Exercise 6

In this exercise, we explore how the limit definition of the derivative more formally shows that $\frac{d}{dx}[\sin(x)] = \cos(x)$. Letting $f(x) = \sin(x)$, note that the limit definition of the derivative tells us that $$\begin{gathered} f'(x) = \lim_{h \to 0} \frac{\sin(x+h) - \sin(x)}{h} \end{gathered}$$.

1. Recall the trigonometric identity for the sine of a sum of angles $\alpha$ and $\beta$: $\sin(\alpha + \beta) = \sin(\alpha)\cos(\beta) + \cos(\alpha)\sin(\beta)$. Use this identity and some algebra to show that $$\begin{gathered} f'(x) = \lim_{h \to 0} \frac{\sin(x)(\cos(h)-1) + \cos(x)\sin(h)}{h} \end{gathered}$$.

2. Next, note that as $h$ changes, $x$ remains constant. Explain why it therefore makes sense to say that $$\begin{gathered} f'(x) = \sin(x) \cdot \lim_{h \to 0} \frac{\cos(h) -1 }{h} + \cos(x) \cdot \lim_{h \to 0} \frac{\sin(h)}{h} \end{gathered}$$.

3. Finally, use small values of $h$ to estimate the values of the two limits in (c): $$\begin{gathered} \lim_{h \to 0} \frac{\cos(h) - 1}{h} \ \ \text{and} \ \ \lim_{h \to 0} \frac{\sin(h)}{h} \end{gathered}$$.

4. What do your results in (b) and (c) thus tell you about $f'(x)$?

5. By emulating the steps taken above, use the limit definition of the derivative to argue convincingly that $\frac{d}{dx}[\cos(x)] = -\sin(x)$.