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

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

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

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

Thus

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

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

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

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

and so on for ever.