Precalculus: Mathematics for Calculus - 6th Edition

6th Edition

ISBN: 9780840068071

Author: Stewart, James, Redlin, Lothar, Watson, Saleem

Publisher: Cengage Learning

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Concept explainers

Fundamentals of Trigonometric Identities

Trigonometric Identities are the equalities that involve trigonometry functions and hold for all the values of variables given in the equation.

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Textbook Question

Chapter 5, Problem 5T

If $cos t = − 8 17$ and if the terminal point determined by t is in Quadrant III, find tan t cot t + csc t.

Expert Solution & Answer

To determine

To find: The value for $tantcott+csct$ .

Answer to Problem 5T

The value of $tantcott+csct$ is $−215$ .

Explanation of Solution

Given:

The given trigonometric function is,

$cost=−817$

Calculation:

Use Pythagoras identities of fundamental identities.

$sin2t+cos2t=1$

Substitute $−817$ in the above equation to find the $sint$ .

$sin2t+(−817)2=1sin2t=1−(−817)2sin2t=1−64289sin2t=289−64289$

Simplify the above equation,

$sin2t=225289sint=225289sint=±1517$

Since the $sint$ is negative in quadrant III.

$sint=−1517$

Use reciprocal identities of fundamental identities.

$tant=sintcost$ . (1)

And,

$csct=1sint$ . (2)

And,

$cott=1tant$ . (3)

Substitute $−817$ for $cost$ and $−1517$ $sint$ in equation (1).

$tant=−15/17−8/17tant=1517×178tant=158$

Substitute $158$ for $tant$ in equation (3).

$cott=115/8cott=815$

Substitute $−1517$ for $sint$ in equation (2).

$csct=−115/17csct=−1715$

Substitute $158$ for $tant$ , $815$ for $cott$ and $−1715$ for $csct$ to find the value of $tantcott+csct$

$tantcott+csct=158×815+(−1715)=1−1715=15−1715=−215$

Hence, the value of $tantcott+csct$ is $−215$ .

Chapter 5, Problem 4T

Chapter 5, Problem 6T

Chapter 5 Solutions

Precalculus: Mathematics for Calculus - 6th Edition

Ch. 5.1 - Prob. 1E Ch. 5.1 - Prob. 2E Ch. 5.1 - Prob. 3E Ch. 5.1 - Prob. 4E Ch. 5.1 - Prob. 5E Ch. 5.1 - Prob. 6E Ch. 5.1 - Prob. 7E Ch. 5.1 - Prob. 8E Ch. 5.1 - Prob. 9E Ch. 5.1 - Prob. 10E

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If cos t = − 8 17 and if the terminal point determined by t is in Quadrant III, find tan t cot t + csc t .

Solution Summary: The author explains how to find the value of mathrmsin2t, if the given trigonometric function is negative in quadrant III.

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Answer to Problem 5T

Explanation of Solution

Chapter 5 Solutions