Why is the derivative of a sine function a cosine function? I am having a hard time coming up with WORDS to explain why is the derivative of a sine function a cosine function? Why? How do I write a few sentences saying why this happens?
I follow example after example solving similar problems but still don't understand enough to put into words.
I would appreciate some help.
Thanks,
Tony.
 A: Consider the curve $f(t)=(\cos t,\sin t)$. This curve moves around a unit circle counter-clockwise, so the derivative of this curve should be tangent at the circle.
The velocity at time $t$ has to have length $1$ because the curve traverses a length of $2\pi$ between $t=0$ and $t=2\pi$, and the behavior is "regular" - the speed is the same at each $t$, only the direction changes.
So the derivative, $f'(t),$ has to be the tangent vector on the circle in the counter-clockwise direction. It is thus perpendicular to $f(t)$ and has to have length $1$. This shows that $f'(t)=(-\sin t,\cos t)$. 
In particular, then, the derivative of $\sin t$ is $\cos t$.
If you want a rigorous proof, you can write:
$$\sin(x+h)=\sin x\cos h + \cos x \sin h$$
So $\sin(x+h)-\sin x = \sin x (\cos h-1) + \cos x \sin h$, dividing by $h$, you'd get:
$$\frac{\sin(x+h)-\sin h}{h} = \sin x \frac{\cos h - 1}{h} +\cos x \frac{\sin h}{h}$$
So now we only need to know the limits:
$$\lim_{h\to 0} \frac{\cos h-1}{h} = 0$$
$$\lim_{h\to 0} \frac{\sin h}{h} = 1$$
In other words, you just need the derivatives at $x=0$.
There are geometric proofs of this case. For example, the distance between $(\cos x,\sin x)$ and $(1,0)$ is $\sqrt{2-2\cos x}$, which is necessarily less than the length of the arc from $(\cos x,\sin x)$ to $(1,0)$, which is of length $|x|$. So:
$$\sqrt{2-2\cos x} \leq |x|\implies\\2-2\cos x \leq x^2\implies \\\left|\frac{\cos x -1}{x}\right|<|x|$$
So $\lim_{x\to 0}\frac{\cos x -1}{x} = 0$.
For $\sin$, we can show that $\sin h \leq h \leq \tan h$ when $h>0$.
A: Notice that $\sin \theta$ has maximums and minimums where $\cos \theta$ has zeroes. Similarly, the slopes of the tangent lines of $\sin \theta$ are 1 and -1 precisely where $\cos \theta$ attains those values.
Now that doesn't explain why $\frac{d}{d\theta} \sin \theta = \cos  \theta$ everywhere, but it's a good way to remember the relationship.
