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Question: How can \[\left( 8,-{{45}^{\circ }} \right)\] be converted into rectangular coordinates?...

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

Explanation

Solution

We are given some coordinates as (8,−45∘)\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 rcos⁡θr\cos \theta and y is given as rsin⁡θr\sin \theta where r is the magnitude and θ\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 (8,−45∘)\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(cos⁡θ+isin⁡θ).z=r\left( \cos \theta +i\sin \theta \right). So, we can write it into coordinate as z=(r,θ∘)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(cos⁡θ+isin⁡θ)=rcos⁡θ+irsin⁡θz=r\left( \cos \theta +i\sin \theta \right)=r\cos \theta +ir\sin \theta we can see that x=rcos⁡θx=r\cos \theta and y=rsin⁡θ.y=r\sin \theta . So, x=rcos⁡θx=r\cos \theta and y=rsin⁡θy=r\sin \theta this is the relation which will help us to convert polar form to rectangular form. As we have (8,−45∘)\left( 8,-{{45}^{\circ }} \right) so it means we have r = 8 and θ=−45∘.\theta =-{{45}^{\circ }}. Using this in x=rcos⁡θ,x=r\cos \theta , we get x=8×cos⁡(−45∘)x=8\times \cos \left( -{{45}^{\circ }} \right) as cos⁡(−θ)=cos⁡θ.\cos \left( -\theta \right)=\cos \theta .
So, x=8cos⁡(45∘)=8×12.x=8\cos \left( {{45}^{\circ }} \right)=8\times \dfrac{1}{\sqrt{2}}. On simplifying this, we get, x=42.x=4\sqrt{2}. Now putting r = 8 and θ=−45∘\theta =-{{45}^{\circ }} in y=rsin⁡θ,y=r\sin \theta , we get,
y=8sin⁡(−45∘)[sin⁡(−θ)=−sin⁡θ]y=8\sin \left( -{{45}^{\circ }} \right)\left[ \sin \left( -\theta \right)=-\sin \theta \right]
⇒y=−8×sin⁡45∘\Rightarrow y=-8\times \sin {{45}^{\circ }}
⇒y=−8×12\Rightarrow y=-8\times \dfrac{1}{\sqrt{2}}
On simplifying, we get,
⇒y=−42\Rightarrow y=-4\sqrt{2}

Hence we get x=42x=4\sqrt{2} and y=−42.y=-4\sqrt{2}. So, the rectangular coordinates are (x,y)=(42,−42).\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 (r,−θ),−θ\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 (x,y)=(42,−42)\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.