PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS)
PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS)
8th Edition
ISBN: 9781337904285
Author: Larson
Publisher: CENGAGE L
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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 26E
To determine

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

Expert Solution & Answer
Check Mark

Answer to Problem 26E

. sin(−23π12)=(6−24) , cos(−23π12)=(6+24) , tan(−23π12)=2−3

Explanation of Solution

Given information:

Given angle,

  (−23π12)

Formula used:

  sin(−θ)=−sinθcos(−θ)=cosθtan(−θ)=−tanθ

And

  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 (−23π12) by using the difference formulas as follows

  sin(−23π12)=−sin(23π12)=−sin(2π3+5π4)=−[sin2π3cos5π4+cos2π3sin5π4]=−[(32)(−12)+(−12)(−12)]=−[−322+(122)]sin(−23π12)=(322−122)=(3−122)

Multiplying numerator and denominator by 2

  sin(−23π12)=(3−1)2(22)2sin(−23π12)=(6−24)

Hence,

  sin(−23π12)=(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(−23π12)=cos(23π12)=cos(2π3+5π4)=cos2π3cos5π4−sin2π3sin5π4=(−12)(−12)−(32)(−12)=122+(322)cos(−23π12)=(322+122)=(3+122)

Multiplying numerator and denominator by 2

  cos(−23π12)=(3+1)2(22)2cos(−23π12)=(6+24)

Hence,

  cos(−23π12)=(6+24) .

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

Taking tangent on both sides as below

  tan(−23π12)=sin(−23π12)cos(−23π12)=(6−24)(6+24)=6−26+2=(6−2)26−2=8−434tan(−23π12)=2−3

Hence,. tan(−23π12)=2−3 .

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Chapter 5.4, Problem 25E
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Chapter 5.4, Problem 27E

Chapter 5 Solutions

PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS)

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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