Lemma 2.8.2
\begin{align*}
\lim_{h \to 0} \frac{\cos h -1}{h} &= 0
\end{align*}
Subsection 2.8.4 Step 2: \(\mathbf{\diff{}{x} \{ \cos x \} \big|_{x=0}}\)
By definition, the derivative of \(\cos x\) evaluated at \(x=0\) is
\begin{align*}
\lim_{h\rightarrow 0}\frac{\cos h-\cos 0}{h}
&=\lim_{h\rightarrow 0}\frac{\cos h - 1}{h}
\end{align*}
Fortunately we don't have to wade through geometry like we did for the previous step. Instead we can recycle our work and massage the above limit to rewrite it in terms of expressions involving \(\frac{\sin h}{h}\text{.}\) Thanks to Lemma 2.8.1 the work is then easy.
We'll show you two ways to proceed — one uses a method similar to “multiplying by the conjugate” that we have already used a few times (see Example 1.4.17 and 2.2.9 ), while the other uses a nice trick involving the double–angle formula 3 See Appendix A.14 if you have forgotten. You should also recall that \(\sin^2\theta + \cos^2\theta = 1\text{.}\) Sorry for nagging. .
Subsubsection 2.8.4.1 Method 1 — Multiply by the “Conjugate”
Start by multiplying the expression inside the limit by 1, written as \(\ds \frac{\cos h + 1}{\cos h +1}\text{:}\)
\begin{align*}
\frac{\cos h - 1}{h}
&= \frac{\cos h - 1}{h} \cdot\frac{\cos h + 1}{\cos h +1}\\
&= \frac{\cos^2 h - 1}{h (1+ \cos h)}
&&\text{$\big($since $(a-b)(a+b)=a^2-b^2\big)$}\\
&= -\frac{\sin^2 h}{h(1+\cos h)}
&&\text{(since $\sin^2 h + \cos^2 h=1$)}\\
&= - \frac{\sin h}{h} \cdot \frac{\sin h}{1 + \cos h}
\end{align*}
Now we can take the limit as \(h \to 0\) via Lemma 2.8.1.
\begin{align*}
\lim_{h \to 0} \frac{\cos h - 1}{h}
&= \lim_{h \to 0} \left( \frac{-\sin h}{h} \cdot \frac{\sin h}{1 + \cos h}
\right)\\
&= -\lim_{h \to 0}\left( \frac{\sin h}{h} \right) \cdot \lim_{h \to0}
\left( \frac{\sin h}{1 + \cos h} \right)\\
&= - 1 \cdot \frac{0}{2}\\
&= 0
\end{align*}
Subsubsection 2.8.4.2 Method 2 — via the Double Angle Formula
The other way involves the double angle formula 4 We hope you looked this up in in Appendix A.14. Nag.,
\begin{gather*}
\cos 2\theta = 1 - 2 \sin^2(\theta) \qquad\text{or}\qquad
\cos 2\theta -1 = - 2 \sin^2(\theta)
\end{gather*}
Setting \(\theta = h/2\text{,}\) we have
\begin{align*}
\frac{\cos h - 1}{h}
&= \frac{-2\big(\sin\tfrac{h}{2}\big)^2}{h}\\
\end{align*}
Now this begins to look like \(\frac{\sin h?}{h}\text{,}\) except that inside the \(\sin(\cdot)\) we have \(h/2\text{.}\) So, setting \(\theta =h/2\text{,}\)
\begin{align*} \frac{\cos h - 1}{h} &= - \frac{\sin^2 \theta}{\theta} = - \theta \cdot\frac{ \sin^2 \theta}{\theta^2}\\ &= - \theta \cdot \frac{\sin \theta}{\theta} \cdot \frac{\sin \theta}{\theta} \end{align*}When we take the limit as \(h \to 0\text{,}\) we are also taking the limit as \(\theta=h/2 \to 0\text{,}\) and so
\begin{align*}
\lim_{h \to 0} \frac{\cos h - 1}{h}
&= \lim_{\theta \to 0} \left[
- \theta \cdot \frac{\sin \theta}{\theta} \cdot \frac{\sin \theta}{\theta}
\right]\\
&= \lim_{\theta \to 0} \left[- \theta \right]
\cdot \lim_{\theta \to 0} \left[\frac{\sin \theta}{\theta}\right]
\cdot \lim_{\theta \to 0} \left[\frac{\sin \theta}{\theta}\right]\\
&= 0 \cdot 1 \cdot 1\\
&= 0
\end{align*}
where we have used the fact that \(\ds \lim_{h \to 0} \frac{\sin h}{h} = 1\) and that the limit of a product is the product of limits (i.e. Lemma 2.8.1 and Theorem 1.4.3).
Thus we have now produced two proofs of the following lemma:
Again, there has been a bit of work to get to here, so we should remind ourselves why we needed it. We were computing
\begin{align*}
\diff{}{x} \{ \cos x \} \Big|_{x=0}
&= \lim_{h \to 0} \frac{\cos h - \cos 0}{h}\\
&= \lim_{h \to 0} \frac{\cos h - 1}{h}\\
&= 0
\end{align*}
Armed with these results we can now build up the derivatives of sine and cosine.