Section 3.5 Derivatives of Trigonometric Functions
References.
- Calculus, Early Transcendentals by Stewart, Section 3.3.
To find the derivatives of trigonometric functions, note first that we always use radian measure which means that the angle \(\theta\) is the length along the arc of the unit circle bounding a sector of angle \(\theta\)
Two calculations are the basis of all others: the derivatives of sine and cosine at the origin:
\begin{align}
\sin'(0) = \lim_{h \to 0} \frac{\sin(0+h)-\sin 0}{h} = \lim_{h \to 0} \frac{\sin h}{h} \amp= 1,\tag{3.5.1}\\
\cos'(0) = \lim_{h \to 0} \frac{\cos(0+h)-\cos 0}{h} = \lim_{h \to 0} \frac{\cos h - 1}{h} \amp= 0.\tag{3.5.2}
\end{align}
These will be calculated with the Squeeze Theorem from Section 2.3 and the trigonometry in this picture:
The line segment \(CB\) has length \(\sin h\text{,}\) so triangle \(\triangle OAB\) has base 1, height \(\sin h\text{,}\) area \(\frac12 \sin h\text{.}\)
The sector \(OAB\) with arc of length \(h\) and sides of length \(1\) has area \(\frac12 h\text{.}\)
The line segment \(AD\) has length \(\tan h\text{,}\) so triangle \(\triangle OAD\) has base 1, height \(\tan h\text{,}\) area \(\frac12 \tan h\text{.}\)
Comparing these areas,
\begin{equation*}
\sin h \leq h \leq \tan h
\quad\text{for angle in the first quadrant.}
\end{equation*}
(For negative angles in the fourth quadrant, the order is reversed.)
Using \(\tan h = \displaystyle\frac{\sin h}{\cos h}\text{,}\) and dividing through by \(h\text{,}\)
\begin{equation*}
\frac{\sin h}{h} \leq 1 \leq \frac{\sin h}{h} \frac{1}{\cos h}, \text{ true also for negative } h.
\end{equation*}
Multiplying the second inequality by \(\cos h\) gives \(\displaystyle \cos h \leq \frac{\sin h}{h}\text{;}\) since we already have \(\displaystyle\frac{\sin h}{h} \leq 1\text{,}\) we now have \(\displaystyle\frac{\sin h}{h}\) squeezed:
\begin{equation*}
\displaystyle \cos h \leq \frac{\sin h}{h} \leq 1
\end{equation*}
Thus, as \(h \to 0\text{,}\) we have \(\cos h \to \cos 0 = 1\) and \(1 \to 1\text{,}\) which combine to force \(\displaystyle\frac{\sin h}{h} \to 1\text{:}\) this confirms Equation (3.5.1).
We can also use this to calculate the derivative of cosine at the origin. The difference quotient used to get the derivative is:
\begin{equation*}
\frac{\cos h - 1}{h}
= \frac{\cos h - 1}{h} \cdot \frac{1 + \cos h}{1+\cos h}
= \frac{\cos^2 h - 1}{h(1+\cos h)}
= \frac{-\sin^2 h}{h(1+\cos h)}
= -\frac{\sin h}{h} \cdot \frac{\sin h}{1+\cos h}.
\end{equation*}
Thus,
\begin{align*}
\lim_{h \to 0} \frac{\cos h - 1}{h}
\amp= \lim_{h \to 0} \left(-\frac{\sin h}{h} \cdot \frac{\sin h}{1+\cos h}\right)\\
\amp= - \lim_{h \to 0} \left(\frac{\sin h}{h}\right) \lim_{h \to 0} \left(\frac{\sin h}{1+\cos h}\right)
= - 1 \cdot \frac{0}{1+1} = 0,
\end{align*}
confirming Equation (3.5.2).
With this we can compute the derivative of \(\sin x\) at any point, using the addition formulas
\begin{align}
\sin(x+y) \amp= \sin x \cos y + \cos x \sin y\tag{3.5.3}\\
\cos(x+y) \amp= \cos x \cos y - \sin x \sin y\tag{3.5.4}
\end{align}
\begin{align*}
\frac{d}{dx}\sin x
\amp= \lim_{h \to 0}\frac{\sin(x+h)-\sin x}{h}\\
\amp= \lim_{h \to 0}\frac{\sin x \cos h + \cos x \sin h - \sin x}{h}, \; \text{using the above sin-of-sum formula}\\
\amp= \lim_{h \to 0}\left[\sin x\frac{\cos h -1}{h} + \cos x \frac{\sin h}{h}\right]\\
\amp= (\sin x) \cdot \lim_{h \to 0}\frac{\cos h -1}{h} + (\cos x) \cdot \lim_{h \to 0}\frac{\sin h}{h}\\
\amp= (\sin x) \cdot 0 + (\cos x) \cdot 1,\; \text{using the above results for the derivatives at zero}\\
\amp= \cos x
\end{align*}
that is,
\begin{equation}
\frac{d}{dx}\sin x = \cos x.\tag{3.5.5}
\end{equation}
With a similar calculation, we get:
\begin{equation}
\frac{d}{dx}\cos x = -\sin x\text{.}\tag{3.5.6}
\end{equation}
Note well where the minus sign is!
Sketch the graphs of \(\sin x\text{,}\) \(\cos x\) and their derivatives, and use it to explain why the minus sign above makes sense.
Checkpoint 3.5.2.
Find the derivative of \(\displaystyle\frac{x^2\sin x}{1+\cos x}\text{.}\)
Checkpoint 3.5.3.
Complete this list of the derivatives of the other four standard trigonometric functions:
- \(\displaystyle \displaystyle\frac{d}{dx}\tan x =\)
- \(\displaystyle \displaystyle\frac{d}{dx}\sec x =\)
- \(\displaystyle \displaystyle\frac{d}{dx}\cot x =\)
- \(\displaystyle \displaystyle\frac{d}{dx}\csc x =\)
Note that the derivative of each complementary “co-” function has a minus sign and swapping of “co-” and “non-co” functions.
Checkpoint 3.5.4.
- Differentiate \(\displaystyle f(x)=\frac{\sec x}{1+\tan x}\)
- For what values of \(x\) does \(f(x)\) have a horizontal tangent?
Exercises Exercises
Study Calculus Volume 1, Section 3.5 2 ; in particular Examples 39 to 44, Checkpoint items 25 to 30, and Exercises 175, 178, 181, 182, 191, 197 and 206.
openstax.org/books/calculus-volume-1/pages/3-5-derivatives-of-trigonometric-functionsopenstax.org/books/calculus-volume-1/pages/3-5-derivatives-of-trigonometric-functions
