How do you find the exact value of #sin ((7pi)/12)#?

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Answer 1 · 2016-04-09T07:49:22

#sin(7pi/12)=sin(pi/3+pi/4)#
#=sin(pi/3)cos( pi/4)+cos(pi/3)sin(pi/4)#
#=sqrt3/2xx1/sqrt2+1/2xx1/sqrt2=(sqrt3+1)/(2sqrt2)#