Trigonometry/Derivative of Sine

To find the derivative of sin(θ).

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

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

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