Question #3ebdb

Question

1364 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-09T11:17:44

$d^n/(dx^n) cosx = cos(x+(npi)/2)$

We have that:

$d/(dx) cosx = -sinx$

$d^2/(dx^2) cosx = d/(dx) (-sinx) = -cosx$

$d^3/(dx^3) cosx = d/(dx) (-cosx) = sinx$

$d^4/(dx^4) cosx = d/(dx) (sinx) = cosx$

And obviously after the fifth order they start repeating.

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

$cos(x+pi/2) =cosxcos(pi/2)- sinxsin(pi/2) = -sinx$

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

$cos(x+3/2pi) =cosxcos(3/2pi)- sinxsin(3/2pi) = sinx$

$cos(x+2pi) =cosx$

So that:

$d^n/(dx^n) cosx = cos(x+(npi)/2)$