To find the derivative of sin(θ).

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

Clearly, the limit of the first term is ${\displaystyle \cos(x)}$ since ${\displaystyle \cos(x)}$ is a continuous function. Write ${\displaystyle k={\frac {h}{2}}}$ ; the second term is then

${\displaystyle {\frac {\sin(k)}{k}}}$ .

Which we proved earlier tends to 1 as ${\displaystyle k\to 0}$ .

And since

${\displaystyle k\to 0{\text{ as }}h\to 0}$ ,

the limit of the second term is 1 too. Thus

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