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Answers without the blur. Sign up and see all textbooks for free! Q63.

Expert-verified Found in: Page 418 ### Precalculus Mathematics for Calculus

Book edition 7th Edition
Author(s) James Stewart, Lothar Redlin, Saleem Watson
Pages 948 pages
ISBN 9781337067508 # Find the values of the trigonometric functions of t from the given information. $\mathrm{sin}t=-\frac{4}{5}$, terminal point of t is in Quadrant IV

The values of the trigonometric functions are $\mathbf{cos}\mathbit{t}\mathbf{=}\frac{\mathbf{3}}{\mathbf{5}}\mathbf{,}\mathbf{tan}\mathbit{t}\mathbf{=}\mathbf{-}\frac{\mathbf{4}}{\mathbf{3}}\mathbf{,}\mathbf{csc}\mathbit{t}\mathbf{=}\mathbf{-}\frac{\mathbf{5}}{\mathbf{4}}\mathbf{,}$$\mathbf{sec}\mathbit{t}\mathbf{=}\frac{\mathbf{5}}{\mathbf{3}}$, and $\mathbf{cot}\mathbit{t}\mathbf{=}\mathbf{-}\frac{\mathbf{3}}{\mathbf{4}}$.

See the step by step solution

## Step 1. State the Trigonometric identities.

The reciprocal identities are $\mathrm{csc}t=\frac{1}{\mathrm{sin}t}$,$\mathrm{sec}t=\frac{1}{\mathrm{cos}t}$,$\mathrm{cot}t=\frac{1}{\mathrm{tan}t}$,$\mathrm{tan}t=\frac{\mathrm{sin}t}{\mathrm{cos}t}$ and $\mathrm{cot}=\frac{\mathrm{cos}t}{\mathrm{sin}t}$.

The Pythagorean identities are:

1. ${\mathrm{sin}}^{2}t+{\mathrm{cos}}^{2}t=1$

2. ${\mathrm{tan}}^{2}t+1={\mathrm{sec}}^{2}t$

3. ${\mathrm{cot}}^{2}t+1={\mathrm{csc}}^{2}t$

## Step 2. Formula used.

Now use Pythagorean Identity ${\mathrm{sin}}^{2}t+{\mathrm{cos}}^{2}t=1$, substitute $-\frac{4}{5}$ for $\mathrm{sin}t$ and simplify for $\mathrm{cos}t$.

${\mathrm{sin}}^{2}t+{\mathrm{cos}}^{2}t=1\phantom{\rule{0ex}{0ex}}{\left(-\frac{4}{5}\right)}^{2}+{\mathrm{cos}}^{2}t=1\phantom{\rule{0ex}{0ex}}\frac{16}{25}+{\mathrm{cos}}^{2}t-\frac{16}{25}=1-\frac{16}{25}\phantom{\rule{0ex}{0ex}}{\mathrm{cos}}^{2}t=\frac{9}{25}\phantom{\rule{0ex}{0ex}}\mathrm{cos}t=±\frac{3}{5}$

Now, t is given to be in quadrant IV, and quadrant IV, $\mathrm{cos}t>0$, therefore, $\mathrm{cos}t=\frac{3}{5}$.

Now substitute $-\frac{4}{5}$ for $\mathrm{sin}t$ for $\frac{3}{5}$and $\mathrm{cos}t$ into reciprocal identity $\mathrm{tan}t=\frac{\mathrm{sin}t}{\mathrm{cos}t}$ and simplify for $\mathrm{tan}t$.

$\mathrm{tan}t=\frac{\mathrm{sin}t}{\mathrm{cos}t}\phantom{\rule{0ex}{0ex}}=\frac{\left(-\frac{4}{5}\right)}{\frac{3}{5}}\phantom{\rule{0ex}{0ex}}=-\frac{4}{3}$

## Step 3. Find other trigonometric functions.

Use the values of $\mathrm{sin}t,\mathrm{cos}t$ and $\mathrm{tan}t$ to find $\mathrm{csc}t,\mathrm{sec}t$ and $\mathrm{cot}t$.

$\mathrm{csc}t=\frac{1}{\mathrm{sin}t}$

$=\frac{1}{\left(-\frac{4}{5}\right)}\phantom{\rule{0ex}{0ex}}=-\frac{5}{4}$

$\mathrm{sec}t=\frac{1}{\mathrm{cos}t}$

$=\frac{1}{\left(\frac{3}{5}\right)}\phantom{\rule{0ex}{0ex}}=\frac{5}{3}$

$\mathrm{cot}t=\frac{1}{\mathrm{tan}t}$

$=\frac{1}{\left(-\frac{4}{3}\right)}\phantom{\rule{0ex}{0ex}}=-\frac{3}{4}$ ### Want to see more solutions like these? 