Chapter 6 · Application of Derivatives

Exercise 6.3 · Q9: Maximum of $\sin x+\cos x$

Exercise 6.3Maximum value of a function · Sections 6.4 and 6.4.1

The question

9. What is the maximum value of the function $\sin x+\cos x$ ⁠?

Exam answer

Since $\cos\frac{\pi}{4} = \sin\frac{\pi}{4} = \frac{1}{\sqrt2}$ ⁠,

\begin{aligned} &\sin x + \cos x \\ &= \sqrt2\left(\sin x\cos\tfrac{\pi}{4} + \cos x\sin\tfrac{\pi}{4}\right) \\ &= \sqrt2\sin\left(x + \tfrac{\pi}{4}\right) \end{aligned}

As $\sin\left(x + \frac{\pi}{4}\right) \le 1$ ⁠, $\sin x + \cos x \le \sqrt2$ ⁠, with equality at $x = \frac{\pi}{4}$ ⁠.

The maximum value of $\sin x + \cos x$ is $\sqrt2$ ⁠.

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

The one idea this problem is built on

Adding a sine wave and a cosine wave of the same period gives another sine wave, only taller and shifted: $\sin x+\cos x=\sqrt2\sin\left(x+\frac{\pi}{4}\right)$ ⁠. A sine never exceeds $1$ ⁠, so this never exceeds $\sqrt2$ ⁠, and it equals $\sqrt2$ when $x=\frac{\pi}{4}$ ⁠. Calculus on one full period gives the same answer.

Given, to find, and the plan

Given: $f(x)=\sin x+\cos x$ for all real $x$ (no interval is stated).
To find: The maximum value of $f$ on $\R$ ⁠.
Plan: Method 1: rewrite $f$ as a single sine, bound it, and find where the bound is reached. Method 2: use periodicity to reduce to the closed interval $[0,2\pi]$ and apply the Working Rule of Section 6.4.1.

Basics you need

1Sine of a sum

$\sin(A+B)=\sin A\cos B+\cos A\sin B$ ⁠, and $\sin\frac{\pi}{4}=\cos\frac{\pi}{4}=\frac{1}{\sqrt2}$ ⁠. Basics: sin and cos on the unit circle

2Range of sine

$\sin t\le1$ ⁠, with equality exactly when $t=\frac{\pi}{2}+2n\pi$ ⁠.

3Periodicity

$\sin(x+2\pi)=\sin x$ and $\cos(x+2\pi)=\cos x$ ⁠, so $f$ repeats every $2\pi$ ⁠: every value of $f$ on $\R$ already occurs for some $x$ in $[0,2\pi]$ ⁠.

4Working Rule (Section 6.4.1)

On a closed interval, compare $f$ at the interior critical points and at the end points. Lesson: absolute extrema on a closed interval

Solution

Method 1: write $f$ as one sine

Expand $\sqrt2\sin\left(x+\frac{\pi}{4}\right)$ with the sum formula:
$\begin{aligned} &\sqrt2\left(\sin x\cos\tfrac{\pi}{4}+\cos x\sin\tfrac{\pi}{4}\right)\\ &=\sqrt2\cdot\tfrac{1}{\sqrt2}\left(\sin x+\cos x\right)\\ &=\sin x+\cos x. \end{aligned}$
So $f(x)=\sqrt2\sin\left(x+\frac{\pi}{4}\right)$ for every real $x$ ⁠.

The factor $\sqrt2$ is chosen so that $\sqrt2\cdot\frac{1}{\sqrt2}=1$ ⁠: it makes the coefficients of $\sin x$ and $\cos x$ both $1$ ⁠. In general $a\sin x+b\cos x$ has the "height" $\sqrt{a^2+b^2}$ ⁠, here $\sqrt{1+1}=\sqrt2$ ⁠.

Sine of a sum. Basics
Bound it and find where the bound is reached

Since $\sin\left(x+\frac{\pi}{4}\right)\le1$ ⁠, we get $f(x)\le\sqrt2$ for all $x$ ⁠. Equality needs $x+\frac{\pi}{4}=\frac{\pi}{2}+2n\pi$ ⁠, i.e. $x=\frac{\pi}{4}+2n\pi$ ⁠. For example
$f\!\left(\tfrac{\pi}{4}\right)=\tfrac{\sqrt2}{2}+\tfrac{\sqrt2}{2}=\sqrt2.$
So the maximum value is $\sqrt2$ .

Both parts of Definition 3 are now in place: $\sqrt2$ is never exceeded, and it is actually attained.

Definition 3, Section 6.4 (printed with $\gt$ ⁠; read $\ge$ ⁠). Lesson: maxima and minima
Method 2: calculus on one period

Because $f$ has period $2\pi$ ⁠, its maximum value on $\R$ is its maximum value on $[0,2\pi]$ ⁠. There, $f'(x)=\cos x-\sin x$ ⁠. If $\cos x=0$ then $\sin x=\pm1$ and $f'\ne0$ ⁠, so at a zero of $f'$ we may divide by $\cos x$ ⁠: $\tan x=1$ ⁠, giving $x=\frac{\pi}{4}$ or $\frac{5\pi}{4}$ in $[0,2\pi]$ ⁠.

$x$ $0$ $\frac{\pi}{4}$ $\frac{5\pi}{4}$ $2\pi$

$\sin x+\cos x$ $1$ $\sqrt2$ $-\sqrt2$ $1$

The largest value is $\sqrt2$ ⁠, at $x=\frac{\pi}{4}$ ⁠, matching Method 1.

The Working Rule needs a closed interval; $\R$ is not one, which is why we first use periodicity. At $\frac{5\pi}{4}$ both sine and cosine equal $-\frac{\sqrt2}{2}$ ⁠, giving $-\sqrt2$ (the minimum value).

Working Rule, Section 6.4.1. Lesson: absolute extrema

$x$	$0$	$\frac{\pi}{4}$	$\frac{5\pi}{4}$	$2\pi$
$\sin x+\cos x$	$1$	$\sqrt2$	$-\sqrt2$	$1$

\sin x+\cos x=\sqrt2\sin\left(x+\frac{\pi}{4}\right)

⁠: a wave of height

\sqrt2

⁠, touching the dashed line

y=\sqrt2

x=\frac{\pi}{4}

(and every

2\pi

after), and

y=-\sqrt2

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

and

-\frac{3\pi}{4}

⁠.

Result

The maximum value of $\sin x+\cos x$ is $\sqrt2$ , attained at $x=\frac{\pi}{4}+2n\pi$ ( $n$ any integer).

(Textbook answer: Maximum value = √2.)

Check it with numbers

$f(0.785398)\approx0.707107+0.707107$ $=1.414214=\sqrt2$ ⁠. ✓

$f(0.7)\approx1.409060$ and $f(0.9)\approx0.783327+0.621610$ $=1.404937$ ⁠, both below $1.414214$ ⁠. ✓

Common mistakes

Answering $2$ because each of $\sin x$ and $\cos x$ can be $1$ ⁠. They are never $1$ at the same $x$ ⁠.
Applying the closed-interval rule directly to $\R$ ⁠. First reduce to one period, or use the single-sine form.
Dividing by $\cos x$ without checking it is non-zero.

What to take away

$a\sin x+b\cos x$ has maximum value $\sqrt{a^2+b^2}$ and minimum value $-\sqrt{a^2+b^2}$ ⁠.
For a periodic function, the maximum over all real numbers is the maximum over one period.

Every fact used, and where it comes from

Fact	Used in	Source
$\sin(A+B)=\sin A\cos B+\cos A\sin B$ ⁠; $\sin\frac{\pi}{4}=\cos\frac{\pi}{4}=\frac{1}{\sqrt2}$	Step 1	Class XI, Chapter 3 (Trigonometric Functions) (Basics)
$\sin t\le1$ ⁠, equality at $t=\frac{\pi}{2}+2n\pi$ ⁠; period $2\pi$	Steps 2, 3	Class XI, Chapter 3
Maximum value of a function	Step 2	Definition 3, Section 6.4 (Lesson: maxima and minima)
Theorems 5, 6 and the Working Rule	Step 3	Section 6.4.1 (Lesson: absolute extrema)

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Exercise 6.3 · Q9: Maximum of sin⁡x+cos⁡x\sin x+\cos xsinx+cosx

Basics you need

Solution

Method 1: write fff as one sine

Bound it and find where the bound is reached

Method 2: calculus on one period

Every fact used, and where it comes from

Exercise 6.3 · Q9: Maximum of $\sin x+\cos x$

Method 1: write $f$ as one sine