How do you evaluate #sin (pi / 12) * cos (3 pi / 4) - cos (pi / 12) * sin (3 pi / 4)#?

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Answer 1 · 2018-06-27T02:57:45

#-sqrt3/2#

Recall

#sin(A-B)=sinAcosB-cosAsinB#

#sin(pi/12)cos((3pi)/4)-cos(pi/12)sin((3pi)/4)=sin(pi/12-(3pi)/4)#

#sin(pi/12-(9pi)/12)#

Recall

#sin(-x)=-sin(x)#

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

#-sqrt3/2#