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Question: How do you convert \(r = 4\sin 2\theta \) to rectangular form?...

How do you convert r=4sin⁡2θr = 4\sin 2\theta to rectangular form?

Explanation

Solution

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⁡2θ\sin 2\theta is given as:
sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta = 2\sin \theta \cos \theta
The relationship between the polar and rectangular coordinates is given as:
x=rcos⁡θx = r\cos \theta , y=rsin⁡θy = r\sin \theta and r2=x2+y2{r^2} = {x^2} + {y^2}

Complete step-by-step solution:
We are given the polar coordinate r=4sin⁡2θr = 4\sin 2\theta .
First, we will apply the trigonometric identity to rewrite the expression sin⁡2θ\sin 2\theta
⇒r=4(2sin⁡θcos⁡θ)\Rightarrow r = 4\left( {2\sin \theta \cos \theta } \right)
⇒r=8sin⁡θcos⁡θ\Rightarrow r = 8\sin \theta \cos \theta...........…….(1)
Now, we will determine the value of sin⁡θ\sin \theta and cos⁡θ\cos \theta using the relationship between the polar and rectangular coordinates.
⇒xr=cos⁡θ\Rightarrow \dfrac{x}{r} = \cos \theta
⇒yr=sin⁡θ\Rightarrow \dfrac{y}{r} = \sin \theta
Now, we will substitute the values of sin⁡θ\sin \theta and cos⁡θ\cos \theta to the equation (1).
⇒r=8×yr×xr\Rightarrow r = 8 \times \dfrac{y}{r} \times \dfrac{x}{r}
⇒r=8xyr2\Rightarrow r = \dfrac{{8xy}}{{{r^2}}}
Now we will substitute the value of r2{r^2}into the expression.
⇒r=8xyx2+y2\Rightarrow r = \dfrac{{8xy}}{{{x^2} + {y^2}}}
Now, on squaring both sides, we get:
⇒r2=(8xyx2+y2)2\Rightarrow {r^2} = {\left( {\dfrac{{8xy}}{{{x^2} + {y^2}}}} \right)^2}
⇒r2=64x2y2(x2+y2)2\Rightarrow {r^2} = \dfrac{{64{x^2}{y^2}}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}}
Now we will substitute the value of r2{r^2}to the left hand side of the expression.
⇒x2+y2=64x2y2(x2+y2)2\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:
⇒(x2+y2)(x2+y2)2=64x2y2\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:
⇒(x2+y2)3=64x2y2\Rightarrow {\left( {{x^2} + {y^2}} \right)^3} = 64{x^2}{y^2}

Hence the polar coordinate in rectangular form is (x2+y2)3=64x2y2{\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.