Question

How do you express `sin(pi/ 6 ) * cos( (3 pi) / 2 )` without using products of trigonometric functions?

Answer 1

`0`

Explanation:

`sin(pi/6) = 1/2`

`cos((3pi)/2) = 0`

So:

`sin(pi/6)*cos((3pi)/2) = 1/2*0 = 0`