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Q 20.

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Found in: Page 457

### Precalculus Enhanced with Graphing Utilities

Book edition 6th
Author(s) Sullivan
Pages 1200 pages
ISBN 9780321795465

# In Problems 9–36, find the exact value of each expression.$\mathrm{csc}\left[{\mathrm{cos}}^{-1}\left(-\frac{\sqrt{3}}{2}\right)\right]$

The value of the expression $\mathrm{csc}\left[{\mathrm{cos}}^{-1}\left(-\frac{\sqrt{3}}{2}\right)\right]=2$.

See the step by step solution

## Step 1. Given information.

The given expression is:

$\mathrm{csc}\left[{\mathrm{cos}}^{-1}\left(-\frac{\sqrt{3}}{2}\right)\right]$

Let's take $\theta ={\mathrm{cos}}^{-1}\left(-\frac{\sqrt{3}}{2}\right)$

$\mathrm{cos}\theta =-\frac{\sqrt{3}}{2}$

$0\le \mathrm{\theta }\le \mathrm{\pi }$ is the domain of the function.

## Step 2. Find the value of θ.

Initially, we have function ${\mathrm{cos}}^{-1}$and the bounds of cos function are going to be in the interval role="math" localid="1646466354693" $\left[0,\mathrm{\pi }\right]$. It represents the range of ${\mathrm{cos}}^{-1}$.

By using the unit circle, we can say that $\theta$ must be equal to $\frac{5\mathrm{\pi }}{6}$, so that we can get $-\frac{\sqrt{3}}{2}$.

${\mathrm{cos}}^{-1}\theta =-\frac{\sqrt{3}}{2}\phantom{\rule{0ex}{0ex}}{\mathrm{cos}}^{-1}\left(-\frac{\sqrt{3}}{2}\right)=\frac{5\mathrm{\pi }}{6}$

## Step 3. Find the exact value of the expression.

$\mathrm{csc}\left[{\mathrm{cos}}^{-1}\left(-\frac{\sqrt{3}}{2}\right)\right]=\mathrm{csc}\left(\frac{5\mathrm{\pi }}{6}\right)\phantom{\rule{0ex}{0ex}}\mathrm{csc}\left(\mathrm{\theta }\right)=\frac{1}{\mathrm{sin}\mathrm{\theta }}\phantom{\rule{0ex}{0ex}}\mathrm{csc}\left(\frac{5\mathrm{\pi }}{6}\right)=\frac{1}{\mathrm{sin}\left(\frac{5\mathrm{\pi }}{6}\right)}$

By using the unit circle, we know that $\mathrm{sin}\left(\frac{5\mathrm{\pi }}{6}\right)=\frac{1}{2}$

role="math" localid="1646466896004"

Therefore, $\mathrm{csc}\left[{\mathrm{cos}}^{-1}\left(-\frac{\sqrt{3}}{2}\right)\right]=2$.