Chapter 6.3, Problem 53AYU

In Problems 43-58, find the exact value of each of the remaining trigonometric functions of $\theta$.

$\sec \theta \text{=}\text{2}$, $\sin \theta \text{<}\text{0}$

To determine

To find: The exact value of each of the remaining trigonometric functions of $\theta$.

Answer to Problem 53AYU

Solution:

$\text{cos}\theta =\frac{1}{2}$, $\text{sin}\theta = -\frac{\sqrt{3}}{2}$, $\text{tan}\theta =-\sqrt{3}$, $\text{csc}\theta =-\frac{2\sqrt{3}}{3}$, $\text{cot}\theta =-\frac{\sqrt{3}}{3}$

Explanation of Solution

Given:

$\text{sec}\theta =2,\text{sin}\theta <0$

Calculation:

Since $\theta$ in quadrant IV, $\text{sin}\theta <0$.

Given $\text{sec}\theta =2$

Therefore $\text{cos}\theta =\frac{1}{2}$

${\text{sin}}^2\theta +{\text{cos}}^2\theta =1$

${\text{sin}}^2\theta =1-{\text{cos}}^2\theta$

${\text{sin}}^2\theta =1-{\left(\frac{1}{2}\right)}^2$

${\text{sin}}^2\theta =1-\frac{1}{4}$

${\text{sin}}^2\theta =\frac{3}{4}$

$\text{sin}\theta =\pm \frac{\sqrt{3}}{2}$

Here $\text{sin}\theta =-\frac{\sqrt{3}}{2}$.

$\text{tan}\theta =\frac{\text{sin}\theta }{\text{cos}\theta }=\frac{-\frac{\sqrt{3}}{2}}{\frac{1}{2}}=-\sqrt{3}$

$\text{csc}\theta =\frac{1}{\text{sin}\theta }=\frac{1}{-\frac{\sqrt{3}}{2}}=-\frac{2}{\sqrt{3}}=-\frac{2\sqrt{3}}{3}$

$\text{cot}\theta =\frac{1}{\text{tan}\theta }=\frac{1}{-\sqrt{3}}=-\frac{\sqrt{3}}{3}$