Trigonometry/Derivative of Cosine

To find the derivative of $\cos (x)$ .

$\frac{d}{dx}\cos (x)=\underset{h\to 0}{\lim }\frac{\cos (x+h)-\cos (x)}{h}=-\underset{h\to 0}{\lim }\frac{2\sin (x+\frac{h}{2})\sin \left(\frac{h}{2}\right)}{h}=-\underset{h\to 0}{\lim }[\sin (x+\frac{h}{2})\frac{\sin \left(\frac{h}{2}\right)}{\frac{h}{2}}]=-\underset{h\to 0}{\lim }[\sin (x+\frac{h}{2})]\cdot \underset{h\to 0}{\lim }\left[\frac{\sin \left(\frac{h}{2}\right)}{\frac{h}{2}}\right]$ .

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

$\frac{d}{dx}[\cos (x)]=-\sin (x)$ .

Thus

$\frac{d}{dx}[\sin (x)]=\cos (x)$

$\frac{d}{dx}[\cos (x)]=-\sin (x)$

$\frac{d}{dx}[-\sin (x)]=-\cos (x)$

$\frac{d}{dx}[-\cos (x)]=\sin (x)$

and so on for ever.