How do you express #sin(pi/ 6 ) * cos( (3 pi) / 2 ) # without using products of trigonometric functions?
1 Answer
Jul 26, 2016
Explanation:
#sin(pi/6) = 1/2#
#cos((3pi)/2) = 0#
So:
#sin(pi/6)*cos((3pi)/2) = 1/2*0 = 0#
