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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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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Question
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Chapter 5.4, Problem 11E

(a)

To determine

To find: the exact value of the expression sin(2π3+5π6) .

(a)

Expert Solution
Check Mark

Answer to Problem 11E

. sin(2π3+5π6)=−334+14 .

Explanation of Solution

Given information:

Given expression

  sin(2π3+5π6)

Calculation:

Find the exact value of sin(2π3+5π6) .

Apply sum formula:

  sin(u+v)=sinucosv+cosusinv

Hence,

  sin(2π3+5π6)=sin2π3cos5π6+cos2π3sin5π6

Now,

  sin2π3=sin(π−π3)=sinπ3=32

And,

  sin5π6=sin(π−π6)=sinπ6=12

Similarly

  cos2π3=cos(π−π3)=−cosπ3=−12

And,

  cos5π6=cos(π−π6)=−cosπ6=−32

Substitute:

  sin2π3=32,sin5π6=12,cos2π3=−12,cos5π6=−32

Then,

  sin2π3cos5π6+cos2π3sin5π6=32(−32)+(−12)(12)=−334−14

Therefore:

  sin(2π3+5π6)=−334+14

(b)

To determine

To find: the exact value of sin2π3+sin5π6 .

(b)

Expert Solution
Check Mark

Answer to Problem 11E

  sin2π3+sin5π6=3+12

Explanation of Solution

Given information:

Given expression

  sin2π3+sin5π6 .

Calculation:

Find the exact value of sin2π3+sin5π6 . So, substitute:

  sin2π3=32,sin5π6=12

Then,

  sin2π3+sin5π6=32+12=3+12

Therefore:

  sin2π3+sin5π6=3+12 .

Previous
Chapter 5.4, Problem 10E
Next
Chapter 5.4, Problem 12E

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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Calculus: Early Transcendental Functions
Calculus
ISBN:9781337552516
Author:Ron Larson, Bruce H. Edwards
Publisher:Cengage Learning
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