Question

Convert $r = \cos \left( \theta \right) - \sin \left( \theta \right)$ to a rectangular equation.

Answer 1

To convert Polar coordinates$\left( {r,\theta } \right)$ to rectangular coordinates $\left( {x,y} \right)$, we have the equation:
$x = r\cos \theta \\\ \Rightarrow y = r\sin \theta$
So by using the above equation and by using the process of substitution we can convert $r = \cos \left( \theta \right) - \sin \left( \theta \right)$ into the rectangular form.

Complete step by step answer:
Given, $r = \cos \left( \theta \right) - \sin \left( \theta \right).........................\left( i \right)$
We know that rectangle coordinates are the Cartesian coordinates seen in the Cartesian plane which is represented by $\left( {x,y} \right)$ and polar coordinates give the position of a point in a plane by using the length $r$ and the angle made to the fixed point $\theta$, and is represented by $\left( {r,\theta } \right).$We know that (i) which is a polar coordinate is to be converted to a rectangular coordinate.For that we can use the formula:
$x = r\cos \theta .................\left( {ii} \right) \\\ \Rightarrow y = r\sin \theta ..................\left( {iii} \right) \\\$
Now we can find the value of $\sin \theta \;{\text{and}}\;\cos \theta$ from equation (ii) and (iii):
$x = r\cos \theta \\\ \Rightarrow \cos \theta = \dfrac{x}{r}......................\left( {iv} \right) \\\$
Similarly we can say that:
$y = r\sin \theta \\\ \Rightarrow \sin \theta = \dfrac{y}{r}......................\left( v \right) \\\$
Substituting the values of $\sin \theta \;{\text{and}}\;\cos \theta$ from (iv) and (v) in (i) we can write:

\Rightarrow r = \dfrac{x}{r} - \dfrac{y}{r} \\\ \Rightarrow r = \dfrac{{x - y}}{r} \\\ \Rightarrow {r^2} = x - y......................\left( {vi} \right) \\\ $$ Now we have to convert $r = \cos \left( \theta \right) - \sin \left( \theta \right)$ into Cartesian form we have to express $r$ in the form of rectangular coordinates.Such that we know from (ii) and (iii): $$x = r\cos \theta \\\ \Rightarrow y = r\sin \theta \\\ \Rightarrow {x^2} + {y^2} = {\left( {r\cos \theta } \right)^2} + {\left( {r\sin \theta } \right)^2} \\\ \Rightarrow {x^2} + {y^2} = {r^2}\left( {{{\cos }^2}\theta + {{\sin }^2}\theta } \right) \\\ \Rightarrow {x^2} + {y^2} = {r^2} \\\ \Rightarrow {r^2} = {x^2} + {y^2}.......................\left( {vii} \right) \\\ $$ Now substituting the value of ${r^2}$in (vi) we can write: $${r^2} = x - y \\\ \Rightarrow{x^2} + {y^2} = x - y \\\ \Rightarrow \left( {{x^2} - x} \right)\left( {{y^2} + y} \right) = 0......................\left( {viii} \right) \\\ $$ Simplifying (viii) by using complete the square method, so taking the coefficient of $x\;{\text{and}}\;y$, dividing by 2 and then taking it’s square: ${\text{for}}\;x,\;{\text{coefficient = }}\left( { - \dfrac{1}{2}} \right) \\\ \Rightarrow{\text{squaring = }}{\left( { - \dfrac{1}{2}} \right)^2} = \left( {\dfrac{1}{4}} \right) \\\ \Rightarrow{\text{for}}\;y,\;{\text{coefficient = }}\left( {\dfrac{1}{2}} \right) \\\ \Rightarrow{\text{squaring = }}{\left( {\dfrac{1}{2}} \right)^2} = \left( {\dfrac{1}{4}} \right) \\\ $ Now inserting the above results to both LHS and RHS of (viii) we can write: $${x^2} - x + \dfrac{1}{4} + {y^2} + y + \dfrac{1}{4} = \dfrac{1}{4} + \dfrac{1}{4}.....................\left( {ix} \right)$$ Simplifying we get: $$\therefore{\left( {x - \dfrac{1}{2}} \right)^2} + {\left( {y + \dfrac{1}{2}} \right)^2} = \dfrac{1}{2}............................\left( x \right)$$ **Therefore, $r = \cos \left( \theta \right) - \sin \left( \theta \right)$ in rectangular form can be written as: $${\left( {x - \dfrac{1}{2}} \right)^2} + {\left( {y + \dfrac{1}{2}} \right)^2} = \dfrac{1}{2}$$.** **Note:** We know that to convert a polar coordinate $\left( {r,\theta } \right)$ to a rectangular coordinate $\left( {x,y} \right)$, we can use the formula: $x = r\cos \theta \\\ \Rightarrow y = r\sin \theta $ In a similar manner convert rectangular coordinate $\left( {x,y} \right)$ to a polar coordinate $\left( {r,\theta } \right)$, we can use the formula: $r = \sqrt {\left( {{x^2} + {y^2}} \right)} \\\ \Rightarrow\theta = {\tan ^{ - 1}}\left( {\dfrac{y}{x}} \right)$ Also while choosing $\theta $ it’s better to choose it in radians since when $\theta $ is in radians the calculations become much easier.