Question

18. ( (2) cos left(40^{circ}+thetaright)-sin left(50^{circ}-thetaright) )
( =cos left(40^{circ}+thetaright)-sin left[90^{circ}-left(40^{circ}+thetaright)right] )
( =cos left(40^{circ}+thetaright)-cos (40+theta)=0 )

# 18. cos(40° + e) - sin(50° - 0) is equal to (1) 1 (2) 0 (3) sin20 (4) cos 20

Solution