Trigonometry/Derivative of Sine

To find the derivative of sin(&theta;).


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

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