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How can $\left( 8,-{{45}^{\circ }} \right)$ be converted into rectangular coordinates?

Last updated date: 20th Jun 2024
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Hint: We are given some coordinates as $\left( 8,-{{45}^{\circ }} \right)$ and we are asked to change it into a rectangle coordinate. To answer this we will learn what are rectangular coordinates and polar coordinates, how they are connected to each other. Then we will use that x is given as $r\cos \theta$ and y is given as $r\sin \theta$ where r is the magnitude and $\theta$ is the argument. We will also learn about complex numbers.

We are given that the coordinates given to us are $\left( 8,-{{45}^{\circ }} \right)$ and if we look closely, we can see that a coordinate is a single number while the other is a number with the degree. So, we are asked to change it into a rectangular coordinate. To answer this we will first learn about complex numbers. Generally, a complex number is represented as z = x + iy. So, we can also write it as z = (x, y). This form of the complex number is called a rectangular coordinate. Another way to write a complex is $z=r\left( \cos \theta +i\sin \theta \right).$ So, we can write it into coordinate as $z=\left( r,{{\theta }^{\circ }} \right)$ and this form is called a complex polar coordinate.
If we compare these two z = x + iy and $z=r\left( \cos \theta +i\sin \theta \right)=r\cos \theta +ir\sin \theta$ we can see that $x=r\cos \theta$ and $y=r\sin \theta .$ So, $x=r\cos \theta$ and $y=r\sin \theta$ this is the relation which will help us to convert polar form to rectangular form. As we have $\left( 8,-{{45}^{\circ }} \right)$ so it means we have r = 8 and $\theta =-{{45}^{\circ }}.$ Using this in $x=r\cos \theta ,$ we get $x=8\times \cos \left( -{{45}^{\circ }} \right)$ as $\cos \left( -\theta \right)=\cos \theta .$
So, $x=8\cos \left( {{45}^{\circ }} \right)=8\times \dfrac{1}{\sqrt{2}}.$ On simplifying this, we get, $x=4\sqrt{2}.$ Now putting r = 8 and $\theta =-{{45}^{\circ }}$ in $y=r\sin \theta ,$ we get,
$y=8\sin \left( -{{45}^{\circ }} \right)\left[ \sin \left( -\theta \right)=-\sin \theta \right]$
$\Rightarrow y=-8\times \sin {{45}^{\circ }}$
$\Rightarrow y=-8\times \dfrac{1}{\sqrt{2}}$
$\Rightarrow y=-4\sqrt{2}$
Hence we get $x=4\sqrt{2}$ and $y=-4\sqrt{2}.$ So, the rectangular coordinates are $\left( x,y \right)=\left( 4\sqrt{2},-4\sqrt{2} \right).$
Note: To check that our solution is correct we can use the knowledge that $\left( r,-\theta \right),-\theta$ always lies in the fourth quadrant. In the rectangular form, the fourth quadrant compromises positive x and negative y. As we can see that in our solution $\left( x,y \right)=\left( 4\sqrt{2},-4\sqrt{2} \right)$ x is positive and y is negative. So, it means we got the correct solution.