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

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.

$$\mathrm{Proof}\:\mathrm{that}\::\:\frac{\mathrm{d}^{\mathrm{n}} }{\mathrm{d}{x}^{{n}} }\left(\mathrm{cos}\:{x}\right)\:=\:\mathrm{cos}\:\left({x}+\frac{{n}\pi}{\mathrm{2}}\right) \\ $$$$\frac{\mathrm{d}^{\mathrm{n}} }{\mathrm{d}{x}^{{n}} }\left(\mathrm{sin}\:{x}\right)\:=\:\mathrm{sin}\:\left({x}+\frac{{n}\pi}{\mathrm{2}}\right) \\ $$$$\mathrm{where}\:\mathrm{n}\in\mathbb{Z}. \\ $$

Commented by prof Abdo imad last updated on 02/Aug/18

let prove by recurrence that cos^((n)) x=cos(x+((nπ)/2)) for n=0 cos^((0)) (x)=cosx=cos(x+((0π)/2)) the relation is true for n=0 let suppose cos^((n)) (x)=cos(x+((nπ)/2)) ⇒ cos^((n+1)) (x)=(cos^((n)) x)^′ =cos(x+((nπ)/2)))^′ =−sin(x+((nπ)/2))=cos(x+((nπ)/2) +(π/2)) =cos(x +(((n+1)π)/2)) the relation is true at term (n+1) the same method prove that sin^((n)) (x)=sin(x +((nπ)/2))

$${let}\:{prove}\:{by}\:{recurrence}\:{that}\:{cos}^{\left({n}\right)} {x}={cos}\left({x}+\frac{{n}\pi}{\mathrm{2}}\right) \\ $$$${for}\:{n}=\mathrm{0}\:\:{cos}^{\left(\mathrm{0}\right)} \left({x}\right)={cosx}={cos}\left({x}+\frac{\mathrm{0}\pi}{\mathrm{2}}\right) \\ $$$${the}\:{relation}\:{is}\:{true}\:{for}\:{n}=\mathrm{0}\:{let}\:{suppose} \\ $$$${cos}^{\left({n}\right)} \left({x}\right)={cos}\left({x}+\frac{{n}\pi}{\mathrm{2}}\right)\:\Rightarrow \\ $$$$\left.{cos}^{\left({n}+\mathrm{1}\right)} \left({x}\right)=\left({cos}^{\left({n}\right)} {x}\right)^{'} ={cos}\left({x}+\frac{{n}\pi}{\mathrm{2}}\right)\right)^{'} \\ $$$$=−{sin}\left({x}+\frac{{n}\pi}{\mathrm{2}}\right)={cos}\left({x}+\frac{{n}\pi}{\mathrm{2}}\:+\frac{\pi}{\mathrm{2}}\right) \\ $$$$={cos}\left({x}\:+\frac{\left({n}+\mathrm{1}\right)\pi}{\mathrm{2}}\right)\:{the}\:{relation}\:{is}\:{true}\:{at}\:{term} \\ $$$$\left({n}+\mathrm{1}\right)\:{the}\:{same}\:{method}\:{prove}\:{that} \\ $$$${sin}^{\left({n}\right)} \left({x}\right)={sin}\left({x}\:+\frac{{n}\pi}{\mathrm{2}}\right) \\ $$$$ \\ $$

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=cosx y_1 =−sinx=cos((Π/2)+x) y_2 =−sin((Π/2)+x)=cos(2.(Π/2)+x) y_3 =−sin(2.(Π/2)+x)=coz(3.(Π/2)+x) ... ...y_n =cos(n.(Π/2)+x) y=sinx y_1 =cosx=sin((Π/2)+x) y_2 =cos((Π/2)+x)=−sinx=sin(2.(Π/2)+x) ... ... y_n =sin(n.(Π/2)+x)

$${y}={cosx} \\ $$$${y}_{\mathrm{1}} =−{sinx}={cos}\left(\frac{\Pi}{\mathrm{2}}+{x}\right) \\ $$$${y}_{\mathrm{2}} =−{sin}\left(\frac{\Pi}{\mathrm{2}}+{x}\right)={cos}\left(\mathrm{2}.\frac{\Pi}{\mathrm{2}}+{x}\right) \\ $$$${y}_{\mathrm{3}} =−{sin}\left(\mathrm{2}.\frac{\Pi}{\mathrm{2}}+{x}\right)={coz}\left(\mathrm{3}.\frac{\Pi}{\mathrm{2}}+{x}\right) \\ $$$$... \\ $$$$...{y}_{{n}} ={cos}\left({n}.\frac{\Pi}{\mathrm{2}}+{x}\right) \\ $$$${y}={sinx} \\ $$$${y}_{\mathrm{1}} ={cosx}={sin}\left(\frac{\Pi}{\mathrm{2}}+{x}\right) \\ $$$${y}_{\mathrm{2}} ={cos}\left(\frac{\Pi}{\mathrm{2}}+{x}\right)=−{sinx}={sin}\left(\mathrm{2}.\frac{\Pi}{\mathrm{2}}+{x}\right) \\ $$$$... \\ $$$$...\:{y}_{{n}} ={sin}\left({n}.\frac{\Pi}{\mathrm{2}}+{x}\right) \\ $$

Commented by rahul 19 last updated on 02/Aug/18

thanks sir ����

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