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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Question
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Chapter 5.4, Problem 15E
To determine

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

Expert Solution & Answer
Check Mark

Answer to Problem 15E

  sin195°=−6+24,cos195°=−6−24,tan195°=2−3 .

Explanation of Solution

Given information:

Given angle,

  195°=225°−30°

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 195° by using the sum formulas as follows

  sin195°=sin(225°−30°)=sin225°cos30°−cos225°sin30°=sin(180°+45°)(32)−cos(180°+45°)(12)=−sin45°(32)−(−cos45°)(12)=−12(32)+12(12)=−3+122=−6+24

Now for cosine we get

  cos195°=cos(225°−30°)=cos225°cos30°+sin225°sin30°=cos(180°+45°)(32)+sin(180°+45°)(12)=−cos45°(32)+(−sin45°)(12)=−12(32)−12(12)=−3−122=−6−24

Also for tangent we get

  tan195°=tan(225°−30°)=tan225°−tan30°1+tan225°tan30°=tan(180°+45°)−131+tan(180°+45°)13=tan45°−131+tan45°13tan195°=1−131+13=3−13+1=3+1−233−1=2−3

Hence, sin195°=−6+24,cos195°=−6−24,tan195°=2−3 .

Previous
Chapter 5.4, Problem 14E
Next
Chapter 5.4, Problem 16E

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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Chain Rule dy:dx = dy:du*du:dx; Author: Robert Cappetta;https://www.youtube.com/watch?v=IUYniALwbHs;License: Standard YouTube License, CC-BY
CHAIN RULE Part 1; Author: Btech Maths Hub;https://www.youtube.com/watch?v=TIAw6AJ_5Po;License: Standard YouTube License, CC-BY