Sin (101pi/6)

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Does anyone know the quick way to solve sin (101pi/6)?

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http://tutorial.math.lamar.edu/Extras/AlgebraTrigReview/TrigFunctions.aspx
Think about reducing 101 by 6, that will yield you 16 and 5/6. Then solve for sin (5/6) to get your answer

Sine and cosin functions are cyclic. In other words they repeat themselves in every 2*pi angle. So
sin (101pi/6)= sin (5pi/6) ->(101*pi/6) - (16pi)

Sin (5pi/6) is in second quadrant so Sin (5pi/6)= Sin (pi/6) = 1/2.
Remember special triangles (30-60-90) (45-45-90) ,if you do not know the value of functions. Dr. Reo forgot to put pi to his answer.
Hopefully this helps.
 
Sine and cosin functions are cyclic. In other words they repeat themselves in every 2*pi angle. So
sin (101pi/6)= sin (5pi/6) ->(101*pi/6) - (16pi)

Sin (5pi/6) is in second quadrant so Sin (5pi/6)= Sin (pi/6) = 1/2.
Remember special triangles (30-60-90) (45-45-90) ,if you do not know the value of functions. Dr. Reo forgot to put pi to his answer.
Hopefully this helps.
Yes, sorry about that.
 
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