Chapter 4, Problem 27CR

To determine

To find:thesix trigonometric functions at the real number $t=-\frac{11\pi }{6}$ if it is possible.

Answer to Problem 27CR

The six trigonometric functions are $\sin \left(\frac{-11\pi }{6}\right)=\frac{1}{2}$ , $\cos \left(\frac{-11\pi }{6}\right)=\frac{\sqrt{3}}{2}$ , $\tan \left(\frac{-11\pi }{6}\right)=\frac{\sqrt{3}}{3}$ , $\csc \left(-\frac{11\pi }{6}\right)=2$ , $\sec \left(-\frac{11\pi }{6}\right)=\frac{2\sqrt{3}}{3}$ , and $\cot \left(\frac{-11\pi }{6}\right)=\sqrt{3}$ .

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 $t=-\frac{11\pi }{6}$ gets in the place of $t=\frac{\pi }{6}$ .

From Figure 4.14 on page 268, it is observed that the real number $t=\frac{\pi }{6}$ corresponding to the point $\left(\frac{\sqrt{3}}{2},\frac{1}{2}\right)$ .

That is, the real number $t=-\frac{11\pi }{6}$ corresponding to the point $\left(\frac{\sqrt{3}}{2},\frac{1}{2}\right)$ on the unit circle.

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

Substitute $t=-\frac{11\pi }{6}$ and $y=\frac{1}{2}$ in $\sin t=y$ , the trigonometric function $\sin \left(\frac{-11\pi }{6}\right)=\frac{1}{2}$ .

Substitute $t=-\frac{11\pi }{6}$ and $x=\frac{\sqrt{3}}{2}$ in $\cos t=x$ , the trigonometric function $\cos \left(\frac{-11\pi }{6}\right)=\frac{\sqrt{3}}{2}$ .

Substitute $t=-\frac{11\pi }{6}$ , $x=\frac{\sqrt{3}}{2}$ , and $y=\frac{1}{2}$ in $\tan t=\frac{y}{x}$ , the trigonometric function is,

  $\begin{array}{c}\tan \left(\frac{-11\pi }{6}\right)=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}\\ =\frac{1}{\sqrt{3}}\\ =\frac{\sqrt{3}}{3}\left[\because \text{ Multiply by }\sqrt{3}\text{ on numerator and denominator}\right]\end{array}$

That is, the trigonometric function $\tan \left(\frac{-11\pi }{6}\right)=\frac{\sqrt{3}}{3}$ .

Substitute $t=-\frac{11\pi }{6}$ and $y=\frac{1}{2}$ in $\csc t=\frac{1}{y}$ , the trigonometric function is,

  $\begin{array}{c}\csc \left(-\frac{11\pi }{6}\right)=\frac{1}{\frac{1}{2}}\\ =2\end{array}$

That is, the trigonometric function $\csc \left(-\frac{11\pi }{6}\right)=2$ .

Substitute $t=-\frac{11\pi }{6}$ and $x=\frac{\sqrt{3}}{2}$ in $\sec t=\frac{1}{x}$ , the trigonometric function is,

  $\begin{array}{c}\sec \left(-\frac{11\pi }{6}\right)=\frac{1}{\frac{\sqrt{3}}{2}}\\ =\frac{2}{\sqrt{3}}\\ =\frac{2}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}\\ =\frac{2\sqrt{3}}{3}\end{array}$

That is, the trigonometric function $\sec \left(-\frac{11\pi }{6}\right)=\frac{2\sqrt{3}}{3}$ .

Substitute $t=-\frac{11\pi }{6}$ , $x=\frac{\sqrt{3}}{2}$ , and $y=\frac{1}{2}$ in $\cot t=\frac{x}{y}$ , the trigonometric function is,

  $\begin{array}{c}\cot \left(\frac{-11\pi }{6}\right)=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}\\ =\frac{\sqrt{3}}{1}\\ =\sqrt{3}\end{array}$

That is, the trigonometric function $\cot \left(\frac{-11\pi }{6}\right)=\sqrt{3}$ .

Therefore, the six trigonometric functions are $\sin \left(\frac{-11\pi }{6}\right)=\frac{1}{2}$ , $\cos \left(\frac{-11\pi }{6}\right)=\frac{\sqrt{3}}{2}$ , $\tan \left(\frac{-11\pi }{6}\right)=\frac{\sqrt{3}}{3}$ , $\csc \left(-\frac{11\pi }{6}\right)=2$ , $\sec \left(-\frac{11\pi }{6}\right)=\frac{2\sqrt{3}}{3}$ , and $\cot \left(\frac{-11\pi }{6}\right)=\sqrt{3}$ .