Derivative: trig functions

Section 14 Derivative: trig functions

Objectives

Derive derivative formulas for sine and cosine using the limit definition.
Derive derivative formulas for tangent, cotangent, secant, and cosecant from the formulas for sine and cosine, using the quotient rule.
Incorporate the new derivative formulas into derivative computations of more complicated functions.

Theorem 14.1. Derivative formulas: sine and cosine.

We have the following derivative formulas:

\begin{align*} \frac{d}{dx}\sin x \amp = \cos x \amp \frac{d}{dx}\cos x\amp =-\sin x\text{.} \end{align*}

Proof.

We prove the derivative formula for \(\sin\text{;}\) the argument for \(\cos\) is very similar. Using the limit definition of the derivative, we have

\begin{align*} \frac{d}{dx}(\sin x) \amp = \lim\limits_{h\to 0}\frac{\sin(x+h)-\sin x}{h}\\ \amp \lim\limits_{h\to 0}\frac{\sin x\cos h+\cos x\sin h-\sin x}{h}\\ \amp = \lim\limits_{h\to 0}\frac{\sin x(\cos h-1)}{h}+\frac{\cos x\sin h}{h}\\ \amp = \sin x\lim\limits_{h\to 0}\frac{\cos h-1}{h}+\cos x \lim\limits_{h\to 0}\frac{\sin h}{h} \\ \amp = \sin x\cdot 0 +\cos x\cdot 1\\ \amp = \cos x\text{,} \end{align*}

where the penultimate step makes use of our previously discussed limits

\begin{align*} \lim\limits_{t\to 0}\frac{\sin t}{t} \amp =1 \amp \lim\limits_{t\to 0}\frac{\cos t-1}{t}=0 \text{.} \end{align*}

Example 14.2. Tangent lines: sinusoidal.

Let \(f(x)=\cos x-\sin x\text{.}\)

Find an equation for the tangent line to the graph of \(f\) at \(x=\pi/3\text{.}\)
Find all points on the graph of \(f\) where the tangent line is horizontal.

Solution.

The tangent line to the graph of \(f\) at \(x=\pi/3\text{,}\) passes through the point

\begin{equation*} P=(\pi/3, f(\pi/3))=(\pi/3, \cos(\pi/3)-\sin(\pi/3))=(\pi/3, 1/2-\sqrt{3}/2). \end{equation*}

The slope of the tangent line here is given by \(f'(\pi/3)\text{.}\) We first compute

\begin{align*} f'(x) \amp = (\cos x)'-(\sin x)' \amp \text{(lin. comb.)}\\ \amp = -\sin x-\cos x\text{.} \end{align*}

Thus the slope of the tangent line is

\begin{equation*} f'(\pi/3)=-\sin(\pi/3)-\cos(\pi/3)=-(\frac{\sqrt{3}+1}{2})\text{,} \end{equation*}

We conclude that an equation for the tangent line here is

\begin{equation*} y-\frac{1-\sqrt{3}}{2}=-\frac{1+\sqrt{3}}{2}(x-\frac{\pi}{3})\text{.} \end{equation*}
The tangent line to a point \(P=(x,f(x))\) on the graph of \(f\) is horizontal when its slope \(f'(x)=0\text{.}\) Thus we need to find all \(x\) such that

\begin{equation*} f'(x)=-\sin x-\cos x=0\text{,} \end{equation*}

or equivalently,

\begin{equation*} \tan x=-1\text{.} \end{equation*}

The solutions to this trig equation are

\begin{equation*} x=\frac{3\pi}{4}+\pi n, \end{equation*}

where \(n\) is an integer. The corresponding points on the graph are

\begin{equation*} P=\left(\frac{3\pi}{4}+\pi n, f\left(\frac{3\pi}{4}+\pi n\right)\right) = \begin{cases} \left(\frac{3\pi}{4}+\pi n, -\sqrt{2}\right) \amp \text{if } n \text{ is even}\\ \left(\frac{3\pi}{4}+\pi n, \sqrt{2}\right) \amp \text{if } n \text{ is odd}\\ \end{cases}\text{.} \end{equation*}

Definition 14.3. Trigonometric functions.

We define the tangent (\(\tan\)), cotangent (\(\cot\)), secant (\(\sec\)), and cosecant (\(\csc\)) functions as follows:

\begin{align*} \tan x \amp = \frac{\sin x}{\cos x} \amp \cot x\amp =\frac{\cos x}{\sin x} \\ \sec x \amp =\frac{1}{\cos x} \amp \csc x\amp =\frac{1}{\sin x} \text{.} \end{align*}

Theorem 14.4. Derivative formulas: trig functions.

We have the following derivative formulas:

\begin{align*} \frac{d}{dx}\sin x \amp = \cos x \amp \frac{d}{dx}\cos x\amp =-\sin x\\ \frac{d}{dx}\tan x \amp = \sec^2 x \amp \frac{d}{dx}\cot x\amp =-\csc^2 x\\ \frac{d}{dx}\sec x \amp = \sec x\tan x \amp \frac{d}{dx}\csc x\amp =-\csc x\cot x\text{.} \end{align*}

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