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Question: How do you convert \[r = 2\sin \left( {3\theta } \right)\] to rectangular form?...

How do you convert r=2sin⁡(3θ)r = 2\sin \left( {3\theta } \right) to rectangular form?

Explanation

Solution

Hint : To solve this question, we need to use the relation between the polar coordinates and Cartesian coordinates. We know that (r,θ)\left( {r,\theta } \right) are the polar coordinates and (x,y)\left( {x,y} \right) are the Cartesian coordinates. We just need to put the values in terms of xx and yy to find our answer.
Formula used:
x=rcos⁡θx = r\cos \theta
y=rsin⁡θy = r\sin \theta
r2=x2+y2{r^2} = {x^2} + {y^2}
tan⁡θ=yx\tan \theta = \dfrac{y}{x}

Complete step-by-step answer :
We are given the equation r=2sin⁡(3θ)r = 2\sin \left( {3\theta } \right) which is in the polar form.
Now, our next step will be to convert it into a simple trigonometric function.
Here, we are given sin⁡3θ\sin 3\theta . Therefore, we have to convert it into sin⁡θ\sin \theta .
We know that sin⁡3θ=3sin⁡θ−sin⁡3θ\sin 3\theta = 3\sin \theta - {\sin ^3}\theta .
Therefore, now we have

r=2(3sin⁡θ−sin⁡3θ) ⇒r=6sin⁡θ−2sin⁡3θ   r = 2\left( {3\sin \theta - {{\sin }^3}\theta } \right) \\\ \Rightarrow r = 6\sin \theta - 2{\sin ^3}\theta \;

Now, we will use the equation y=rsin⁡θy = r\sin \theta and therefore sin⁡θ=yr\sin \theta = \dfrac{y}{r}.

⇒r=6(yr)−2(y3r3) ⇒r4=6yr2−2y3   \Rightarrow r = 6\left( {\dfrac{y}{r}} \right) - 2\left( {\dfrac{{{y^3}}}{{{r^3}}}} \right) \\\ \Rightarrow {r^4} = 6y{r^2} - 2{y^3} \;

Here, we have taken LCM. Now, we will use the equation r2=x2+y2{r^2} = {x^2} + {y^2}.

⇒(x2+y2)2=6y(x2+y2)−2y3 ⇒(x2+y2)2=2y(x2−y2)   \Rightarrow {\left( {{x^2} + {y^2}} \right)^2} = 6y\left( {{x^2} + {y^2}} \right) - 2{y^3} \\\ \Rightarrow {\left( {{x^2} + {y^2}} \right)^2} = 2y\left( {{x^2} - {y^2}} \right) \;

Thus, (x2+y2)2=2y(x2−y2){\left( {{x^2} + {y^2}} \right)^2} = 2y\left( {{x^2} - {y^2}} \right) is the rectangular form of r=2sin⁡(3θ)r = 2\sin \left( {3\theta } \right) .
So, the correct answer is “ (x2+y2)2=2y(x2−y2){\left( {{x^2} + {y^2}} \right)^2} = 2y\left( {{x^2} - {y^2}} \right)”.

Note : In this type of question, where we are asked to convert the polar equation into Cartesian one, we need to keep in mind three important steps:
First, we should rewrite any trigonometric functions in terms of the basic trigonometric functions.
Second, we have to change the equation in rectangular coordinates by using the relations between polar and Cartesian coordinates.
Third, we need to substitute the values and then just simplify the equation to get to the final answer.