# Trigonometry/Derivative of Cosine

To find the derivative of cos(θ).

$\frac{d}{d\theta} \cos(\theta) = \lim_{h \rightarrow 0} \frac{\cos(\theta+h)-\cos(\theta)}{h} = -\lim_{h \rightarrow 0} \frac{2\sin(\theta+\frac{h}{2})\sin(\frac{h}{2})}{h} = -\lim_{h \rightarrow 0} {\left(\sin\left(\theta+\frac{h}{2}\right)\right)}\frac{\sin(\frac{h}{2})}{\frac{h}{2}}$.

As in the proof of Derivative of Sine, the limit of the first term is $\displaystyle\sin(\theta)$ and the limit of the second term is 1. Thus

$\frac{d}{d\theta} \cos(\theta) = -\sin(\theta)$.

Thus

$\frac{d}{d\theta} \sin(\theta) = \cos(\theta)$
$\frac{d}{d\theta} \cos(\theta) = -\sin(\theta)$
$\frac{d}{d\theta} -\sin(\theta) = -\cos(\theta)$
$\frac{d}{d\theta} -\cos(\theta) = \sin(\theta)$

and so on for ever.