# Derivative of sin(x)

## Statement

The derivative of $\sin(x)$ is:

$$\tfrac{d}{dx}\big(\sin(x)\big) = \cos(x)$$

## Proof

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

Use the angle sum formula for sine to expand $\sin(x + h)$:

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

Factor out $\sin(x)$.

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

Split the limit.

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

Factor out all factors without $h$.

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

Evaluate these limits. Use that the second limit is equal to $1$ and $\cos(0) = 1$.

\begin{aligned}\tfrac{d}{dx}\big(\sin(x)\big) &= \sin(x) \cdot \lim_{h \to 0} \left(\frac{1 - 1}{h}\right) + \cos(x) \cdot 1 \newline&= \sin(x) \cdot 0 + \cos(x) \cdot 1 \newline&= \cos(x)\end{aligned}