Chapter 4, Problem 91RE

To determine

To Find:An algebraic expression, which is equivalent to given expression $\tan (\arccos (\frac{x}{2}))$ .

Answer to Problem 91RE

The algebraic expression of $\tan (\arccos (\frac{x}{2}))$ is $\frac{\sqrt{16-x^2}}{x}$ .

Explanation of Solution

Given information:

To draw right angle triangle.

Formula used:

Pythagoras theorem that is $p^2+b^2=h^2$

Where p is perpendicular is base,h is height.

And trigonometric formula, $\tan ({\tan }^{-1}(\frac{1}{x}))=x$

Also, by right angle triangle, $\tan x=\frac{p}{b}$

  $x={\tan }^{-1}(\frac{p}{b})$

Where p is perpendicular and b is base.

Calculation:

Consider the statement “given trigonometric expression”

Recall that by the Pythagoras formula on right angle triangle

  

Therefore, given that perpendicular is p

Base is $x$

Height is $2$

So by Pythagoras formula,

  $\begin{array}{l}\sqrt{p^2+b^2}=h\\ \sqrt{p^2+x^2}=2\\ p^2+x^2=4\\ p^2=4-x^2\\ p=\sqrt{4-x^2}\end{array}$

Where p is perpendicular and h is height

Now, recall the trigonometric formula of $x^{*}={\tan }^{-1}(\frac{p}{b})$

Therefore, ${\cos }^{-1}\frac{x}{2}={\tan }^{-1}(\frac{\sqrt{4-x^2}}{x})$

Recall the trigonometric formula of $\tan ({\tan }^{-1}x)=x$

Therefore, $\tan ({\tan }^{-1}(\frac{\sqrt{4-x^2}}{x})=\frac{\sqrt{4-x^2}}{x}$

So, the algebraic value of trigonometric function $\tan (\arccos (\frac{x}{2})=\frac{\sqrt{4-x^2}}{x}$ .