Chapter 4, Problem 89RE

To determine

To Find:Exact value oftrigonometric expression $\sec (\arcsin \frac{-1}{4})$ .

Answer to Problem 89RE

The exact value of trigonometric expression $\sec (\arcsin \frac{-1}{4})$ is $1.788$ .

Explanation of Solution

Given information:

To draw right angle triangle.

Formula used:

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

Where p is perpendicular is base,h is height.

And trigonometric formula

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

Calculation:

Consider the statement “given trigonometric expression”

Recall that by the Pythagoras formula on right angle triangle

  

Therefore, given that perpendicular is $-1$

Base is $b$

Height is $4$

So by Pythagoras formula,

  $\begin{array}{l}\sqrt{p^2+b^2}=h\\ \sqrt{{(-1)}^2+b^2}=4\\ \sqrt{1+b^2}=4\\ 1+b^2=16\\ b^2=15\\ b=\sqrt{15}\end{array}$

Where b is height

Now, recall the trigonometric formula of ${\cos }^{-1}x={\sec }^{-1}\frac{1}{x}$

  ${\cos }^{-1}\frac{\sqrt{15}}{4}={\sec }^{-1}\frac{4}{\sqrt{15}}$

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

Therefore, $\sec ({\sec }^{-1}\frac{4}{\sqrt{15}})=\frac{4}{\sqrt{15}}=1.788$

So, the exact value of trigonometric function $\sec ({\sec }^{-1}\frac{4}{\sqrt{15}})is1.788$ .