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Question #d532a

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Apr 6, 2017

#sin((8pi)/3)=sqrt(3)/3#
#cos((8pi)/3)=-1/2#

Explanation:

#(8pi)/3=2pi and (2pi)/3#

#2pi# is full period and you can remove it.

SO, You have:
#sin((2pi)/3)#

You can use this:
#sin(pi-alpha)=sinalpha#
#cos(pi-alpha)=-cosalpha#

#sin((2pi)/3)=sin(pi-pi/3)=sin(pi/3)=sqrt(3)/3#
#cos((2pi)/3)=cos(pi-pi/3)=-cos(pi/3)=-1/2#

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