Why does #tan(x+(pi/2))=cot(x)#? or #cos((pi/2)-x)=sin(x)# and is it like that with #(3pi)/2# as well?

Please explain what's so special with #pi/2# and #(3pi)/2#

1 Answer
Jun 3, 2018

a. #tan (x + pi/2) = sin (x + pi/2)/(cos (x + pi/2))#
Reminder of trig identities:
#sin (x + pi/2) = cos x#
#cos (x + pi/2) = - sin x#
Therefor,
#tan (x + pi/2) = (cos x)/(-sin x) = - cot x#
To fully understand the property of complementary arcs, we prove theses below trig identities:
#cos (pi/2 + x) = cos (pi/2).cos x - sin (pi/2).sin x = - sin x#
#sin (pi/2 + x) = sin x.cos (pi/2) + cos x.sin (pi/2) = cos x#
#cos (pi/2 - x) = cos (pi/2.cos x + sin (pi/2).sin x = sin x#
#sin (pi/2 - x) = sin (pi/2).cos x + sin x.cos (pi/2) = cos x#