## Proof: The Derivative of sin(x) in Radians

Using the formula for differentiation by first principles:

$\frac{d}{dx} \sin x = \lim_{h \to 0} \frac{\sin(x+h)-\sin x}{h}$

We can expand $$\sin(x+h)$$ using the addition formulae, giving:

$\frac{d}{dx} \sin x = \lim_{h \to 0} \frac{\sin x \cos h + \sin h \cos x -\sin x}{h}$

We know that $$sin x \approx x$$ and $$\cos x \approx 1-\frac{x^2}{2}$$ (proofs of these can be found here). Therefore we can use these to substitute for $$\sin h$$ and $$\cos h$$ like so:

$\frac{d}{dx} \sin x = \lim_{h \to 0} \frac{\left(1-\frac{h^2}{2}\right)\sin x + h \cos x -\sin x}{h}$

$\frac{d}{dx} \sin x = \lim_{h \to 0} \frac{\sin x-\left(\frac{h^2}{2}\right)\sin x + h \cos x -\sin x}{h}$

$\frac{d}{dx} \sin x = \lim_{h \to 0} \frac{-\left(\frac{h^2}{2}\right)\sin x + h \cos x}{h}$

$\frac{d}{dx} \sin x = \lim_{h \to 0} \left( -\left(\frac{h}{2}\right)\sin x + \cos x \right)$

$\frac{d}{dx} \sin x = \cos x$

$Q.E.D$

A similar argument can be used to show that $$\frac{d}{dx} (\cos x)=-\sin x$$ .