We will use two important facts to write a proof for the derivative of $\cos x$. Firstly, we will use the angle sum formula

$$\cos(x+y) = \cos x\cos y – \sin x\sin y$$

This formula is part of precalculus and is common knowledge. The second formula we will use is

$$\lim_{h\to0}\frac{\sin h} h = 1$$

This is a well-known limit which which we could prove geometrically. The proof relies on drawing the unit circle, but we will not go into it here. Another way to show that the formula is true is by using L’Hopital’s rule by taking the derivative of the numerator and the denominator. However, in order to do that, you should already know that the derivative of $\sin x$ is $\cos x$. Below is the proof that uses L’Hopital’s rule and not the geometric proof of

$$\lim_{h\to0}\frac{\sin h} h = 1$$

Using the limit definition of derivative:

\[ \begin{align*} &\lim_{h \to 0} \dfrac{\cos(x + h) – \cos(x)}{h}\\ &=\lim_{h \to 0} \dfrac{\cos x \cos h – \sin x \sin h – \cos x}{h}\\ &=\cos x \lim_{h \to 0} \dfrac{\cos h – 1}{h} – \sin x \lim_{h \to 0} \dfrac{\sin h}{h} \end{align*} \] Putting $h = 0$ in the first limit and using L’Hospital’s rule on the second limit, we get: \[ \begin{align*} &\cos x \lim_{h \to 0} \dfrac{\cos h – 1}{h} = 0\\ &\sin x \lim_{h \to 0} \dfrac{\sin h}{h} = \sin x\\ &\lim_{h \to 0} \dfrac{\cos(x + h) – \cos(x)}{h} = 0 – \sin x\\ \end{align*} \]

We get the final answer that $\dfrac{d}{dx} \cos x = \boxed{-\sin x}$.