Answer

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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.

**Complete step by step answer:**

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,

\[y=8\sin \left( -{{45}^{\circ }} \right)\left[ \sin \left( -\theta \right)=-\sin \theta \right]\]

\[\Rightarrow y=-8\times \sin {{45}^{\circ }}\]

\[\Rightarrow y=-8\times \dfrac{1}{\sqrt{2}}\]

On simplifying, we get,

\[\Rightarrow y=-4\sqrt{2}\]

**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.

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