Nice $\epsilon$-$\delta$ proof that the derivative of $\sin(x)$ is $\cos(x)$? Looked around a bit and all I see are proofs using the limit definition of a derivative. This is not for an assignment, I could just use the limit definition if I wanted to, but I was wondering how you could go about proving this using the epsilon-delta definition of a derivative ($\forall \epsilon >0$, $\exists \delta >0$ such that if $0< |x-c| <\delta$, then $\left | \frac{f(x) - f(c)}{x-c} - f'(c) \right | < \epsilon$).
Edit: Just to be clear I am explicitly looking for an epsilon-delta formulation of the proof. Was just trying to prove this without throwing a bunch of trig identities and limit theorems at the problem.
 A: I know this post is somewhat old, but I'll post a thorough answer anyway so it may benefit future readers.
To prove this, we will want to use the limits $\frac {\sin x}x \to 1$ and $\frac {1 - \cos x}x \to 0$ when $x \to 0$. By the $\epsilon-\delta$ definition, we can write (quantifiers left out, assuming reader is familiar with the definition):
$$ (1):  0 < |x - 0| < \delta_1 \implies \left|\frac {\sin x}x  - 1\right| < \frac {\epsilon}2  \\
  (2):  0 < |x - 0| < \delta_2 \implies \left|\frac {1 - \cos x}x  - 0\right| < \frac {\epsilon}2   \\$$
Now let $\delta = \min \{ \delta_1, \delta_2 \}$. What we then want to show is that
$$  0 < |x - 0| < \delta \implies \left|\frac {\sin(x+h) - \sin x}x  - \cos x\right| <\epsilon $$
Using $\sin (x +h) = \sin x \cos h + \sin h \cos x$ and some algebra inside the absolute value sign, we can write
$$ \left|\frac {\sin(x+h) - \sin x}x  - \cos x\right| = \left| \frac { \sin x (\cos h - 1) }h + \frac { \cos x (\sin h  - h) }h  \right|  $$
By the triangle inequality, we have
$$\left| \frac { \sin x (\cos h - 1) }h + \frac { \cos x (\sin h  - h) }h  \right| \leq  | \sin x | \left| \frac {\cos h - 1}h  \right| + | \cos x | \left| \frac {\sin h - h}h \right|$$
Trivially, $|\sin x | \leq 1$ and $| \cos x | \leq 1$. Therefore
$$    | \sin x | \left| \frac {\cos h - 1}h  \right| + | \cos x | \left| \frac {\sin h - h}h \right| \leq \left| \frac {\cos h - 1}h  \right| + \left| \frac {\sin h - h}h \right|  $$
But by $(1)$ and $(2)$ we finally have
$$  \left| \frac {\cos h - 1}h  \right| + \left| \frac {\sin h - h}h \right| < \frac {\epsilon}2 + \frac {\epsilon}2 = \epsilon  $$
A: It may depend on what limits you know, but
$$\frac{\sin(x+h)-\sin x}{h}=\frac{\sin x\cos h+\sin h\cos x-\sin x}{h}=$$
$$=\sin x\frac{\cos h-1}{h}+\cos x\frac{\sin h}{h}\xrightarrow[h\to 0]{}\sin x\cdot 0+\cos x\cdot 1 =\cos x$$
You can also try other trigonometric identities with an equivalent definition fro the derivative:
$$\frac{\sin x-\sin x_0}{x-x_0}=2\frac{\sin\frac{x-x_0}{2}\cos\frac{x+x_0}{2}}{x-x_0}=$$
$$=\frac{\sin\frac{x-x_0}{2}}{\frac{x-x_0}{2}}\cos\frac{x+x_0}{2}\xrightarrow[x\to x_0]{}1\cdot \cos\frac{2x_0}{2}=\cos x_0$$
