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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Trigonometric Identities
Trigonometry in mathematics deals with the right-angled triangle’s angles and sides. By trigonometric identities, we mean the identities we use whenever we need to express the various trigonometric functions in terms of an equation.
Inverse Trigonometric Functions
Inverse trigonometric functions are the inverse of normal trigonometric functions. Alternatively denoted as cyclometric or arcus functions, these inverse trigonometric functions exist to counter the basic trigonometric functions, such as sine (sin), cosi…
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Chapter 5.4, Problem 18E
To determine

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

Expert Solution & Answer
Check Mark

Answer to Problem 18E

. sin(−19π12)=(6+24) , cos(−19π12)=(6−24) , tan(−19π12)=2+3

Explanation of Solution

Given information:

Given angle,

  −19π12=2π3−9π4

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(−19π12)=sin(2π3−9π4)=sin2π3cos9π4−cos2π3sin9π4=(32)(12)−(−12)(12)=322+(122)sin(−19π12)=(322+122)=(3+122)

Multiplying numerator and denominator by 2

  sin(−19π12)=(3+1)2(22)2sin(−19π12)=(6+24)

Hence,

  sin(−19π12)=(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(−19π12)=cos(2π3−9π4)=cos2π3cos9π4+sin2π3sin9π4=(−12)(12)+(32)(12)=−122+(322)cos(−19π12)=(322−122)=(3−122)

Multiplying numerator and denominator by 2

  cos(−19π12)=(3−1)2(22)2cos(−19π12)=(6−24)

Hence,

  cos(−19π12)=(6−24) .

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

Taking tangent on both sides as below

  tan(−19π12)=sin(−19π12)cos(−19π12)=(6+24)(6−24)=6+26−2=(6+2)26−2=8+434tan(−19π12)=2+3

Hence,. tan(−19π12)=2+3 .

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Chapter 5.4, Problem 17E
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Chapter 5.4, Problem 19E

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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