`sin(-(8pi)/3)` Evalute the trigonometric function using its period as an aid

Textbook Question

Chapter 4, 4.2 - Problem 36 - Precalculus (3rd Edition, Ron Larson).
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kspcr111 | In Training Educator

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Given

`sin(-8pi/3)`

= `-sin(8pi/3)`

= `-sin(2pi +2pi/3)`

= `-sin(2pi/3)`

= `-sin(pi - pi/3)`

= `-sin(pi/3)`

= `-sqrt(3)/2`

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