TRIUMPHS Project 14.6.1.
The Derivatives of the Sine and Cosine Functions by Dominic Klyve
5
https://digitalcommons.ursinus.edu/triumphs_calculus/1/Section 14.6 Derivatives of the Trigonometric Functions, via Limits
Theorem 14.6.2. Derivative of \(\sin(x)\).
Suppose \(\alpha(\theta)=\sin(\theta)\text{.}\) Then \(\alpha^\prime(\theta)=\cos(\theta). \)
Proof.
Showing that the derivative of \(\sin(\theta)\) is \(\cos(\theta)\) is mostly straightforward but we’re going to hit a snag partway through. We’ll proceed for a bit to see where the trouble is. Start with the limit definition:
\begin{equation*}
\alpha^\prime(\theta) =
\limit{h}{0} { \frac{\sin(\theta+h)-\sin(\theta)}{h}}
\end{equation*}
In the numerator we see the expression \(\sin(\theta+h)\text{.}\) Recall the sum formula for the Sine:
\begin{equation*}
\sin(A+B)=\sin(A)\cos(B)+\cos(A)\sin(B).
\end{equation*}
Taking \(A=\theta\) and \(B=h\) we have:
\begin{align*}
\alpha^\prime(x) \amp = \limit{h}{0}{\frac{\textcolor{red}{\sin(\theta)}\cos(h)+\cos(\theta)\sin(h)-\textcolor{red}{\sin(\theta)}}{h}}\\
\end{align*}
Next, if we factor \textcolor{red}{\(\sin(\theta)\)} out of the terms where it appears and rearrange the numerator a bit we have:
\begin{align*}
\amp =
\limit{h}{0}{\frac{\textcolor{red}{\sin(\theta)}(\cos(h)-1)+\cos(\theta)\sin(h)}{h}}.\\
\amp =
\limit{h}{0}{\left(\frac{\sin(\theta)(\cos(h)-1)}{h}+\frac{\cos(\theta)\sin(h)}{h}\right)}.\\
\end{align*}
By Theorem 14.1.1:
\begin{align*}
\amp =
\limit{h}{0}{\frac{\sin(\theta)(\cos(h)-1)}{h}}+\limit{h}{0}{\frac{\cos(\theta)\sin(h)}{h}}\\
\end{align*}
and by Corollary 14.1.13:
\begin{align*}
\amp =
\sin(\theta)\underbrace{\left(\limit{h}{0}{\frac{(\cos(h)-1)}{h}}\right)}_{=0}+\cos(\theta)\underbrace{\left(\limit{h}{0}{\frac{\sin(h)}{h}}\right)}_{=1}.
\end{align*}
If the values of the two limits are \(0\) and \(1\) respectively as we’ve indicated we can conclude that \(\alpha^\prime(\theta)=\cos(\theta)\text{.}\) Unfortunately this proof cannot be considered complete until we have shown that these last two limits are what we claim they are. We will do this via the two lemmas below.
It is tempting to use L’Hôpital’s Rule to evaluate the two limits at the end of Proof 14.6.1, especially since it is so very easy to do. See Drill 14.6.3 below.
Drill 14.6.3.
Use L’Hôpital’s Rule to show
- \(\displaystyle \tlimit{h}{0}{\frac{(\cos(h)-1)}{h}}=0\)
- \(\tlimit{h}{0}{\frac{\sin(h)}{h}}=1\text{.}\)
Sadly, using Drill 14.6.3 to finish the proof of Theorem 14.6.2 is an example of circular reasoning. We can’t use the fact that the derivative of \(\sin(x)\) is \(\cos(x)\) to prove that the derivative of \(\sin(x)\) is \(\cos(x)\text{.}\) So we will have to find a way to evaluate these limits without using L’Hôpital’s Rule.
Lemma 14.6.4.
\(\limit{h}{0}{\frac{\sin(h)}{h}}=1\)
Proof.
There are two cases:
Case 1 \(\theta\ge0\text{:}\) We will use the Squeeze Theorem. Recall that in Section 6.1 we observed that the lengths of certain line segments associated with the unit circle in the first quadrant are equal to the trigonometric functions. Figure 14.6.5 shows the relationship between \(\theta\text{,}\) \(\sin(\theta)\text{,}\) and \(\tan(\theta)\text{.}\) Notice in particular that
\begin{equation*}
\sin(\theta)\le\theta\le\tan(\theta) \text{.}
\end{equation*}
Dividing each expression in the inequality by \(\sin(\theta)\) almost does the trick:
\begin{equation*}
1\le\frac{\theta}{\sin(\theta)}\le\frac{1}{\cos(\theta)}.
\end{equation*}
In the center we now have the reciprocal of what we need, so we need to invert each expression. However, keep in mind that these are not equations they are inequalities. When we invert an inequality we must reverse its sense. This gives
\begin{equation*}
1\ge\frac{\sin(\theta)}{\theta}\ge\frac{\cos(\theta)}{1},
\end{equation*}
and this is true on the interval \(\left[0,\frac{\pi}{2}\right]\text{.}\) Since \(\cos (\theta
)\) is a continuous function we see that
\begin{equation*}
\rlimit{\theta}{0}{1}=\rlimit{\theta}{0}{\cos(\theta)}=1
\end{equation*}
the Squeeze Theorem applies, and we conclude that
\begin{equation*}
\rlimit{\theta}{0}{\frac{\sin(\theta)}{\theta}}=1.
\end{equation*}
Case 2 \(\theta\lt0\text{:}\) For this case notice that \(\sin(-\theta)=-\sin(\theta)\) so that \(\frac{\sin(-\theta)}{-\theta}=\frac{\sin(\theta)}{\theta}\text{.}\) We make the substitution \(\theta=-\phi\) where \(\phi\gt0\text{.}\) Therefore when \(\theta\lt0\) we have
\begin{equation*}
\llimit{\theta}{0}{\frac{\sin(\theta)}{\theta}}
=\rlimit{\phi}{0}{\frac{\sin(-\phi)}{-\phi}}
=\rlimit{\phi}{0}{\frac{\sin(\phi)}{\phi}} =1
\end{equation*}
by Case 1.
Problem 14.6.6.
Show that \(\tlimit{h}{0}{\frac{(\cos(h)-1)}{h}}=0\text{.}\)\\
Hint.
It is tempting to model this proof on the proof of Lemma 14.6.4. While this can be done, it is delicate. It is simpler to multiply by \(1\) in the form \(\frac{\cos(h)+1}{\cos(h)+1}\text{.}\) Try that instead.
Once Problem 14.6.6 has been solved the proof that \(\dfdx{(\sin(x))}{x}= \cos(x)\) is complete.
Drill 14.6.7.
Prove that \(\dfdx{(\cos(\theta))}{\theta} =
-\sin(\theta)\text{,}\) using the proof of Theorem 14.6.2 as a guide.
We can now use Problem 14.5.6 to find the derivatives of \(\tan(\theta)\text{,}\) \(\cot(\theta)\text{,}\) \(\sec(\theta)\text{,}\) and \(\csc(\theta)\) as well. Since this is exactly what we did in Section 6.3 we have the derivatives of all of the trigonometric functions.
