Trigonometry/Derivative of Cosine

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To find the derivative of \cos(x) .

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

Thus

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

and so on for ever.