Chapter 7.2, Problem 32AYU

To determine

To calculate: The exact value of expression $\cot \left[{\cos }^{-1}\left(-\frac{\sqrt{3}}{3}\right)\right]$ .

Answer to Problem 32AYU

The exact value is $-\frac{1}{\sqrt{2}}$ .

Explanation of Solution

Given information:

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

Formula used:

Equation $y={\cos }^{-1}x$ means $\cos y=x$ .

Theorem :- For any angle $\theta$ , in standard position, let $P=(x,y)$ be the point on the terminal side of $\theta$ , that is on the circle $x^2+y^2=r^2$ . Then

  $\begin{array}{l}\sin \theta =\frac{y}{r}\\ \cos \theta =\frac{x}{r}\\ \tan \theta =\frac{y}{x},x\ne 0\end{array}$$\begin{array}{l}\cos ec\theta =\frac{r}{y},y\ne 0\\ \sec \theta =\frac{r}{x},x\ne 0\\ \cot \theta =\frac{x}{y},y\ne 0\end{array}$ .

Calculation:

Consider ,

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

Let

  $\begin{array}{l}\theta ={\cos }^{-1}\left(-\frac{\sqrt{3}}{3}\right)\\ \cos \theta =-\frac{\sqrt{3}}{3}\end{array}$

where $-\pi <\theta <\pi$ .

Since $\cos \theta <0$ , it follows $\frac{\pi }{2}<\theta \le \pi$ .

This implies $\theta$ lies in quadrant II.

Now,

  $\cos \theta =\frac{x}{r}=-\frac{\sqrt{3}}{3}$

  $=x=-\sqrt{3};r=3$

Using equation of circle we have,

  $\begin{array}{l}=x^2+y^2=r^2\\ ={\left(-\sqrt{3}\right)}^2+y^2={\left(3\right)}^2\\ =y^2+3=9\\ =y^2=9-3\\ =y^2=6\\ =y=\pm \sqrt{6}\end{array}$

Thus we have,

  $=x=-\sqrt{3};y=\sqrt{6};r=3$ .

Now,

  $\cot \left[{\cos }^{-1}\left(-\frac{\sqrt{3}}{3}\right)\right]=\cot \theta =\frac{-\sqrt{3}}{\sqrt{6}}=-\frac{1}{\sqrt{2}}$ ( where $\cot \frac{x}{y}$ )

Hence , the exact value of $\cot \left[{\cos }^{-1}\left(-\frac{\sqrt{3}}{3}\right)\right]$ is $-\frac{1}{\sqrt{2}}$ .