Chapter 7.2, Problem 32AYU

In Problems 9-36, find the exact value of each expression.

$\cot \left[{\cos }^{-1}\left(-\frac{\sqrt{3}}{3}\right)\right]$

To determine

To find: The exact value of $\cot \left({\cos }^{-1}\left(-\surd 3/3\right)\right)$.

Answer to Problem 32AYU

Solution:

The exact value of $\cot \left({\cos }^{-1}\left(-\surd 3/3\right)\right)\text{ = }-\surd 3/\surd 6$.

Explanation of Solution

Given:

$\cot \left({\cos }^{-1}\left(-\surd 3/3\right)\right)$

Calculation:

Let $\theta =\left({\cos }^{-1}\left(-\surd 3/3\right)\right)$   -----1

$\cos \theta \text{ = }-\surd 3/3$ where $\text{0 }\le \theta \le \pi$.

From equation 1,

$\cot \left({\cos }^{-1}\left(-\surd 3/3\right)\right)\text{ = cot}\theta$

We need to find $\text{cot}\theta$.

$\cos \theta \text{ = }-\surd 3/3\text{ = }x/r$

$x=-\surd 3$ and $r=\text{ 3}$

$r\text{ = }d\left(\text{O},\text{ P}\right)\text{ = 3}$

Equation of the circle $=x^2+y^2=\text{ 9}$.

using the formula,

$y=\surd (r^2-x^2)=\surd \left(3^2\right)-(-\surd 3^2))$

$y\text{ = }\surd 6$

$y\text{ = }\surd 6$ and $x=-\surd 3$

Therefore, $\theta$ is in II quadrant.

$\cot \theta =x/y=-\surd 3/\surd 6$

$\cot \left({\cos }^{-1}(-\surd 3/3)\right)=\cot \theta =-\surd 3/\surd 6$