91Ó°ÊÓ

The equations $\mathrm{θ}=\frac{\mathrm{π}}{2}$and $\mathrm{θ}=-\frac{\mathrm{π}}{2}$

Consider the equations $\mathrm{θ}=\frac{\mathrm{π}}{2}$and $\mathrm{θ}=-\frac{\mathrm{π}}{2}$.

The objective is to show that $\mathrm{θ}=\frac{\mathrm{π}}{2}$and $\mathrm{θ}=-\frac{\mathrm{π}}{2}$are same in polar coordinate system.

For the value of $\mathrm{θ}$, the polar equation $\mathrm{θ}=\frac{\mathrm{π}}{2}$describes the set of points with polar angle $\frac{\mathrm{π}}{2}$.

As the angle is positive it moves in the counterclockwise direction.

When $\mathrm{θ}=\frac{\mathrm{π}}{2}$the value of $r$can be either positive or negative.

Thus, the graph for $\mathrm{θ}=\frac{\mathrm{π}}{2}$is line through the pole.

When $\mathrm{θ}=-\frac{\mathrm{π}}{2}$the value of $r$can be either positive or negative. and $\mathrm{θ}=-\frac{\mathrm{π}}{2}$lies on the same line that of $\mathrm{θ}=\frac{\mathrm{π}}{2}$which passes through the pole.

As the angle is negative it moves in the clockwise direction.

The graphical representation of $\mathrm{θ}=\frac{\mathrm{π}}{2}$and $\mathrm{θ}=-\frac{\mathrm{π}}{2}$

That means the equations $\mathrm{θ}=\frac{\mathrm{π}}{2}$and $\mathrm{θ}=-\frac{\mathrm{π}}{2}$are identical in polar coordinate system. Hence the explanation.