Question

How to prove that nth derivative of sinx is sin[x + (npi/2)]

Answer 1

\[\sin(x+\frac{\pi}{2})=\cos(x)\]

Answer 2

\[\sin(x+\pi)=-\sin(x)\]

Answer 3

\[\sin(x+\frac{3\pi}{2})=-\cos(x)\]

Answer 4

etc

Answer 5

sorry but is this proof that nth derivative of sinx is sin(x+npi/2)?

Answer 6

no, but it is the idea behind it

Answer 7

if \(n=0,4,8,12,...\) then \(\sin(x+\frac{n\pi}{2})=\sin(x)\) and also \(\sin^{(n)}(x)=\sin(x)\)

Answer 8

if \(n=1,5,9,...\) then
\[\sin(x+\frac{n\pi}{2})=\cos(x)\] and also \(\sin^{(n)}(x)=\cos(x)\)

Answer 9

you can be more precise, for example say if \(n=4k\) then ... and if \(n=4k+1\) then ... and so on

Answer 10

thanks al lot sir.