Chapter 3.P, Problem 1P

Show that $\left|\sin x-\cos x\right|\le \sqrt{2}$ for all $x$

To determine

To show:

$\left|\mathit{sin}x-\mathit{cos}x\right|\le \sqrt{2}$ for all $x$

Answer to Problem 1P

Solution:

$\left|\mathit{sin}x-\mathit{cos}x\right|\le \sqrt{2}$

Explanation of Solution

1) Concept:

Use Closed Interval Method to find the absolute maximum and absolute minimum values for the given function.

2) The Closed Interval Method:

To find the absolute maximum and minimum values of a continuous function $f$ on a closed interval $\left[a, b\right]$:

i) Find the values of $f$ at the critical numbers of $f$ in $\left(a, b\right)$.

ii) Find the values of $f$ at the endpoints of the interval.

iii) The largest of the values from the above steps is the absolute maximum value and the smallest of these values is the absolute minimum value.

3) Formula:

$\left|\mathrm{z}\right|\le \mathrm{a}\mathrm{ }\Rightarrow -\mathrm{a}\le \mathrm{z}\le \mathrm{a}$

4) Calculation:

To show that $\left|\mathit{sin}x-\mathit{cos}x\right|\le \sqrt{2}$ for all $x$

Using formula, above equation becomes

$-\sqrt{2}\le \mathit{sin}x-\mathit{cos}x\le \sqrt{2}$

Now, let $f\left(x\right)=\mathit{sin}x-\mathit{cos}x$

The function $f$ is periodic with period $2\pi$.

Take derivative of $f$.

$f^{\text{'}}\left(x\right)=\mathit{cos}x-(-sinx)$

Simplify.

$f^{\text{'}}\left(x\right)=\mathit{cos}x+sinx$

To find critical numbers, equate $f^{\text{'}}\left(x\right)=0$.

$\mathit{cos}x+sinx=0$

Subtract $sinx$

$\mathit{cos}x+sinx-sinx=-sinx$

Simplify.

$\mathit{cos}x=-sinx$

Divide by $\mathit{cos}x$

$\tan x=-1$

This is only true when,$x=\frac{3\pi }{4}$ and $\frac{7\pi }{4}$

Therefore, critical numbers are $\frac{3\pi }{4}, \frac{7\pi }{4}$

To find the absolute maximum and minimum values

substitute critical numbers and endpoints in the function $f$.

Substitute.$x=\frac{3\pi }{4}$ in $f\left(x\right)=\mathit{sin}x-\mathit{cos}x$

$f\left(\frac{3\pi }{4}\right)=\mathit{sin}\frac{3\pi }{4}-\mathit{cos}\frac{3\pi }{4}$

Simplify.

$f\left(\frac{3\pi }{4}\right)=\sqrt{2}>0$

Substitute.$x=\frac{7\pi }{4}$ in $f\left(x\right)=\mathit{sin}x-\mathit{cos}x$

$f\left(\frac{7\pi }{4}\right)=\mathit{sin}\left(\frac{7\pi }{4}\right)-\mathit{cos}(\frac{7\pi }{4})$

Simplify.

$f\left(\frac{7\pi }{4}\right)=-\sqrt{2}<0$

Substitute, $x=0$ in $f\left(x\right)=\mathit{sin}x-\mathit{cos}x$

$f\left(0\right)=\mathit{sin}0-\mathit{cos}0$

Simplify.

$f\left(0\right)=-1<0$

Substitute.

 $x=2\pi$ in $f\left(x\right)=\mathit{sin}x-\mathit{cos}x$

$f\left(2\pi \right)=\mathit{sin}\left(2\pi \right)-\mathit{cos}(2\pi )$

Simplify.

$f\left(2\pi \right)=-1<0$

By Closed Interval Method,

$f$ has absolute maximum value $\sqrt{2}$ and absolute minimum value $-\sqrt{2}$.

Therefore, $-\sqrt{2}\le \mathit{sin}x-\mathit{cos}x\le \sqrt{2}$ for all $x$.

Therefore, $\left|\mathit{sin}x-\mathit{cos}x\right|\le \sqrt{2}$ for all $x$.

Conclusion:

Therefore, $\left|\mathit{sin}x-\mathit{cos}x\right|\le \sqrt{2}$ for all $x$.