Mathematics Asked by cxlim on November 29, 2020

I’m not quite sure how does $f^{(n)}(x)=sin(x+frac{npi}{2})$ for $f(x)=sin(x)$. Taking the initial derivatives I get,

begin{align}

f'(x)&=cos(x)\

f”(x)&=-sin(x)\

f”'(x)&=-cos(x)\

f^{4}(x)&=sin(x)

end{align}

I’m not quite understanding this relation.

Just see how sin line change its value when rotating 90degrees counter clock wise and compare with the fact that every 4 derivatives you back to the same place (ups!, you got the same function).

Using maths:

The usage of $sin(x+nfrac{pi}{2})=sin(x)cos(nfrac{pi}{2})+cos(x)sin(frac{pi}{2})$ is a must, because the value of functions with argument $frac{pi}{2}$ takes $1,0$ or $-1$, so when one takes value $0$, the other can take values $1$ or $-1$ which are the neccesary for having the right derivative. Of course, this is not casuality, this is because the derivate for sin is a idempotent operator of order $4$ (i.e $T^4=T$).

Also, since $sin$ is periodic, its derivative will be periodic too:

$$f'(x+c) = displaystylelim_{hto0}frac{(f((x+h)+c)-f(x+c))}{h}$$ $$= displaystylelim_{hto0}frac{f(x+h)-f(x)}{h} = f'(x)$$

Answered by Luis Felipe on November 29, 2020

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