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How do you evaluate #sin ((23pi) / 3) #?

1 Answer

May 22, 2016

#-sqrt3/2#

Explanation:

Trig table and trig unit circle -->
#sin ((23pi)/3) = sin (-pi/3 + (24pi)/3) = sin (-pi/3 + 8pi) =#
#= sin (-pi/3) = - sin (pi/3) = - sqrt3/2#

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