Chapter 4, Problem 2P

Show that $|\sin x-\cos x|⩽\sqrt{2}$ for all x.

To determine

To prove: For all values of x, $\left|\sin x-\cos \right|\le \sqrt{2}$.

Explanation of Solution

Proof:

Find absolute minimum and absolute maximum of the function $f\left(x\right)=\sin x-\cos x$ over the interval $\left[0,2\pi \right]$.

Observe that $\sin x\text{and}\cos x$ are both periodic functions of period $2\pi$.

Therefore, $f\left(x\right)=\sin x-\cos x$ is also a periodic function of period $2\pi$.

So, find the critical points of the function $f$ on the interval $\left[0,2\pi \right]$.

Differentiate $f\left(x\right)$ with respect to x .

$f^{\prime }\left(x\right)=\cos x+\sin x$.

Set $f^{\prime }\left(x\right)=0$ and obtain the critical points over the interval $\left[0,2\pi \right]$.

$\begin{array}{l}\cos x+\sin x=0\\ \cos x=-\sin x\\ \tan x=-1\\ x=\frac{3\pi }{4}\text{or }\frac{7\pi }{4}\end{array}$

Thus, the critical points are $\frac{3\pi }{4}\text{and }\frac{7\pi }{4}$.

Apply the extreme values of the given interval and the critical number in $f\left(x\right)$.

Substitute $x=0$ in $f\left(x\right)$,

$\begin{array}{c}f\left(0\right)=\sin 0-\cos 0\\ =0-1\\ =-1\end{array}$

Substitute $x=2\pi$ in $f\left(x\right)$,

$\begin{array}{c}f\left(2\pi \right)=\sin 2\pi -\cos 2\pi \\ =0-1\\ =-1\end{array}$

Substitute $x=\frac{3\pi }{4}$ in $f\left(x\right)$,

$\begin{array}{c}f\left(\frac{3\pi }{4}\right)=\sin \left(\frac{3\pi }{4}\right)-\cos \left(\frac{3\pi }{4}\right)\\ =\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\\ =\frac{2}{\sqrt{2}}\\ =\sqrt{2}\end{array}$

Substitute $x=\frac{7\pi }{4}$ in $f\left(x\right)$,

$\begin{array}{c}f\left(\frac{7\pi }{4}\right)=\sin \left(\frac{7\pi }{4}\right)-\cos \left(\frac{7\pi }{4}\right)\\ =-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\\ =-\frac{2}{\sqrt{2}}\\ =-\sqrt{2}\end{array}$

Since the largest functional value is the absolute maximum and the smallest functional value is the absolute minimum, the absolute maximum of $f\left(x\right)$ is $\sqrt{2}$ and the absolute minimum of $f\left(x\right)$ is $-\sqrt{2}$.

Thus, over the interval $\left[0,2\pi \right]$, $-\sqrt{2}\le \sin x-\cos x\le \sqrt{2}$.

Can be expressed as, $\left|\sin x-\cos x\right|\le \sqrt{2}$.

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