91Ó°ÊÓ

The polar coordinates are$\left(3,\frac{\mathrm{Ï€}}{6}\right),\left(-3,\frac{\mathrm{Ï€}}{6}\right),\left(3,-\frac{\mathrm{Ï€}}{6}\right)$and$\left(-3,-\frac{\mathrm{Ï€}}{6}\right)$

Let us consider the polar coordinates$\left(3,\frac{\mathrm{Ï€}}{6}\right),\left(-3,\frac{\mathrm{Ï€}}{6}\right),\left(3,-\frac{\mathrm{Ï€}}{6}\right)$and $\left(-3,-\frac{\mathrm{Ï€}}{6}\right)$.

The objective is to convert the polar coordinates to the rectangular coordinates.

In the point $\left(3,\frac{\mathrm{π}}{6}\right)$, then $r=3,\mathrm{θ}=\frac{\mathrm{π}}{6}$.

Use the equations $x=r\cos \mathrm{θ},y=r\sin \mathrm{θ}$

Take $x=r\cos \mathrm{θ}$and substitute $r=3,\mathrm{θ}=\frac{\mathrm{π}}{6}$then

$x=3\cos \frac{\mathrm{π}}{6}$$x=3×\frac{\sqrt{3}}{2}\left[Since\left.\cos \frac{\mathrm{π}}{6}=\frac{\sqrt{3}}{2}\right]\right]$

Then the value is $x=\frac{3\sqrt{3}}{2}$.

Now take $y=r\sin \mathrm{θ}$and substitute $r=3,\mathrm{θ}=\frac{\mathrm{π}}{6}$then

$$\begin{gathered} y=3\sin \frac{\mathrm{π}}{6} \\ y=3×\frac{1}{2}\left[since\sin \frac{\mathrm{π}}{6}=\frac{1}{2}\right] \end{gathered}$$

Thus the rectangular coordinate $(x,y)=\left(\frac{3\sqrt{3}}{2},\frac{3}{2}\right)$.

Therefore, the rectangular coordinates are$(x,y)=\left(\frac{3\sqrt{3}}{2},\frac{3}{2}\right)$.

for the point $\left(-3,\frac{\mathrm{π}}{6}\right)$then $r=-3,\mathrm{θ}=\frac{\mathrm{π}}{6}$.

Use the equations,

Take $x=r\cos \mathrm{θ}$and substitute $r=-3,\mathrm{θ}=\frac{\mathrm{π}}{6}$then

$$\begin{gathered} x=-3\cos \frac{\mathrm{π}}{6} \\ x=-3×\frac{\sqrt{3}}{2}\left[\text{since}\cos \frac{\mathrm{π}}{6}=\frac{\sqrt{3}}{2}\right] \end{gathered}$$

Then the value is

$x=\frac{-3\sqrt{3}}{2}$

Now take $y=r\sin \mathrm{θ}$and substitute $r=-3,\mathrm{θ}=\frac{\mathrm{π}}{6}$then

$y=-3\sin \frac{\mathrm{Ï€}}{6}$

Thus,

$y=-3×\frac{1}{2}\left[since\sin \frac{\mathrm{π}}{6}=\frac{1}{2}\right]$

the rectangular coordinate $(x,y)=\left(\frac{-3\sqrt{3}}{2},\frac{-3}{2}\right)$.

Therefore, the rectangular coordinates are $(x,y)=\left(\frac{-3\sqrt{3}}{2},\frac{-3}{2}\right)$

For the point $\left(3,-\frac{\mathrm{π}}{6}\right)$then $r=3,\mathrm{θ}=-\frac{\mathrm{π}}{6}$.

Use the equations,

Take $x=r\cos \mathrm{θ}$and substitute $r=3,\mathrm{θ}=-\frac{\mathrm{π}}{6}$then

$$\begin{gathered} x=3\cos \left(-\frac{\mathrm{π}}{6}\right) \\ x=3×\frac{\sqrt{3}}{2}\left[\text{since}\cos \left(-\frac{\mathrm{π}}{6}\right)=\cos \frac{\mathrm{π}}{6}=\frac{\sqrt{3}}{2}\right] \end{gathered}$$

Then the value is $x=\frac{3\sqrt{3}}{2}$

Now take $y=r\sin \mathrm{θ}$and substitute $r=3,\mathrm{θ}=-\frac{\mathrm{π}}{6}$then

$y=3\sin \left(-\frac{\mathrm{Ï€}}{6}\right)$

Thus,

$y=-3·\frac{1}{2}\left[since\sin \frac{\mathrm{π}}{6}=\frac{1}{2}\right]$

The rectangular coordinate $(x,y)=\left(\frac{3\sqrt{3}}{2},\frac{-3}{2}\right)$.

Therefore, the rectangular coordinates are $(x,y)=\left(\frac{3\sqrt{3}}{2},\frac{-3}{2}\right)$.

for the point $\left(-3,-\frac{\mathrm{π}}{6}\right)$then $r=-3,\mathrm{θ}=-\frac{\mathrm{π}}{6}$.

Use the equations,

Take $x=r\cos \mathrm{θ}$and substitute $r=-3,\mathrm{θ}=-\frac{\mathrm{π}}{6}$then

width="344">x=-3cos-π6x=-3×32sincecos-π6=cosπ6=32

Then the value is

$x=\frac{-3\sqrt{3}}{2}$

Now take $y=r\sin \mathrm{θ}$and substitute $r=-3,\mathrm{θ}=-\frac{\mathrm{π}}{6}$then

$y=-3\sin \left(-\frac{\mathrm{Ï€}}{6}\right)$

Thus,

$y=-3·-\frac{1}{2}\left[\sin ce\right.$$\left.\sin \left(-\frac{\mathrm{π}}{6}\right)=-\sin \frac{\mathrm{π}}{6}=-\frac{\sqrt{3}}{2}\right]$

$y=\frac{3}{2}$

The rectangular coordinate $(x,y)=\left(\frac{-3\sqrt{3}}{2},\frac{3}{2}\right)$.

Thus, the rectangular coordinates are $(x,y)=\left(\frac{-3\sqrt{3}}{2},\frac{3}{2}\right)$.

Therefore, the answer is $\left(\frac{3\sqrt{3}}{2},\frac{3}{2}\right),\left(\frac{-3\sqrt{3}}{2},\frac{-3}{2}\right),\left(\frac{3\sqrt{3}}{2},\frac{-3}{2}\right),\left(\frac{-3\sqrt{3}}{2},\frac{3}{2}\right)$

The graphical representation is as follows,