Chapter 7, Problem 11CRT

(a)

To determine

To find:The domain and range of the function $f\left(x\right)={\cos }^{-1}x$ and sketch its graph.

Answer to Problem 11CRT

The domain and range of the function $f\left(x\right)={\cos }^{-1}x$ is $\left[-1,1\right]$ and $\left[0,\pi \right]$ respectively. The graph of the function $f\left(x\right)={\cos }^{-1}x$ is shown in Figure(1).

Explanation of Solution

Given information:

The function is $f\left(x\right)={\cos }^{-1}x$ .

Graph:

The graph of the function $f\left(x\right)={\cos }^{-1}x$ is shown below:

  

Figure(1)

Interpretation:

In the figure (1), the positive value yields a first quadrant angle and the negative value yields a second quadrant angle. Since the cosine function is positive in the first quadrant and negative in the second quadrant. So, the domain and range of the function $f\left(x\right)={\cos }^{-1}x$ is $\left[-1,1\right]$ and $\left[0,\pi \right]$ .

Therefore, the domain and range of the function $f\left(x\right)={\cos }^{-1}x$ is $\left[-1,1\right]$ and $\left[0,\pi \right]$ , respectively.

(b)

To determine

To find:The exact value of ${\cos }^{-1}\left(\cos \left(\frac{7\pi }{6}\right)\right)$ .

Answer to Problem 11CRT

The exact value of ${\cos }^{-1}\left(\cos \left(\frac{7\pi }{6}\right)\right)$ is $\frac{5\pi }{6}$ .

Explanation of Solution

Given information:

The expression is ${\cos }^{-1}\left(\cos \left(\frac{7\pi }{6}\right)\right)$ .

Calculation:

Firstly find the value of $\cos \frac{7\pi }{6}$ , As per identity $\cos \left(\pi +\theta \right)=-\cos \theta$ .

Substitute $\frac{\pi }{6}$ for $\theta$ in the identity.

  $\begin{array}{c}\cos \left(\pi +\theta \right)=-\cos \theta \\ \cos \left(\pi +\frac{\pi }{6}\right)=-\cos \frac{\pi }{6}\\ \cos \left(\frac{7\pi }{6}\right)=-\cos \frac{\pi }{6}\end{array}$

Substitute $\frac{\sqrt{3}}{2}$ for $\cos \frac{\pi }{6}$ .

  $\begin{array}{c}\cos \left(\frac{7\pi }{6}\right)=-\cos \frac{\pi }{6}\\ \cos \left(\frac{7\pi }{6}\right)=-\frac{\sqrt{3}}{2}\end{array}$

Substitute this value in the given expression.

  $\begin{array}{c}{\cos }^{-1}\left(\cos \left(\frac{7\pi }{6}\right)\right)={\cos }^{-1}\left(-\frac{\sqrt{3}}{2}\right)\\ =\pi -{\cos }^{-1}\left(\frac{\sqrt{3}}{2}\right)\\ =\pi -\frac{\pi }{6}\\ =\frac{5\pi }{6}\end{array}$

Therefore, the exact value of ${\cos }^{-1}\left(\cos \left(\frac{7\pi }{6}\right)\right)$ is $\frac{5\pi }{6}$ .

(c)

To determine

To represent:The expression $\tan \left({\cos }^{-1}x\right)$ as an algebraic function of $x$ .

Answer to Problem 11CRT

The expression $\tan \left({\cos }^{-1}x\right)$ can be written as $\sqrt{1-x^2}$ as an algebraic function of $x$ .

Explanation of Solution

Given information:

The expression is $\tan \left({\cos }^{-1}x\right)$ .

Calculation:

Take ${\cos }^{-1}x=y$ , then $\cos y=x$ .

The value of adjacent is $x$ and hypotenuse is $1$ . Use the Pythagoras theorem.

  $\begin{array}{c}\text{opp}=\sqrt{{\left(\text{hyp}\right)}^2-{\left(\text{adj}\right)}^2}\\ =\sqrt{1-x^2}\end{array}$

Use the formula for tangent.

  $\begin{array}{c}\tan y=\frac{\text{opp}}{\text{adj}}\\ =\frac{\sqrt{1-x^2}}{1}\\ =\sqrt{1-x^2}\end{array}$

As $\tan y=\sqrt{1-x^2}$ , then $y={\tan }^{-1}\left(\sqrt{1-x^2}\right)$ .

Substitute $y={\tan }^{-1}\left(\sqrt{1-x^2}\right)$ in the given equation.

  $\begin{array}{c}\tan \left({\cos }^{-1}x\right)=\tan \left({\tan }^{-1}\left(\sqrt{1-x^2}\right)\right)\\ =\sqrt{1-x^2}\end{array}$

Therefore, the expression $\tan \left({\cos }^{-1}x\right)$ can be written as $\sqrt{1-x^2}$ as an algebraic function of $x$ .