\begin{frame} \frametitle{Derivatives of Trigonometric Functions} We investigate \alert{$\lim_{\phi\to 0}\frac{\sin \phi}{\phi}$} \pause\hfill \textcolor{gray}{(for simplicity assume $0 < \phi < \pi/2$)} \pause \begin{center} \begin{tikzpicture}[default,scale=.9] \coordinate (a) at (0,0); \coordinate (b) at (40:3); \coordinate (c) at (0:3); \coordinate (d) at ({cos(40)*3},0); \coordinate (e) at ($(c) + (0,{(tan 20)*3})$); \coordinate (f) at ($(c) + (0,{(tan 40)*3})$); \draw (a) -- node[above,yshift=.5mm] {$1$} (b); \draw (a) -- node[below] {$1$} (c); \begin{scope}[cgreen] \draw (b) arc (40:0:3); \node at (20:2.8) {$\phi$}; \end{scope} \node[anchor=east] at (a) {$A$}; \node[anchor=south,xshift=-1mm] at (b) {$B$}; \node[anchor=north] at (c) {$C$}; \mpause[1]{ \begin{scope}[cred] \node[anchor=north] at (d) {$D$}; \draw (b) -- node[pos=.65,left] {$\sin \phi$} (d); \end{scope} } \mpause[4]{ \begin{scope}[cblue] \draw (c) -- (e) -- node[anchor=west,at start] {$E$} (b); \end{scope} \begin{scope}[gray] \draw ($(c) + (-2mm,0)$) -- ++(0,2mm) -- ++(2mm,0); \draw ($(b) + (-140:2mm)$) -- ++(-50:2mm) -- ++(40:2mm); \end{scope} } \mpause[6]{ \begin{scope}[cblue] \draw (c) -- (f) -- node[anchor=west,at start] {$F$} (b); \end{scope} \begin{scope}[gray] \draw ($(b) + (-140:-2mm)$) -- ++(-50:2mm) -- ++(40:-2mm); \end{scope} } \mpause[9]{ \begin{scope}[xshift=60mm,yshift=10mm] \draw (0,0) -- node[below] {$a$} (3,0) -- node[right] {$b$} (3,1.5) -- cycle; \draw (26:1.2) arc (26:0:1.2); \node at (13:.9) {$\alpha$}; \mpause[10]{ \node[align=center] at (1.5,-.7) {$\tan \alpha = \frac{b}{a}$}; } \mpause[11]{ \node[align=center] at (1.5,-1.2) {$b = a \cdot \tan \alpha$}; } \end{scope} } \end{tikzpicture} \end{center}\vspace{-2ex} \begin{talign} &\mpause[2]{\sin \phi = |BD| < \phi} \mpause[3]{\quad\implies\quad \frac{\sin \phi}{\phi} < 1} \\ & \mpause[5]{ \phi < |CE| + |EB| } \mpause[7]{\;\text{ \& }\; |EB| < |EF|} \mpause[8]{\quad\implies\quad \phi < |CE| + |EF| = |CF|} \\ &\mpause[12]{\phi < |CF|} \mpause[13]{ = 1\cdot \tan \phi} \mpause[14]{= \frac{\sin \phi}{\cos \phi}} \mpause[15]{\quad\implies\quad \alert{\cos \phi < \frac{\sin \phi}{\phi} < 1}} \end{talign}% \pause[+19]% We use the Squeeze Theorem:\vspace{-1ex} \begin{talign} \lim_{\phi \to 0} \cos \phi = 1 = \lim_{\phi \to 0} 1 && \mpause[1]{\implies} && \mpause[1]{\alert{\lim_{\phi\to 0}\frac{\sin \phi}{\phi} = 1}} \end{talign} \end{frame}