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

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

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.