Recently I showed my students how to prove that the derivative of $\sin(x) = \cos(x)$, using the limit definition of the derivative, trigonometric identities, and the fact that $\lim_{h \to 0} \frac{sin(h)}{h} = 1$.
I'm trying to think of another way for them to think about the derivative of $\sin(x)$. Can I say something about the periodicity - the derivative is a function that is still $2\pi$-periodic? That doesn't sound enlightening. Can I say that the derivative changes the evenness / oddness of the function, e.g. the derivative of sin(x), an odd function, is cos(x), an even function? That sounds a little more interesting but still not something that would grab their attention.
Is there a deeper way of thinking about the derivative of $\sin(x)$?
(This is for a first term course in Calculus, so no power series, please.)
Apropos "deeper way":
1) $f(x) = f(-x),$ even fct.
Examples: $y=x^2$, $y=cos(x)$
$ f'(x) = -f'(-x),$ chain rule, odd fct.
2) $f(x)=-f(-x)$, odd fct.
Examples: $y=x^3,$ $ y=sin(x)$.
$f'(x) = f'(-x)$, chain rule, even fct.
3) Example, periodic fct:
$f(x) = f(x +2πk)$, $k \in \mathbb{N}$.
$f'(x)=f'(x+2πk)$.
4) Draw $\sin$ and $\cos$ curve, $0 \le x\le π/2$,
(in one diagram, superimpose).
Choose any $x_0$ in this interval.
Find the derivative of the $\sin$ fct at $x_0$ by inspection. ($\cos(x_0)$ on $\cos$ curve).
5) By inspection:
Find the derivative of $\sin(x)$ at the point of intersection of the 2 curves.
Given that at the point of intersection, $x_0=π/4$, $\sin(x_0) =(1/2)√2$, find the derivative of $\sin$ at this point.
Helps a little?