How do you express #cos(pi/ 2 ) * cos (( 17 pi) / 12 ) # without using products of trigonometric functions?
1 Answer
Feb 24, 2016
0
Explanation:
Consider
# cos(pi/2) = 0# hence
# cos(pi/2) . cos((17pi)/12) = 0 xx cos((17pi)/12) = 0#
