin (π) = - cos x$ $cos(x+pi) =cosxcos(pi)- sinxsin(pi) = -cosx$

$cos (x + \frac{3}{2} π) = cos x cos (\frac{3}{2} π) - sin x sin (\frac{3}{2} π) = sin x$ $cos(x+3/2pi) =cosxcos(3/2pi)- sinxsin(3/2pi) = sinx$

$cos (x + 2 π) = cos x$ $cos(x+2pi) =cosx$

So that:

$\frac{d^{n}}{{d x}^{n}} cos x = cos (x + \frac{n π}{2})$ $d^n/(dx^n) cosx = cos(x+(npi)/2)$