Chapter C, Problem 35E

To determine

To calculate:

The value of $x$ to satisfy the following inequality equation:

  $-1<\tan x<1$

Answer to Problem 35E

The values of $x$ for the interval $\left[0,2\pi \right]$ to satisfy eq. $-1<\tan x<1$ are $\left[0\le x\le \frac{\pi }{4}\right],\left[\frac{3\pi }{4}\le x\le \frac{5\pi }{4}\right],\left[\frac{7\pi }{4}\le x\le 2\pi \right]$ .

Explanation of Solution

Given information:

  $\begin{array}{l}-1<\tan x<1\\ \left[0,2\pi \right]\end{array}$

Calculation:

Know that:

  $\begin{array}{l}-1<\tan x<1\\ -1<\frac{\sin x}{\cos x}<1\end{array}$

That means, for the tangent value of $\pm 1$ , the value of sine and cosine value are same with the different signs. It will occurs at only $\pm \frac{1}{\sqrt{2}}$ , angle is odd multiple of $\frac{\pi }{4}$ .

So, $x$ for the interval $\left[0,2\pi \right]$ are $\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4}$ . These angles break into 4 sections as:

In the first quadrant $0<\tan x<1$ , $x$ value of interval $\left[0\le x\le \frac{\pi }{4}\right]$ , it is greater than $-1$ that means the inequality equation is not satisfied with this range.

In the second quadrant $-1<\tan x<0$ , $x$ value of interval $\left[\frac{3\pi }{4}\le x\le \pi \right]$

In the third quadrant $0<\tan x<1$ , $x$ value of interval $\left[\pi \le x\le \frac{5\pi }{4}\right]$

In the fourth quadrant $-1<\tan x<0$ , $x$ value of interval $\left[\frac{7\pi }{4}\le x\le 2\pi \right]$

So, the unit circle with co-ordinates $\left(x,y\right)=\left(\cos x,\sin x\right)$ and radius $1$ :

  

Fig. Unit circle

Therefore, the values of $x$ for the interval $\left[0,2\pi \right]$ to satisfy eq. $-1<\tan x<1$ are $\left[0\le x\le \frac{\pi }{4}\right],\left[\frac{3\pi }{4}\le x\le \frac{5\pi }{4}\right],\left[\frac{7\pi }{4}\le x\le 2\pi \right]$ .