# A calculus of the absurd

#### 10.10 Derivatives of trigonometric functions

##### 10.10.1 Derivative of $$\sin (x)$$

What is the derivative of $$\sin (x)$$? First, we can use the definition of the limit and a little algebra.

\begin{align} \frac {d}{dx}[\sin (x)] &= \lim _{h\to 0} \frac {\sin (x+h) - \sin (x)}{h} \\ &= \lim _{h\to 0} \frac {\sin (x)\cos (h) + \sin (h)\cos (x) - \sin (x)}{h} \end{align}

We want to rewrite this in terms of the two limits we found in the previous section!

\begin{align*} \lim _{h\to 0} \frac {\sin (x)\cos (h) + \sin (h)\cos (x) - \sin (x)}{h} &= \lim _{h\to 0} \left [\frac {\sin (x)(\cos (h)-1)}{h} + \frac {\sin (h)}{h}\cos (x)\right ] \\ &= \lim _{h\to 0} \left [-\sin (x)\frac {(1-\cos (h))}{h} + \frac {\sin (h)}{h}\cos (x)\right ] \\ &= \lim _{h\to 0} \left [0 + 1\cos (x)\right ] \\ &= \cos (x) \end{align*}