Trigonometry/Derivative of Cosine

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


 * $$\frac{d}{dx}\cos(x)=\lim_{h\to0}\frac{\cos(x+h)-\cos(x)}{h}=-\lim_{h\to0}\frac{2\sin\left(x+\frac{h}{2}\right)\sin\left(\frac{h}{2}\right)}{h}=-\lim_{h\to0}\left[\sin\left(x+\frac{h}{2}\right)\frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}}\right]=-\lim_{h\to0}\left[\sin\left(x+\frac{h}{2}\right)\right]\cdot\lim_{h\to0}\left[\frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}}\right]$$.

As in the proof of Derivative of Sine, the limit of the first term is $$\sin(x)$$ and the limit of the second term is 1. Thus
 * $$\frac{d}{dx}\big[\cos(x)\big]=-\sin(x)$$.

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


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


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


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

and so on for ever.