Question 689854
{{{cot(theta)=-cot(-theta)}}} because
{{{cot(theta)=cos(theta)/sin(theta)}}},
{{{cot(-theta)=cos(-theta)/sin(-theta)}}} and,
as {{{sin(theta)=-sin(-theta)}}} and {{{cos(theta)=cos(-theta)}}}, then
{{{cot(theta)=cos(theta)/sin(theta)=cos(-theta)/(-sin(-theta))=-cos(-theta)/sin(-theta)=-cot(-theta)}}}
{{{drawing(300,300,-10,10,-10,10,
circle(0,0,7),line(-6.36,-6.36,6.36,6.36),
line(-7.79,-4.5,7.79,4.5),line(-9,0,0,0),
arrow(0,0,9.5,0),locate(9.6,0.5,x),
line(-4.5,7.79,4.5,-7.79),line(0,-9,0,0),
arrow(0,0,0,9.5),locate(-0.6,11,y),
line(-4.5,-7.79,4.5,7.79),line(-6.36,6.36,6.36,-6.36),
line(-7.79,4.5,7.79,-4.5),
locate(7.1,0,1),locate(7.1,1,A),
locate(0.1,7,1),locate(0,7.9,E),locate(0.1,-7.1,"E'"),
locate(6.3,3.7,B),locate(6.3,-2.7,"B'"),
locate(5.2,5.4,C),locate(5.2,-4.4,"C'"),
locate(3.2,7,D),locate(3.2,-6.2,"D'"),
locate(1.5,0,O),locate(-8.5,0,-1),locate(-7.5,1,I),
locate(-3.6,7,F),locate(-3.6,-6.2,"F'"),
blue(line(6.06,-3.5,6.06,3.5)),blue(rectangle(6.06,0,5.66,0.4)),
locate(5.3,-0.1,P)
)}}} Angles measuring {{{30^o}}} (AOB), {{{45^o}}} (AOC), {{{60^o}}} (AOD), ... are represented.
 
For the {{{30^o}}} or {{{pi/6}}} angle AOB:
{{{sin(AOB)=1/2=PB}}}, and {{{cos(AOB)=sqrt(3)/2=OP}}} are the x and y coordinates of point B
{{{cot(AOB)=sqrt(3)}}}
The reflection of angle AOB is the {{{-30^o}}} or {{{-pi/6}}} angle AOB':
{{{sin(-30^o)=-1/2}}}, {{{cos(-30^o)=sqrt(3)/2}}}, and {{{cot(-30^o)=-sqrt(3)}}}
 
For the {{{45^o}}} angle AOC and its reflection, the {{{-45^o}}} angle AOC':
{{{sin(45^o)=sqrt(2)/2}}}, {{{cos(45^o)=sqrt(2)/2}}},
{{{sin(-45^o)=-sqrt(2)/2}}}, and {{{cos(-45^o)=sqrt(2)/2}}}, so
{{{cot(45^o)=1}}} {{{cot(-45^o)=-1}}}
 
For the {{{60^o}}} angle AOD and its reflection, the {{{-60^o}}} angle AOD':
{{{sin(60^o)=sqrt(3)/2}}}, {{{cos(60^o)=1/2}}},
{{{sin(-60^o)=sqrt(3)/2}}}, and {{{cos(-60^o)=1/2}}}, so
{{{cot(60^o)=1/sqrt(3)=sqrt(3)/3}}} and {{{cot(-60^o)=-1/sqrt(3)=-sqrt(3)/3}}}
 
For the {{{90^o}}} angle AOE and its reflection, the {{{-90^o}}} angle AOE':
{{{sin(90^o)=1}}}, {{{cos(90^o)=0}}},
{{{sin(-90^o)=1}}}, and {{{cos(-90^o)=0}}}, so
{{{cot(90^o)=0}}} {{{cot(-90^o)=0}}}