# Why does $f^{(n)}(x)=sin(x+frac{npi}{2})$ for $f(x)=sin(x)$?

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