A Sine

\[f’(x)=\lim_{ \epsilon\to 0} \frac {f(x+\epsilon)-f(x)}{\epsilon}\]

\[(\sin x)’=\lim_{ \epsilon\to 0} \frac {\sin(x+\epsilon)-\sin (x)}{\epsilon}\]

Using sum-to-product identity:

\[ {\sin(x+\epsilon)-\sin (x)}=2\sin \frac {x+\epsilon -x}2 \cos \frac {x+\epsilon +x}2 \]

\[ {\sin(x+\epsilon)-\sin (x)}=2\sin \frac {\epsilon}2 \cos (x+\frac {\epsilon }2) \]

\[(\sin x)’=\lim_{ \epsilon\to 0} \frac {2\sin \frac {\epsilon}2 \cos (x+\frac {\epsilon }2)}{\epsilon}\]

\[(\sin x)’=\lim_{ \epsilon\to 0} \frac {\sin \frac {\epsilon}2 \cos (x+\frac {\epsilon }2)}{\frac \epsilon 2}\]

\[(\sin x)’=\lim_{\epsilon \to 0}\cos (x+\frac {\epsilon }2)\lim_{ \epsilon\to 0} \frac {\sin \frac {\epsilon}2 }{\frac \epsilon 2}\]

\[\lim_{\epsilon \to 0}\cos (x+\frac {\epsilon }2)=\cos x\]

\[\lim_{ \epsilon\to 0} \frac {\sin \frac {\epsilon}2 }{\frac \epsilon 2}=1\]

so:

\[(sin x)’=\cos x\]