cos(pi)=-1cos(π)=−1
color(white)("XXX")XXXwith an angle of piπ
color(white)("XXX")XXXthe adjacent arm has the same magnitude as the hypotenuse
color(white)("XXX")XXXbut is negative.
color(white)("XXX")XXXTherefore cos(pi)= (-"hypotenuse")/"hypotenuse" =-1cos(π)=−hypotenusehypotenuse=−1
"arctan"(-1)=pi/4arctan(−1)=π4
color(white)("XXX")XXXNote that as a function the range of "arctan"arctan
color(white)("XXX")XXXis limited to [-pi/2,+pi/2][−π2,+π2]
color(white)("XXX")"arctan(-1)"XXXarctan(-1) means that
color(white)("XXX")XXXthe "opposite"/"adjacent"oppositeadjacent for the angle must be (-1)(−1)
color(white)("XXX")XXXor, in terms of x and yxandy coordinates:
color(white)("XXX")y/x = -1 rarr y=-xXXXyx=−1→y=−x
color(white)("XXX")XXXand we have an equilateral right-angled triangle
color(white)("XXX")XXXbelow the X-axis
