#cos(pi)=-1#
#color(white)("XXX")#with an angle of #pi#
#color(white)("XXX")#the adjacent arm has the same magnitude as the hypotenuse
#color(white)("XXX")#but is negative.
#color(white)("XXX")#Therefore #cos(pi)= (-"hypotenuse")/"hypotenuse" =-1#
#"arctan"(-1)=pi/4#
#color(white)("XXX")#Note that as a function the range of #"arctan"#
#color(white)("XXX")#is limited to #[-pi/2,+pi/2]#
#color(white)("XXX")"arctan(-1)"# means that
#color(white)("XXX")#the #"opposite"/"adjacent"# for the angle must be #(-1)#
#color(white)("XXX")#or, in terms of #x and y# coordinates:
#color(white)("XXX")y/x = -1 rarr y=-x#
#color(white)("XXX")#and we have an equilateral right-angled triangle
#color(white)("XXX")#below the X-axis
