## Proof of the Week for 20120820

Theorem: $\frac{d}{dx} \sin x = \cos x$.

Proof: Let $f(x) = \sin x$.
$f'(x) = \lim_{\Delta x \to 0} \frac{\sin(x + \Delta x) - \sin x}{\Delta x}$
$= \lim_{\Delta x \to 0} \frac{2}{\Delta x} \sin\left(\frac{x + \Delta x - x}{2}\right) \cos\left(\frac{x + \Delta x + x}{2}\right)$
$= \lim_{\Delta x \to 0} \frac{\sin\left(\frac{\Delta x}{2}\right)}{\frac{\Delta x}{2}} \cos \left(x + \frac{\Delta x}{2} \right)$
$= \lim_{\Delta x \to 0} \frac{\sin\left(\frac{\Delta x}{2}\right)}{\frac{\Delta x}{2}} \lim_{\Delta x \to 0} \cos \left(x + \frac{\Delta x}{2} \right)$
$= \cos x$

Notes:
1. We used the following trigonometric identity: $\sin \theta - \sin \phi = 2\sin \left(\frac{\theta - \phi}{2}\right) \cos \left(\frac{\theta + \phi}{2}\right)$
2. We used the result: $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$
3. We also invoked the continuity of the cosine function in the last step of the proof.