Precalculus with Limits: A Graphing Approach
Precalculus with Limits: A Graphing Approach
7th Edition
ISBN: 9781305071711
Author: Ron Larson
Publisher: Brooks/Cole
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Chapter 5.4, Problem 17E
To determine

To find: the exact values of the sine, cosine and tangent of the angle.

Expert Solution & Answer
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Answer to Problem 17E

. sin(−11π12)=−(6−24) , cos(−11π12)=−(6+24) , tan(−11π12)=2−3

Explanation of Solution

Given information:

Given angle,

  −11π12=π4−7π6

Formula used:

  sin(u−v)=sinucosv−cosusinvcos(u−v)=cosucosv+sinusinvtan(u−v)=tanu−tanv1+tanutanv

Calculation:

Find the value of sine, cosine and tangent of the angle −11π12 by using the difference formulas as follows

  sin(−11π12)=sin(π4−7π6)=sinπ4cos7π6−cosπ4sin7π6=(12)(−32)−(12)(−12)=−322+(122)sin(−11π12)=−(322−122)=−(3−122)

Multiplying numerator and denominator by 2

  sin(−11π12)=−(3−1)2(22)2sin(−11π12)=−(6−24)

Hence,

  sin(−11π12)=−(6−24) .

Find exact values of the cosine of the above angle as below

Taking cosine on both sides as below

Using the sum formula cos(u−v)=cosucosv+sinusinv , we get

  cos(−11π12)=cos(π4−7π6)=cosπ4cos7π6+sinπ4sin7π6=(12)(−32)+(12)(−12)=−322−(122)cos(−11π12)=−(322+122)=−(3+122)

Multiplying numerator and denominator by 2

  cos(−11π12)=−(3+1)2(22)2cos(−11π12)=−(6+24)

Hence,

  cos(−11π12)=−(6+24) .

Find the exact values of the tangent of the above angle as below

Taking tangent on both sides as below

  tan(−11π12)=sin(−11π12)cos(−11π12)=−(6−24)−(6+24)=6−26+2=(6−2)26−2=8−434tan(−11π12)=2−3

Hence,. tan(−11π12)=2−3 .

Previous
Chapter 5.4, Problem 16E
Next
Chapter 5.4, Problem 18E

Chapter 5 Solutions

Precalculus with Limits: A Graphing Approach

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