Chapter 4, Problem 28CR

To determine

To find:thesix trigonometric functions at the real number $t=-\frac{2\pi }{3}$ if it is possible.

Answer to Problem 28CR

The six trigonometric functions are $\sin \left(\frac{-2\pi }{3}\right)=-\frac{\sqrt{3}}{2}$ , $\cos \left(-\frac{2\pi }{3}\right)=-\frac{1}{2}$ , $\tan \left(-\frac{2\pi }{3}\right)=\sqrt{3}$ , $\csc \left(-\frac{2\pi }{3}\right)=-\frac{2\sqrt{3}}{3}$ , $\sec \left(-\frac{2\pi }{3}\right)=-2$ , and $\cot \left(\frac{-2\pi }{3}\right)=\frac{\sqrt{3}}{3}$ .

Explanation of Solution

Definition used:

Let t be a real number and let $\left(x,y\right)$ be the point on the unit circle corresponding to t.

  $\begin{array}{l}\sin t=y\cos t=x\tan t=\frac{y}{x},x\ne 0\\ \csc t=\frac{1}{y},y\ne 0\sec t=\frac{1}{x}\text{,}x\ne 0\cot t=\frac{x}{y},y\ne 0\end{array}$

Calculation:

For every real number t, there exist the corresponding to the point $\left(x,y\right)$ on the unit circle.

Moving clockwise around the unit circle, the value of $t=-\frac{2\pi }{3}$ gets in the place of $t=\frac{4\pi }{3}$ .

From Figure 4.14 on page 268, it is observed that the real number $t=\frac{4\pi }{3}$ corresponding to the point $\left(-\frac{1}{2},-\frac{\sqrt{3}}{2}\right)$ .

That is, the real number $t=-\frac{2\pi }{3}$ corresponding to the point $\left(-\frac{1}{2},-\frac{\sqrt{3}}{2}\right)$ on the unit circle.

Therefore, the values of $x=-\frac{1}{2}\text{ and }y=-\frac{\sqrt{3}}{2}$ .

The six trigonometric function at the real number is computed as follows,

Substitute $t=-\frac{2\pi }{3}$ and $y=-\frac{\sqrt{3}}{2}$ in $\sin t=y$ , the trigonometric function $\sin \left(\frac{-2\pi }{3}\right)=-\frac{\sqrt{3}}{2}$ .

Substitute $t=-\frac{2\pi }{3}$ and $x=-\frac{1}{2}$ in $\cos t=x$ , the trigonometric function $\cos \left(-\frac{2\pi }{3}\right)=-\frac{1}{2}$ .

Substitute $t=-\frac{2\pi }{3}$ , $x=-\frac{1}{2}$ , and $y=-\frac{\sqrt{3}}{2}$ in $\tan t=\frac{y}{x}$ , the trigonometric function is,

  $\begin{array}{c}\tan \left(-\frac{2\pi }{3}\right)=\frac{-\frac{\sqrt{3}}{2}}{-\frac{1}{2}}\\ =\sqrt{3}\end{array}$

That is, the trigonometric function $\tan \left(-\frac{2\pi }{3}\right)=\sqrt{3}$ .

Substitute $t=-\frac{2\pi }{3}$ and $y=-\frac{\sqrt{3}}{2}$ in $\csc t=\frac{1}{y}$ , the trigonometric function is,

  $\begin{array}{c}\csc \left(-\frac{2\pi }{3}\right)=\frac{1}{-\frac{\sqrt{3}}{2}}\\ =-\frac{2}{\sqrt{3}}\\ =-\frac{2}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}\\ =-\frac{2\sqrt{3}}{3}\end{array}$

That is, the trigonometric function $\csc \left(-\frac{2\pi }{3}\right)=-\frac{2\sqrt{3}}{3}$ .

Substitute $t=-\frac{2\pi }{3}$ and $x=-\frac{1}{2}$ in $\sec t=\frac{1}{x}$ , the trigonometric function is,

  $\begin{array}{c}\sec \left(-\frac{2\pi }{3}\right)=\frac{1}{-\frac{1}{2}}\\ =-2\end{array}$

That is, the trigonometric function $\sec \left(-\frac{2\pi }{3}\right)=-2$ .

Substitute $t=-\frac{2\pi }{3}$ , $x=-\frac{1}{2}$ , and $y=-\frac{\sqrt{3}}{2}$ in $\cot t=\frac{x}{y}$ , the trigonometric function is,

  $\begin{array}{c}\cot \left(\frac{-2\pi }{3}\right)=\frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}}\\ =\frac{1}{2}\times \frac{2}{\sqrt{3}}\\ =\frac{1}{\sqrt{3}}\\ =\frac{\sqrt{3}}{3}\end{array}$

That is, the trigonometric function $\cot \left(\frac{-2\pi }{3}\right)=\frac{\sqrt{3}}{3}$ .

Therefore, the six trigonometric functions are $\sin \left(\frac{-2\pi }{3}\right)=-\frac{\sqrt{3}}{2}$ , $\cos \left(-\frac{2\pi }{3}\right)=-\frac{1}{2}$ , $\tan \left(-\frac{2\pi }{3}\right)=\sqrt{3}$ , $\csc \left(-\frac{2\pi }{3}\right)=-\frac{2\sqrt{3}}{3}$ , $\sec \left(-\frac{2\pi }{3}\right)=-2$ , and $\cot \left(\frac{-2\pi }{3}\right)=\frac{\sqrt{3}}{3}$ .