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How do you convert $\left( -4,3\pi \right)$ from polar to Cartesian coordinates?

Answer
VerifiedVerified
547.2k+ views
Hint: To convert the given polar coordinates $\left( r,\theta \right)$ into Cartesian coordinates $\left( x,y \right)$ we will use the following relations:
$\begin{align}
  & x=r\cos \theta \\
 & y=r\sin \theta \\
\end{align}$
Then by substituting the values and simplify the obtained equations we get the desired answer.

Complete step-by-step answer:
We have been given a polar coordinates $\left( -4,3\pi \right)$.
We have to convert the given polar coordinates into Cartesian coordinates.
Now, we know that $\left( x,y \right)$ are Cartesian coordinates and $\theta $ is the angle the ray joining the origin and the point makes with positive x-axis.
Now, we know that the relation between polar coordinates $\left( r,\theta \right)$ (here, r is called the radical coordinate and $\theta $ is called the angular coordinate) and Cartesian coordinates $\left( x,y \right)$ is given by
$\begin{align}
  & x=r\cos \theta \\
 & y=r\sin \theta \\
\end{align}$
So by substituting the values we will get
$\Rightarrow x=-4\cos 3\pi $
Now, we know that $\cos 3\pi =-1$ so substituting the value we will get
$\begin{align}
  & \Rightarrow x=-4\left( -1 \right) \\
 & \Rightarrow x=4 \\
\end{align}$
Now, consider $y=r\sin \theta $
So by substituting the values we will get
$\Rightarrow y=-4\sin 3\pi $
Now, we know that $\sin 3\pi =0$ so substituting the value we will get
$\begin{align}
  & \Rightarrow y=-4\left( 0 \right) \\
 & \Rightarrow y=0 \\
\end{align}$
Hence we get the required Cartesian coordinates of the given polar coordinates $\left( -4,3\pi \right)$ as $\left( 4,0 \right)$.

Note: We can use the relation $r=\sqrt{{{x}^{2}}+{{y}^{2}}}$,$\theta =atan2\left( y,x \right)$ to convert the Cartesian coordinates into polar coordinates. The point to be remembered is that the angle is always measured in degree and the value of radical coordinate is always positive. We can convert the negative angle to positive by adding $2\pi $.