How do you find #\sin \frac { 11\pi } { 6}#?

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Answer 1 · 2018-06-24T06:08:15

The answer is #=-0.5#

We need

#sin(a+b)=sinacosb+sinbcosa#

#sin(pi)=0#

#cos(pi)=-1#

#sin(5/6pi)=sin(1/6pi)=1/2#

Therefore, one way is

#sin(11/6pi)=sin(pi+5/6pi)#

#=sin(pi)cos(5/6pi)+cos(pi)sin(5/6pi)#

#=0*(-sqrt3/2)-1*1/2#

#=-1/2#

#=-0.5#

Answer 2 · 2018-06-24T06:13:30

#color(indigo)(sin ((11pi) / 6) = sin (-pi/6) = -1/2 = -0.5#

#sin ((11pi) / 6)#

#=> sin (-2pi + (11pi)/6)#

#color(indigo)(=> sin (-pi/6) = sin (-30^@) = -1/2 = -0.5#