Question

What is the Cartesian form of $\left( -8,\dfrac{-15\pi }{4} \right)$?

Answer 1

Hint: Compare the given coordinates with general notation of the polar coordinates given as $\left( r,\theta \right)$. Now, to find the values of x and y coordinates which represent the Cartesian coordinates by using the relations \[y=r\sin \theta \] and $x=r\cos \theta $. Use the properties of the sine and cosine functions given as $\sin \left( -\theta \right)=-\sin \theta $ and $\cos \left( -\theta \right)=\cos \theta $ to simplify the relation. Write $\left( \dfrac{15\pi }{4} \right)=\left( 4\pi -\dfrac{\pi }{4} \right)$ and use the formulas $\sin \left( 2n\pi -\theta \right)=-\sin \theta $ and $\cos \left( 2n\pi -\theta \right)=\cos \theta $ where n is any integer, use the value $\sin \left( \dfrac{\pi }{4} \right)=\cos \left( \dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}}$ to get the answer.

Complete step by step solution:
Here we have been provided with the coordinates in polar form as $\left( -8,\dfrac{-15\pi }{4} \right)$ and we are asked to convert it into the Cartesian form.
Now, any point in polar coordinates is given by the general notation \[\left( r,\theta \right)\], where r is called the radius vector of the point and $\theta $ is the angle subtended by the vector with the positive x axis. On comparing \[\left( r,\theta \right)\] with $\left( -8,\dfrac{-15\pi }{4} \right)$ we have $r=-8$ and $\theta =\left( -\dfrac{15\pi }{4} \right)$. The relation between the Cartesian coordinates (x, y) and the polar coordinate is given as: -
$y=r\sin \theta ..................\left( i \right)$
$x=r\cos \theta .................\left( ii \right)$
Let us find the Cartesian coordinates by considering the above two relations one by one.
(1) For y coordinate, substituting the given values in relation (i) we get,
$\Rightarrow y=\left( -8 \right)\times \sin \left( -\dfrac{15\pi }{4} \right)$
Using the property of the sine function given as $\sin \left( -\theta \right)=-\sin \theta $ we get,
$\begin{align}
  & \Rightarrow y=8\times \sin \left( \dfrac{15\pi }{4} \right) \\
 & \Rightarrow y=8\times \sin \left( 4\pi -\dfrac{\pi }{4} \right) \\
\end{align}$
Using the formula $\sin \left( 2n\pi -\theta \right)=-\sin \theta $ where $n\in $ integers we get,
$\Rightarrow y=-8\times \sin \left( \dfrac{\pi }{4} \right)$
We know that $\sin \left( \dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}}$, so we get,
$\begin{align}
  & \Rightarrow y=-8\times \dfrac{1}{\sqrt{2}} \\
 & \therefore y=-4\sqrt{2} \\
\end{align}$
(2) For x coordinate, substituting the given values in relation (ii) we get,
$\Rightarrow x=\left( -8 \right)\times \cos \left( -\dfrac{15\pi }{4} \right)$
Using the property of the sine function given as $\cos \left( -\theta \right)=\cos \theta $ we get,
$\begin{align}
  & \Rightarrow x=\left( -8 \right)\times \cos \left( \dfrac{15\pi }{4} \right) \\
 & \Rightarrow x=\left( -8 \right)\times \cos \left( 4\pi -\dfrac{\pi }{4} \right) \\
\end{align}$
Using the formula $\cos \left( 2n\pi -\theta \right)=\cos \theta $ where $n\in $ integers we get,
$\Rightarrow x=\left( -8 \right)\times \cos \left( \dfrac{\pi }{4} \right)$
We know that $\cos \left( \dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}}$, so we get,
$\begin{align}
  & \Rightarrow x=-8\times \dfrac{1}{\sqrt{2}} \\
 & \therefore x=-4\sqrt{2} \\
\end{align}$
Hence, the Cartesian form of the given point in polar form is \[\left( x,y \right)=\left( -4\sqrt{2},-4\sqrt{2} \right)\].

Note: Note that if we will find the locus of the point then we will get the equation $y=x$ which clearly represents a straight line. You must remember the relationship between the Cartesian coordinates and the polar coordinates to solve the above question. Also, remember the important relation: - \[{{x}^{2}}+{{y}^{2}}={{r}^{2}}\] as it will be used further in the coordinate geometry of circles.