Trigonometric and Inverse Trigonometric Functions

Lessons

Senior 5 · Mathematics

Trigonometric and Inverse Trigonometric Functions

5 lessons

Lesson 3 of 5

Derivatives of the six trigonometric functions

25 min

How fast does a sine change?

In S4 you differentiated polynomials, rational functions and roots with the sum, product, quotient and chain rules. Waves, rotations and oscillations need the derivatives of the trigonometric functions too: the velocity of a vibrating string is the derivative of its displacement, and an engineer who wants the steepest point of a road profile shaped like a sine wave needs the slope of the sine. Figure 1 already suggests the answer: the slope of $\sin x$ behaves exactly like $\cos x$ .

Figure 1. The slope of

y = \sin x

at each point (red tangents) equals the value of

\cos x

there (orange dots): steepest upward at 0, flat at

\frac{\pi}{2}

, steepest downward at

π \pi

Important

The derivative of $\sin x$ from first principles ( $x$ in radians)

\frac{\sin(x + h) - \sin x}{h} = \frac{\sin x\cos h + \cos x\sin h - \sin x}{h} = \sin x\cdot\frac{\cos h - 1}{h} + \cos x\cdot\frac{\sin h}{h}.

From Lesson 2, $\frac{\sin h}{h} \to 1$ and $\frac{\cos h - 1}{h} \to 0$ as $h \to 0$ . So $(\sin x)' = \cos x$ . In the same way (Practice question 3), $(\cos x)' = -\sin x$ .

The other four, by the quotient rule

(\tan x)' = \left(\frac{\sin x}{\cos x}\right)' = \frac{\cos x\cos x - \sin x(-\sin x)}{\cos^2 x} = \frac{1}{\cos^2 x} = \sec^2 x = 1 + \tan^2 x.

Similarly $(\cot x)' = -\csc^2 x$ , $(\sec x)' = \left((\cos x)^{-1}\right)' = \frac{\sin x}{\cos^2 x} = \sec x\tan x$ , and $(\csc x)' = -\csc x\cot x$ .

$f (x)$	$f^{'} (x)$	with the chain rule, $u = u (x)$
$\sin x$	$\cos x$	$(\sin u)' = u'\cos u$
$\cos x$	$-\sin x$	$(\cos u)' = -u'\sin u$
$\tan x$	$\sec^2 x = 1 + \tan^2 x$	$(\tan u)' = u'\sec^2 u$
$\cot x$	$-\csc^2 x = -(1 + \cot^2 x)$	$(\cot u)' = -u'\csc^2 u$
$\sec x$	$\sec x\tan x$	$(\sec u)' = u'\sec u\tan u$
$\csc x$	$-\csc x\cot x$	$(\csc u)' = -u'\csc u\cot u$

Each formula holds on the domain of the function. Note the pattern: the three "co-" functions have a minus sign.

Definition

Derivative of $f$ at $x$ (from S4): $f'(x) = \lim_{h \to 0}\frac{f(x + h) - f(x)}{h}$ , when this limit exists. It is the gradient of the tangent to the graph at $(x, f (x))$ .

Example: since $(\sin x)' = \cos x$ , the tangent to $y = \sin x$ at the origin has gradient $\cos 0 = 1$ : it is the line $y = x$ , which is why $\sin x \approx x$ for small $x$ .

Worked example 1: the basic rules

Differentiate (a) $y = 3\sin 2x - 4\cos\frac{x}{2}$ (b) $y = x^2\tan x$ (c) $y = \frac{\sin x}{1 + \cos x}$ .

Solution

Chain rule: $y' = 3(2\cos 2x) - 4\left(-\frac{1}{2}\sin\frac{x}{2}\right) = 6\cos 2x + 2\sin\frac{x}{2}$ .
Product rule: $y' = 2x\tan x + x^2\sec^2 x$ .
Quotient rule: $y' = \frac{\cos x(1 + \cos x) - \sin x(-\sin x)}{(1 + \cos x)^2} = \frac{\cos x + \cos^2 x + \sin^2 x}{(1 + \cos x)^2} = \frac{1 + \cos x}{(1 + \cos x)^2} = \frac{1}{1 + \cos x}$ .

Worked example 2: composites

Differentiate (a) $y = \sec(3x - 1)$ (b) $y = \sin^3 2x$ (c) $y = \cos(x^2)$ .

Solution

$y' = 3\sec(3x - 1)\tan(3x - 1)$ .
$y = (\sin 2x)^3$ : $y' = 3\sin^2 2x\cdot(2\cos 2x) = 6\sin^2 2x\cos 2x$ .
$y' = -2x\sin(x^2)$ . Compare $\cos^2 x$ , whose derivative is $-2\sin x\cos x = -\sin 2x$ : $\cos(x^2)$ and $\cos^2 x$ are different functions.

Worked example 3: tangent line and turning points

(a) Find the tangent to $y = \sin x$ at $x = \frac{\pi}{6}$ . (b) Find where $f(x) = x + 2\cos x$ is increasing on $[0, 2\pi]$ , and its turning points.

Solution

The point is $\left(\frac{\pi}{6}, \frac{1}{2}\right)$ and the gradient is $\cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}$ : $y = \frac{1}{2} + \frac{\sqrt{3}}{2}\left(x - \frac{\pi}{6}\right)$ .
$f'(x) = 1 - 2\sin x \ge 0 \iff \sin x \le \frac{1}{2}$ (a trigonometric inequality, Unit 1). So $f$ increases on $\left[0, \frac{\pi}{6}\right]$ and $\left[\frac{5\pi}{6}, 2\pi\right]$ and decreases on $\left[\frac{\pi}{6}, \frac{5\pi}{6}\right]$ . Local maximum $f\left(\frac{\pi}{6}\right) = \frac{\pi}{6} + \sqrt{3} \approx 2.26$ ; local minimum $f\left(\frac{5\pi}{6}\right) = \frac{5\pi}{6} - \sqrt{3} \approx 0.89$ .

Warning

Radians, and signs. $(\sin x)' = \cos x$ needs $x$ in radians; in degrees, $\frac{d}{dx}\sin x^\circ = \frac{\pi}{180}\cos x^\circ$ . The derivative of $\cos x$ is $-\sin x$ , not $\sin x$ . And do not forget the inner derivative: $(\sin 5x)' = 5\cos 5x$ .

Tip

Simplify before or after. $(\sin x\cos x)'$ by the product rule is $\cos^2 x - \sin^2 x$ ; writing $\sin x\cos x = \frac{1}{2}\sin 2x$ first gives $\cos 2x$ at once. The identities of Unit 1 often shorten a derivative.

3Blue1Brown: Derivative formulas through geometry (Essence of calculus, chapter 3). Watch the second half, where a tiny step around the unit circle shows why the derivative of sin is cos, without any limits on paper.

✦

Practice

The last two are harder.

Differentiate (a) $5\sin x - 2\cos x$ (b) $\cos 3x$ (c) $\tan(x^2)$ (d) $\sin x\cos x$ .
Differentiate (a) $x\sin x$ (b) $\frac{\cos x}{x}$ (c) $\cot 2x$ (d) $\csc x$ .
Prove from first principles that $(\cos x)' = -\sin x$ .
Show that $(\sec x)' = \sec x\tan x$ .
Find $\frac{dy}{dx}$ for $y = \sin^2(3x + 1)$ , and simplify.
Find the equation of the tangent to $y = \tan x$ at $x = \frac{\pi}{4}$ .
Find the turning points of $y = \sin x + \cos x$ on $[0, 2\pi]$ .
Find the stationary points of $f(x) = \sin^2 x - \cos x$ on $[0, 2\pi]$ , and the greatest value of $f$ .

Show answersHide answers

(a) $5\cos x + 2\sin x$ (b) $-3\sin 3x$ (c) $2x\sec^2(x^2)$ (d) $\cos^2 x - \sin^2 x = \cos 2x$ .
(a) $\sin x + x\cos x$ (b) $\frac{-x\sin x - \cos x}{x^2}$ (c) $-2\csc^2 2x$ (d) $-\csc x\cot x$ .
$\frac{\cos(x + h) - \cos x}{h} = \cos x\cdot\frac{\cos h - 1}{h} - \sin x\cdot\frac{\sin h}{h} \to \cos x\cdot 0 - \sin x\cdot 1 = -\sin x$ .
$\sec x = (\cos x)^{-1}$ : $(\sec x)' = -(\cos x)^{-2}(-\sin x) = \frac{\sin x}{\cos^2 x} = \frac{1}{\cos x}\cdot\frac{\sin x}{\cos x} = \sec x\tan x$ .
$y' = 2\sin(3x + 1)\cdot 3\cos(3x + 1) = 3\sin(6x + 2)$ .
The point $\left(\frac{\pi}{4}, 1\right)$ , gradient $\sec^2\frac{\pi}{4} = 2$ : $y = 1 + 2\left(x - \frac{\pi}{4}\right)$ .
$y' = \cos x - \sin x = 0 \iff \tan x = 1$ : maximum $\sqrt{2}$ at $x = \frac{\pi}{4}$ , minimum $-\sqrt{2}$ at $x = \frac{5\pi}{4}$ .
$f'(x) = 2\sin x\cos x + \sin x = \sin x(2\cos x + 1) = 0$ : $x = 0, \pi, 2\pi$ or $x = \frac{2\pi}{3}, \frac{4\pi}{3}$ . $f(0) = f(2\pi) = -1$ , $f(\pi) = 1$ , $f\left(\frac{2\pi}{3}\right) = f\left(\frac{4\pi}{3}\right) = \frac{3}{4} + \frac{1}{2} = \frac{5}{4}$ . The greatest value is $\frac{5}{4}$ .

Sources and credits

Written by Ganzaa: the explanations, the derivations, the graph, the worked examples, the practice items and the quiz questions.
Video: Derivative formulas through geometry | Chapter 3, Essence of calculus, 3Blue1Brown, embedded from YouTube.

1How fast does a sine change?

2The other four, by the quotient rule

How fast does a sine change?

The other four, by the quotient rule