How do you evaluate #-sin ((4pi)/3)#?

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Answer 1 · 2016-05-24T06:53:32

#"solution :"sqrt3/2#

#-sin((4pi)/3)= ?#

#((4pi)/3)=pi+pi/3#

#sin(a+b)=sin a*cos b+cos a*sin b#

#sin(pi+pi/3)=sin pi*cos pi/3+cos pi*sin pi/3#

#sin pi=0#

#cos pi=-1#

#sin( pi/3)=sqrt3/2#

#cos(pi/3)=1/2#

#-sin((4pi)/3)=0*1/2-1*sqrt3/2#

#-sin((4pi)/3)=-sqrt3/2#

#"solution :"sqrt3/2#