How do you evaluate sin (pi / 12) * cos (3 pi / 4) - cos (pi / 12) * sin (3 pi / 4)sin(π12)cos(3π4)cos(π12)sin(3π4)?

1 Answer
Jun 27, 2018

-sqrt3/232

Explanation:

Recall

sin(A-B)=sinAcosB-cosAsinBsin(AB)=sinAcosBcosAsinB

sin(pi/12)cos((3pi)/4)-cos(pi/12)sin((3pi)/4)=sin(pi/12-(3pi)/4)sin(π12)cos(3π4)cos(π12)sin(3π4)=sin(π123π4)

sin(pi/12-(9pi)/12)sin(π129π12)

Recall

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

sin(-(2pi)/3)=-sin((2pi)/3)sin(2π3)=sin(2π3)

-sqrt3/232