To find the derivative of sin(θ).

- .

Clearly, the limit of the first term is since is a continuous function. Write ; the second term is then

- .

Which we proved earlier tends to 1 as .

And since

- ,

the limit of the second term is 1 too. Thus

- .

To find the derivative of sin(θ).

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

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

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

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

And since

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

the limit of the second term is 1 too. Thus

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