## The Derivative of the Sine Function

$$\dfrac{d}{dx} [\sin x] = \cos x$$
Proof:

Certainly, by the limit definition of the derivative, we know that

$$\dfrac{d}{dx} [\sin x] = \displaystyle{\lim_{h \rightarrow 0} \dfrac{\sin (x+h) - \sin (x)}{h}}$$

Recalling the trigonometric identity $\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$, we can replace the $\sin(x + h)$ above to arrive at $$\dfrac{d}{dx} [\sin x] = \displaystyle{\lim_{h \rightarrow 0} \dfrac{\sin x \cos h + \cos x \sin h - \sin x}{h}}$$

Remember, we know $$\displaystyle{\lim_{\theta \rightarrow 0} \dfrac{\sin \theta}{\theta} = 1}$$

Seeing all of the components of a similar limit in our expression for the derivative, (i.e., there is a $\sin h$ in the numerator, an $h$ in the denominator, and both of these are inside a limit as $h \rightarrow 0$), we use algebra and the limit laws to reveal this known limit in our expression:

$$\begin{array}{rcl} \dfrac{d}{dx} [\sin x] & = & \displaystyle{\lim_{h \rightarrow 0} \left[ \dfrac{\cos x \sin h}{h} + \dfrac{\sin x \cos h - \sin x}{h} \right]}\\\\ & = & \displaystyle{\lim_{h \rightarrow 0} \left[ \cos x \cdot \dfrac{\sin h}{h} + \dfrac{\sin x \cos h - \sin x}{h} \right]}\\\\ & = & \displaystyle{\lim_{h \rightarrow 0} \cos x \cdot \lim_{h \rightarrow 0} \dfrac{\sin h}{h} + \lim_{h \rightarrow 0} \dfrac{\sin x \cos h - \sin x}{h}} \end{array}$$

Note, the first limit in the last line above is of an expression that does not depend on $h$, and hence effectively the limit of a constant. The second is the one we sought to reveal -- one which we know equals $1$. Thus,

$$\dfrac{d}{dx} [\sin x] = \displaystyle{\cos x + \lim_{h \rightarrow 0} \dfrac{\sin x \cos h - \sin x}{h}}$$

Pulling out the common factor of $\sin x$ in the remaining limit and splitting the resulting product with the limit laws again, we see another familiar limit -- one we which we know equals zero...

$$\begin{array}{rcl} \dfrac{d}{dx} [\sin x] & = & \displaystyle{\cos x + \lim_{h \rightarrow 0} \dfrac{\sin x (\cos h - 1)}{h}} \\\\ & = & \displaystyle{\cos x + \left[ \lim_{h \rightarrow 0} \sin x \right] \cdot \left[ \lim_{h \rightarrow 0} \dfrac{\cos h - 1}{h} \right]} \\\\ & = & \displaystyle{\cos x + \sin x \cdot 0}\\\\ & = & \cos x \end{array}$$

QED.