Chapter 6 · Application of Derivatives

Exercise 6.3 · Q8: Where does $\sin 2x$ peak on $[0,2\pi]$ ⁠?

Exercise 6.3Absolute extrema on a closed interval · Section 6.4.1

The question

8. At what points in the interval $[0,2\pi]$ ⁠, does the function $\sin 2x$ attain its maximum value?

Exam answer

Let $f(x) = \sin 2x$ ⁠. Then $f'(x) = 2\cos 2x$ ⁠.

$f'(x) = 0$ gives $\cos 2x = 0$ ⁠, so in $[0, 2\pi]$ ⁠, $x = \frac{\pi}{4}$ ⁠, $\frac{3\pi}{4}$ ⁠, $\frac{5\pi}{4}$ ⁠, $\frac{7\pi}{4}$ ⁠.

\begin{aligned} f(0) &= 0 \\ f\left(\tfrac{\pi}{4}\right) &= 1 \\ f\left(\tfrac{3\pi}{4}\right) &= -1 \\ f\left(\tfrac{5\pi}{4}\right) &= 1 \\ f\left(\tfrac{7\pi}{4}\right) &= -1 \\ f(2\pi) &= 0 \end{aligned}

The maximum value $1$ is attained at $x = \dfrac{\pi}{4}$ and $x = \dfrac{5\pi}{4}$ ⁠.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

Sine can never exceed $1$ ⁠, so the maximum value of $\sin 2x$ is $1$ provided it is reached. It is reached whenever the angle $2x$ is $\frac{\pi}{2}$ plus a whole number of turns. As $x$ runs over $[0,2\pi]$ ⁠, the angle $2x$ runs over $[0,4\pi]$ ⁠, which is two full turns, so the peak is hit twice.

Given, to find, and the plan

Given: $f(x)=\sin 2x$ on the closed interval $[0,2\pi]$ ⁠.
To find: All points of $[0,2\pi]$ at which $f$ attains its maximum value.
Plan: (1) Use $\sin t\le1$ to get the maximum value. (2) Solve $\sin 2x=1$ for $x\in[0,2\pi]$ ⁠, translating the interval for $x$ into one for $2x$ ⁠. (3) Confirm with the Working Rule of Section 6.4.1 (critical points and end points).

Basics you need

1Range of sine and where it equals 1

$-1\le\sin t\le1$ ⁠, and $\sin t=1$ exactly when $t=\frac{\pi}{2}+2n\pi$ ( $n$ an integer). Basics: sin and cos on the unit circle

2Changing the variable in an interval

If $0\le x\le2\pi$ ⁠, then doubling every part gives $0\le2x\le4\pi$ ⁠. Basics: interval notation

3Working Rule (Section 6.4.1)

On a closed interval, compare $f$ at the critical points inside and at the end points; the largest value is the absolute maximum. Lesson: absolute extrema on a closed interval

Solution

The maximum value is at most $1$

For every $x$ ⁠, $\sin 2x\le1$ ⁠. So no value of $f$ exceeds $1$ ⁠; if $f$ equals $1$ somewhere in $[0,2\pi]$ ⁠, then $1$ is the maximum value and the points where it happens are exactly the points asked for.

Definition 3: the maximum value is a value $f(c)$ with $f(c)\ge f(x)$ for all $x$ in the interval (the book prints $\gt$ ⁠; since $x=c$ is itself in the interval, it must be read as $\ge$ ⁠). A value equal to $1$ beats or ties everything.

$\sin t\le1$ ⁠. Basics
Solve $\sin 2x=1$ with $0\le x\le 2\pi$

Let $t=2x$ ⁠. Then $0\le t\le4\pi$ ⁠, and $\sin t=1$ means $t=\frac{\pi}{2}+2n\pi$ ⁠. The values in $[0,4\pi]$ are
$t=\tfrac{\pi}{2}\quad\text{and}\quad t=\tfrac{\pi}{2}+2\pi=\tfrac{5\pi}{2}$
(the next one, $\frac{9\pi}{2}$ ⁠, exceeds $4\pi=\frac{8\pi}{2}$ ⁠, and $n\lt0$ gives negative $t$ ⁠). Dividing by $2$ ⁠:
$x=\tfrac{\pi}{4}\quad\text{and}\quad x=\tfrac{5\pi}{4}.$
The common slip is to solve $\sin 2x=1$ only for $2x$ in $[0,2\pi]$ and find one answer. Because the angle is $2x$ ⁠, it goes round the circle twice while $x$ goes round once.

Doubling an interval. Basics: interval notation

Confirm with the Working Rule

$f'(x)=2\cos 2x$ (chain rule, inner function $2x$ ⁠). $f'(x)=0$ when $\cos 2x=0$ ⁠, i.e. $2x=\frac{\pi}{2},\frac{3\pi}{2},\frac{5\pi}{2},\frac{7\pi}{2}$ in $[0,4\pi]$ ⁠, so $x=\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}$ ⁠. With the end points:

$x$	$0$	$\frac{\pi}{4}$	$\frac{3\pi}{4}$	$\frac{5\pi}{4}$	$\frac{7\pi}{4}$	$2\pi$
$2x$	$0$	$\frac{\pi}{2}$	$\frac{3\pi}{2}$	$\frac{5\pi}{2}$	$\frac{7\pi}{2}$	$4\pi$
$\sin 2x$	$0$	$1$	$-1$	$1$	$-1$	$0$

The largest value, $1$ ⁠, occurs at $x=\frac{\pi}{4}$ and $x=\frac{5\pi}{4}$ and nowhere else on the list. So $\sin 2x$ attains its maximum value $1$ at $x=\frac{\pi}{4}$ and $x=\frac{5\pi}{4}$ .

By Theorems 5 and 6 every point of absolute maximum in

[0,2\pi]

is on this list (an interior one must have

f'=0

⁠), so no other point can give the value

1

⁠. The two methods agree.

Working Rule, Section 6.4.1. Lesson: absolute extrema

\sin 2x

[0,2\pi]

makes two full waves. It touches the dashed line

y=1

x=\frac{\pi}{4}

and

x=\frac{5\pi}{4}

⁠, and reaches

-1

\frac{3\pi}{4}

and

\frac{7\pi}{4}

⁠.

Result

The maximum value of $\sin 2x$ on $[0,2\pi]$ is $1$ ⁠, attained at $x=\frac{\pi}{4}$ and $x=\frac{5\pi}{4}$ .

(Textbook answer: At x = π/4 and 5π/4.)

Check it with numbers

$\frac{\pi}{4}\approx0.785398$ ⁠: $\sin(1.570796)=1.000000$ ⁠. ✓ $\frac{5\pi}{4}\approx3.926991$ ⁠: $\sin(7.853982)=1.000000$ ⁠. ✓

Nearby: $\sin(2\times0.7)=\sin1.4\approx0.985450\lt1$ ⁠, $\sin(2\times0.9)=\sin1.8\approx0.973848\lt1$ ⁠. ✓

Common mistakes

Giving only $x=\frac{\pi}{4}$ ⁠. The angle $2x$ covers $[0,4\pi]$ ⁠, two full turns, so there are two peaks.
Answering $x=\frac{\pi}{2}$ ⁠. That solves $\sin x=1$ ⁠, not $\sin 2x=1$ ⁠; in fact $\sin(2\cdot\frac{\pi}{2})=\sin\pi=0$ ⁠.
Including $\frac{3\pi}{4}$ and $\frac{7\pi}{4}$ ⁠. They are critical points too, but they give the minimum value $-1$ ⁠.

What to take away

For $\sin(kx)$ on an interval for $x$ ⁠, first find the interval for the angle $kx$ ⁠.
Known bounds ( $\sin\le1$ ⁠) give the maximum value; solving for where it is reached gives the points.
The Working Rule is a reliable cross-check: it lists every possible point of absolute maximum.

Every fact used, and where it comes from

Fact	Used in	Source
$\sin t\le1$ ⁠; $\sin t=1\iff t=\frac{\pi}{2}+2n\pi$ ⁠; $\cos t=0$ at odd multiples of $\frac{\pi}{2}$	Steps 1–3	Class XI, Chapter 3 (Trigonometric Functions) (Basics)
Maximum value of a function on an interval	Step 1	Definition 3, Section 6.4 (Lesson: maxima and minima)
Chain rule: $(\sin 2x)'=2\cos 2x$	Step 3	Chapter 5 (Continuity and Differentiability)
Theorems 5, 6 and the Working Rule	Step 3	Section 6.4.1 (Lesson: absolute extrema)

Found an error, or want more?

Tell us if something on this page is wrong or unclear, or if you would like a fuller explanation, another method or a new feature. Send feedback on this problem →

Exercise 6.3 · Q8: Where does sin⁡2x\sin 2xsin2x peak on [0,2π][0,2\pi][0,2π]⁠?

Basics you need

Solution

The maximum value is at most 111

Solve sin⁡2x=1\sin 2x=1sin2x=1 with 0≤x≤2π0\le x\le 2\pi0≤x≤2π

Confirm with the Working Rule

Every fact used, and where it comes from

Exercise 6.3 · Q8: Where does $\sin 2x$ peak on $[0,2\pi]$ ⁠?

The maximum value is at most $1$

Solve $\sin 2x=1$ with $0\le x\le 2\pi$