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Question: How do you convert \[\left( 6,\dfrac{\pi }{4} \right)\] into rectangular coordinates?...

How do you convert (6,π4)\left( 6,\dfrac{\pi }{4} \right) into rectangular coordinates?

Explanation

Solution

Consider the given coordinate (6,π4)\left( 6,\dfrac{\pi }{4} \right) as polar coordinates (r,θ)\left( r,\theta \right) and hence compare the values of r and θ\theta . Now, assume these points in the rectangular coordinates system, also known as cartesian coordinates system, as (x, y). Use the relationship: - x=rcosθx=r\cos \theta and y=rsinθy=r\sin \theta to find the values of x and y and get the answer. Use the value: - cos(π4)=sin(π4)=12\cos \left( \dfrac{\pi }{4} \right)=\sin \left( \dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}}.

Complete step by step answer:
Here, we have been provided with the point (6,π4)\left( 6,\dfrac{\pi }{4} \right) and we are asked to convert it into rectangular coordinates.
Now, the given point is of the form (r,θ)\left( r,\theta \right) so we can say that we have been provided with the polar coordinates of a point in a plane. In Mathematics, rectangular coordinates are also known as the cartesian coordinates. Any point in the cartesian system of coordinates is denoted by (x, y) where ‘x’ represents the distance of the point from y – axis and ‘y’ represents the distance of the point from x – axis, used with signs.
Let us come to the question. On comparing (6,π4)\left( 6,\dfrac{\pi }{4} \right) with (r,θ)\left( r,\theta \right) we have: -
r=6,θ=π4\Rightarrow r=6,\theta =\dfrac{\pi }{4}
Assuming the respective point in rectangular coordinates as (x, y) and using the relations: - x=rcosθx=r\cos \theta and y=rsinθy=r\sin \theta , we have,
(i) x – coordinate: -
x=6×cos(π4)\Rightarrow x=6\times \cos \left( \dfrac{\pi }{4} \right)
Substituting, cosπ4=12\cos \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}, we get,
x=6×12\Rightarrow x=6\times \dfrac{1}{\sqrt{2}}
x=32\Rightarrow x=3\sqrt{2}
(ii) y – coordinate: -
y=6×sin(π4)\Rightarrow y=6\times \sin \left( \dfrac{\pi }{4} \right)
Substituting sinπ4=12\sin \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}, we get,

& \Rightarrow y=6\times \dfrac{1}{\sqrt{2}} \\\ & \Rightarrow y=3\sqrt{2} \\\ \end{aligned}$$ **Hence, the required point in the rectangular system of coordinates is $$\left( 3\sqrt{2},3\sqrt{2} \right)$$.** **Note:** One may note that the point we have obtained will lie in the first quadrant because both x and y – coordinates are positive. You must remember the different forms in which a point is represented in a plane and relationship between them like: - polar form, parametric form etc. Here, in the above question, ‘r’ represents the radius vector of a point and ‘$$\theta $$’ represents the angle subtended by the radius vector with positive x – axis. You must remember the relations: - $$x=r\cos \theta $$ and $$y=r\sin \theta $$ to solve the question.