Proof that : (d^n /dx^n )(cos x) = cos (x+((nπ)/2)) (d^n /dx^n )(sin x) = sin (x+((nπ)/2)) where n∈Z.


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Question Number 41151 by rahul 19 last updated on 02/Aug/18

$Proof that : \frac{d^{n}}{d x^{n}} (\cos x) = \cos (x + \frac{n π}{2})$ $\frac{d^{n}}{d x^{n}} (\sin x) = \sin (x + \frac{n π}{2})$ $where n \in Z .$
Commented by prof Abdo imad last updated on 02/Aug/18

$l e t p r o v e b y r e c u r r e n c e t h a t {c o s}^{(n)} x = c o s (x + \frac{n π}{2})$ $f o r n = 0 {c o s}^{(0)} (x) = c o s x = c o s (x + \frac{0 π}{2})$ $t h e r e l a t i o n i s t r u e f o r n = 0 l e t s u p p o s e$ ${c o s}^{(n)} (x) = c o s (x + \frac{n π}{2}) \Rightarrow$ ${{c o s}^{(n + 1)} (x) = {({c o s}^{(n)} x)}^{^{'}} = c o s (x + \frac{n π}{2}))}^{^{'}}$ $= - s i n (x + \frac{n π}{2}) = c o s (x + \frac{n π}{2} + \frac{π}{2})$ $= c o s (x + \frac{(n + 1) π}{2}) t h e r e l a t i o n i s t r u e a t t e r m$ $(n + 1) t h e s a m e m e t h o d p r o v e t h a t$ ${s i n}^{(n)} (x) = s i n (x + \frac{n π}{2})$
Commented by rahul 19 last updated on 03/Aug/18
thanks prof Abdo ��
	Answered by tanmay.chaudhury50@gmail.com last updated on 02/Aug/18

	$y = c o s x$ $y_{1} = - s i n x = c o s (\frac{Π}{2} + x)$ $y_{2} = - s i n (\frac{Π}{2} + x) = c o s (2 . \frac{Π}{2} + x)$ $y_{3} = - s i n (2 . \frac{Π}{2} + x) = c o z (3 . \frac{Π}{2} + x)$ $. . .$ $. . . y_{n} = c o s (n . \frac{Π}{2} + x)$ $y = s i n x$ $y_{1} = c o s x = s i n (\frac{Π}{2} + x)$ $y_{2} = c o s (\frac{Π}{2} + x) = - s i n x = s i n (2 . \frac{Π}{2} + x)$ $. . .$ $. . . y_{n} = s i n (n . \frac{Π}{2} + x)$
	Commented by rahul 19 last updated on 02/Aug/18
	thanks sir ��
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