Chapter 4.4, Problem 67E

To determine

The sine, cosine, and tangent of the angle $-\frac{17\pi }{6}$ without using a calculator.

Answer to Problem 67E

The sine, cosine, and tangent of the angle $-\frac{17\pi }{6}$ without using a calculator is

  $\begin{array}{l}\sin \left(-\frac{17\pi }{6}\right)=-\frac{1}{2}\\ \cos \left(-\frac{17\pi }{6}\right)=-\frac{\sqrt{3}}{2}\\ \tan \left(-\frac{17\pi }{6}\right)=\frac{1}{\sqrt{3}}\end{array}$

Explanation of Solution

Given info:

  $\theta =-\frac{17\pi }{6}$

Formula used:

Let be $\theta$ an angle in standard position. Its reference angle $\theta \text{'}$ is the acute angleformed by the terminal side of $\theta$ and the horizontal axis.

In quadrant I all trigonometric functions are positive.

In quadrant II sine functions are positive.

In quadrant III tangent functions are positive.

In quadrant IV cosine functions are positive.

Calculation:

We have

  $\begin{array}{l}\theta =-\frac{17\pi }{6}\\ \theta =4\pi -\frac{17\pi }{6}=\frac{7\pi }{6}\text{(coterminal angle)}\\ \theta \text{'}=\frac{7\pi }{6}-\pi =\frac{\pi }{6}\end{array}$

Here $\theta =\frac{7\pi }{6}$ lies in third quadrant,

In quadrant IIItangent functions are positive.

Hence

  $\begin{array}{l}\sin \left(-\frac{17\pi }{6}\right)=-\sin \frac{\pi }{6}=-\frac{1}{2}\\ \cos \left(-\frac{17\pi }{6}\right)=-\cos \frac{\pi }{6}=-\frac{\sqrt{3}}{2}\\ \tan \left(-\frac{17\pi }{6}\right)=\tan \frac{\pi }{6}=\frac{1}{\sqrt{3}}\end{array}$

Conclusion:

Thus,sine, cosine, and tangent of the angle $-\frac{17\pi }{6}$ without using a calculator is

  $\begin{array}{l}\sin \left(-\frac{17\pi }{6}\right)=-\frac{1}{2}\\ \cos \left(-\frac{17\pi }{6}\right)=-\frac{\sqrt{3}}{2}\\ \tan \left(-\frac{17\pi }{6}\right)=\frac{1}{\sqrt{3}}\end{array}$