How do you convert $r = 4\sin 2\theta $ to rectangular form?
Answer
600.6k+ views
Hint: Given the value of polar coordinates. We have to convert the polar form into a rectangular form. First, we will apply the trigonometric identities to the expression. Then, we will apply the relationship between the polar and rectangular coordinates. Then, we will substitute the values in the form x and y into the equation. Then, simplify the equation.
Formula used: The trigonometric identity for $\sin 2\theta $ is given as:
$\sin 2\theta = 2\sin \theta \cos \theta $
The relationship between the polar and rectangular coordinates is given as:
$x = r\cos \theta $, $y = r\sin \theta $ and ${r^2} = {x^2} + {y^2}$
Complete step-by-step solution:
We are given the polar coordinate $r = 4\sin 2\theta $.
First, we will apply the trigonometric identity to rewrite the expression $\sin 2\theta $
$ \Rightarrow r = 4\left( {2\sin \theta \cos \theta } \right)$
$ \Rightarrow r = 8\sin \theta \cos \theta $...........…….(1)
Now, we will determine the value of $\sin \theta $ and $\cos \theta $ using the relationship between the polar and rectangular coordinates.
$ \Rightarrow \dfrac{x}{r} = \cos \theta $
$ \Rightarrow \dfrac{y}{r} = \sin \theta $
Now, we will substitute the values of $\sin \theta $ and $\cos \theta $ to the equation (1).
$ \Rightarrow r = 8 \times \dfrac{y}{r} \times \dfrac{x}{r}$
$ \Rightarrow r = \dfrac{{8xy}}{{{r^2}}}$
Now we will substitute the value of ${r^2}$into the expression.
$ \Rightarrow r = \dfrac{{8xy}}{{{x^2} + {y^2}}}$
Now, on squaring both sides, we get:
$ \Rightarrow {r^2} = {\left( {\dfrac{{8xy}}{{{x^2} + {y^2}}}} \right)^2}$
$ \Rightarrow {r^2} = \dfrac{{64{x^2}{y^2}}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}}$
Now we will substitute the value of ${r^2}$to the left hand side of the expression.
$ \Rightarrow {x^2} + {y^2} = \dfrac{{64{x^2}{y^2}}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}}$
Now, we will cross multiply the terms, we get:
$ \Rightarrow \left( {{x^2} + {y^2}} \right){\left( {{x^2} + {y^2}} \right)^2} = 64{x^2}{y^2}$
On simplifying the expression, we get:
$ \Rightarrow {\left( {{x^2} + {y^2}} \right)^3} = 64{x^2}{y^2}$
Hence the polar coordinate in rectangular form is ${\left( {{x^2} + {y^2}} \right)^3} = 64{x^2}{y^2}$
Note: In such types of questions students mainly get confused in applying the formula. As they don't know which formula they have to apply. So, when the trigonometric function is given, then the student must apply the trigonometric identity to rewrite the expression. The students can make mistakes while representing the polar coordinates in rectangular form using the relationships between them.
Formula used: The trigonometric identity for $\sin 2\theta $ is given as:
$\sin 2\theta = 2\sin \theta \cos \theta $
The relationship between the polar and rectangular coordinates is given as:
$x = r\cos \theta $, $y = r\sin \theta $ and ${r^2} = {x^2} + {y^2}$
Complete step-by-step solution:
We are given the polar coordinate $r = 4\sin 2\theta $.
First, we will apply the trigonometric identity to rewrite the expression $\sin 2\theta $
$ \Rightarrow r = 4\left( {2\sin \theta \cos \theta } \right)$
$ \Rightarrow r = 8\sin \theta \cos \theta $...........…….(1)
Now, we will determine the value of $\sin \theta $ and $\cos \theta $ using the relationship between the polar and rectangular coordinates.
$ \Rightarrow \dfrac{x}{r} = \cos \theta $
$ \Rightarrow \dfrac{y}{r} = \sin \theta $
Now, we will substitute the values of $\sin \theta $ and $\cos \theta $ to the equation (1).
$ \Rightarrow r = 8 \times \dfrac{y}{r} \times \dfrac{x}{r}$
$ \Rightarrow r = \dfrac{{8xy}}{{{r^2}}}$
Now we will substitute the value of ${r^2}$into the expression.
$ \Rightarrow r = \dfrac{{8xy}}{{{x^2} + {y^2}}}$
Now, on squaring both sides, we get:
$ \Rightarrow {r^2} = {\left( {\dfrac{{8xy}}{{{x^2} + {y^2}}}} \right)^2}$
$ \Rightarrow {r^2} = \dfrac{{64{x^2}{y^2}}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}}$
Now we will substitute the value of ${r^2}$to the left hand side of the expression.
$ \Rightarrow {x^2} + {y^2} = \dfrac{{64{x^2}{y^2}}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}}$
Now, we will cross multiply the terms, we get:
$ \Rightarrow \left( {{x^2} + {y^2}} \right){\left( {{x^2} + {y^2}} \right)^2} = 64{x^2}{y^2}$
On simplifying the expression, we get:
$ \Rightarrow {\left( {{x^2} + {y^2}} \right)^3} = 64{x^2}{y^2}$
Hence the polar coordinate in rectangular form is ${\left( {{x^2} + {y^2}} \right)^3} = 64{x^2}{y^2}$
Note: In such types of questions students mainly get confused in applying the formula. As they don't know which formula they have to apply. So, when the trigonometric function is given, then the student must apply the trigonometric identity to rewrite the expression. The students can make mistakes while representing the polar coordinates in rectangular form using the relationships between them.
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