Chapter 4.7, Problem 66E

Solution

(a)

To calculate: explaining in algebraic expression.

  $\sqrt{1-4x^2}$

Given information: $\cos ({\sin }^{-1}2x)$

Calculation: the acute angle is ${\sin }^{-1}2x$ . Since $\sin =\frac{opp}{hyp}$ then $opp=2x$ and $hyp=1$ , if $s$ is the adjacent side, then:

  $\begin{array}{l}$ {(2x)}^2+s^2=1^2$4x^2+s^2=1\\ s^2=1-4x^2\\ s=\sqrt{1-4x^2}\end{array}$

Draw the right triangle as shown:

  

Find $\cos ({\sin }^{-1}2x)$ :

  $\begin{array}{l}\cos ({\sin }^{-1}2x)=\frac{adj}{hyp}\\ \cos ({\sin }^{-1}2x)=\frac{\sqrt{1-4x^2}}{1}=\sqrt{1-4x^2}\end{array}$

(b)

To calculate: explaining in algebraic expression.

  $1+x^2$

Given information: ${\sec }^2({\tan }^{-1}x)$

Calculation: the acute angle is ${\tan }^{-1}2x$ . Since $\tan =\frac{opp}{hyp}$ then $opp=x$ and $adj=1$ , if $s$ is the hypotenuse, then:

  $\begin{array}{l}{1}^2+x^2=s^2\\ 1+x^2=s^2\\ \sqrt{1-x^2}=s\end{array}$

Draw the right triangle as shown:

  

Find $\sec ({\tan }^{-1}x)$ :

  $\begin{array}{l}\sec ({\tan }^{-1}x)=\frac{hyp}{adj}\\ \sec ({\tan }^{-1}x)=\frac{\sqrt{1-x^2}}{1}=\sqrt{1-x^2}\end{array}$

This follows that:

  $\begin{array}{l}{\sec }^2({\tan }^{-1}x)=({\sqrt{1+x^2)}}^2\\ {\sec }^2({\tan }^{-1}x)=1+x^2\end{array}$

(c)

To calculate: explaining in algebraic expression.

  $\sqrt{1-x}$

Given information: $\sin ({\cos }^{-1}\sqrt{x})$

Calculation: the acute angle is ${\cos }^{-1}\sqrt{x}$ . Since $\cos =\frac{adj}{hyp}$ then $adj=\sqrt{x}$ and $hyp=1$ , if $s$ is the opposite side, then:

  $\begin{array}{l}{s}^2+{\left(\sqrt{x}\right)}^2=1^2\\ s^2+x=1\\ s^2=1-x\\ s=\sqrt{1-x}\end{array}$

Draw the right triangle as shown:

  

Find $\sin ({\cos }^{-1}\sqrt{x})$ :

  $\begin{array}{l}\sin ({\cos }^{-1}\sqrt{x})=\frac{opp}{hyp}\\ \sin ({\cos }^{-1}x)=\frac{\sqrt{1-x}}{1}=\sqrt{1-x}\end{array}$

For $0\le x\le 1$

(d)

To calculate: explaining in algebraic expression.

  $-x^2-1$

Given information: $-{\csc }^2({\cot }^{-1}x)$

Calculation: the acute angle is ${\cot }^{-1}2x$ . Since $\cot =\frac{adj}{opp}$ then $adj=x$ and $opp=1$ , if $s$ is the hypotenuse, then:

  $\begin{array}{l}{x}^2+1^2=s^2\\ x^2+1=s^2\\ \sqrt{x^2+1}=s\end{array}$

Draw the right triangle as shown:

  

Find $cse({\cot }^{-1}x)$ :

  $\begin{array}{l}\csc ({\cot }^{-1}x)=\frac{hyp}{opp}\\ cse({\cot }^{-1}x)=\frac{\sqrt{x^2+1}}{1}=\sqrt{x^2+1}\end{array}$

This follows that:

  $\begin{array}{l}-{\csc }^2({\cot }^{-1}x)=({\sqrt{x^2+1)}}^2\\ -{\csc }^2({\cot }^{-1}x)=(x^2+1)\\ -{\csc }^2({\cot }^{-1}x)=-x^2-1\end{array}$

(e)

To calculate: explaining in algebraic expression.

  $\sqrt{x^4-1}$

Given information: $\tan ({\sec }^{-1}{x}^2)$

Calculation: the acute angle is ${\sec }^{-1}{x}^2$ . Since $\sec =\frac{hyp}{adj}$ then $hyp=x^2$ and $adj=1$ , if $s$ is the opposite side, then:

  $\begin{array}{l}$ s^2+1^2={(x^2)}^2$s^2+1=x^2\\ s^2=x^4-1\\ \sqrt{x^4-1}=s\end{array}$

Draw the right triangle as shown:

  

Find $\tan ({\sec }^{-1}{x}^2)$ :

  $\begin{array}{l}\tan ({\sec }^{-1}{x}^2)=\frac{opp}{adj}\\ \tan ({\sec }^{-1}{x}^2)=\frac{\sqrt{x^4-1}}{1}=\sqrt{x^4-1}\end{array}$