How do you evaluate #sin ((7pi)/12)#?

1 Answer
Jothi S.
Jul 27, 2016

#( (sqrt(2)+sqrt(6))/4)#

Explanation:

#sin(7pi/12)=sin(pi/4+pi/3)#
Use the formula #sin (a+b)=sina cosb+cosasinb#
# sin(pi/4+pi/3)=sin(pi/4)cos(pi/3)+cos(pi/4)sin(pi/3).....> 1#
#sin(pi/4)=sqrt(2)/2;cos (pi/4)=sqrt2/2#
#sin(pi/3)=sqrt(3)/2;cos(pi/3)=1/2#
Plug these values on equation 1

#sin(pi/4+pi/3)=(sqrt(2)/2)(1/2)+(sqrt(2)/2)*(sqrt(3)/2)#

#sin(pi/4+pi/3)=(sqrt(2)+sqrt(6))/4#

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