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Question: How do you convert \( r = 2\cos 2\theta \) into rectangular form?...

How do you convert r=2cos⁡2θr = 2\cos 2\theta into rectangular form?

Explanation

Solution

In this problem, we have given an equation in the polar form. Here we are asked to convert the given polar form equation into the rectangular form. To convert the given equation into the equation into the rectangular form we need to use some identities and by using that identity we can get a required solution.

Complete step-by-step solution:
The polar form equation is r=2cos⁡2θr = 2\cos 2\theta .
Now, the identities that we are going to use are,
x=rcos⁡θx = r\cos \theta
y=rsin⁡θy = r\sin \theta
Now just squaring and adding the x,yx,y values, we get
x2=r2cos⁡2θ{x^2} = {r^2}{\cos ^2}\theta and y2=r2sin⁡2θ{y^2} = {r^2}{\sin ^2}\theta
Then, x2+y2=r2cos⁡2θ+r2sin⁡2θ{x^2} + {y^2} = {r^2}{\cos ^2}\theta + {r^2}{\sin ^2}\theta
Also, r2{r^2} is common in both the terms, so take r2{r^2} as common.
⇒x2+y2=r2(cos⁡2θ+sin⁡2θ)−−−−−(1)\Rightarrow {x^2} + {y^2} = {r^2}({\cos ^2}\theta + {\sin ^2}\theta ) - - - - - (1)
Now we know that, cos⁡2θ+sin⁡2θ=1{\cos ^2}\theta + {\sin ^2}\theta = 1
Substitute the value of cos⁡2θ+sin⁡2θ{\cos ^2}\theta + {\sin ^2}\theta in equation (1), we get
⇒x2+y2=r2\Rightarrow {x^2} + {y^2} = {r^2}
Also we can write cos⁡2θ=cos⁡2θ−sin⁡2θ\cos 2\theta = {\cos ^2}\theta - {\sin ^2}\theta , so we can substitute this value in the given polar equation. Then we get,
r=2cos⁡2θ=2(cos⁡2θ−sin⁡2θ)−−−−−(∗)r = 2\cos 2\theta = 2({\cos ^2}\theta - {\sin ^2}\theta ) - - - - - (*)
Now, further simplification using the identities…
x2−y2=r2cos⁡2θ−r2sin⁡2θ{x^2} - {y^2} = {r^2}{\cos ^2}\theta - {r^2}{\sin ^2}\theta , now take the r2{r^2} common outside.
⇒x2−y2=r2(cos⁡2θ−sin⁡2θ)\Rightarrow {x^2} - {y^2} = {r^2}({\cos ^2}\theta - {\sin ^2}\theta )
On rewriting we get
⇒x2−y2r2=cos⁡2θ−sin⁡2θ−−−−−(2)\Rightarrow \dfrac{{{x^2} - {y^2}}}{{{r^2}}} = {\cos ^2}\theta - {\sin ^2}\theta - - - - - (2)
Now use (*) in equation (2), we get
r=2r2(x2−y2)r = \dfrac{2}{{{r^2}}}({x^2} - {y^2})
Now, substitute the value of r2{r^2} , we get
⇒r=2(x2−y2)(x2+y2)\Rightarrow r = \dfrac{{2({x^2} - {y^2})}}{{({x^2} + {y^2})}}
Also, we know that x2+y2=r2⇒x2+y2=r{x^2} + {y^2} = {r^2} \Rightarrow \sqrt {{x^2} + {y^2}} = r
⇒x2+y2=2(x2−y2)(x2+y2)\Rightarrow \sqrt {{x^2} + {y^2}} = \dfrac{{2({x^2} - {y^2})}}{{({x^2} + {y^2})}}
Taking cross multiply and we get
⇒(x2+y2)32=2(x2−y2)\Rightarrow {\left( {{x^2} + {y^2}} \right)^{\dfrac{3}{2}}} = 2\left( {{x^2} - {y^2}} \right)
On rewriting we get
⇒(x2+y2)32−2(x2−y2)=0\Rightarrow {\left( {{x^2} + {y^2}} \right)^{\dfrac{3}{2}}} - 2\left( {{x^2} - {y^2}} \right) = 0 .

**This is the rectangular form of the given polar equation.
(x2+y2)32−2(x2−y2)=0{\left( {{x^2} + {y^2}} \right)^{\dfrac{3}{2}}} - 2\left( {{x^2} - {y^2}} \right) = 0 **

Additional Information: In two dimensions, the Cartesian coordinate (x, y) specifies the location of a point P in the plane. Another two- dimensional coordinate system is polar coordinate system is polar coordinates. As r ranges from 00 to infinity and θ\theta ranges from 0 to 2π0{\text{ to 2}}\pi , the point P specified by the polar coordinate (r,θ)(r,\theta ) covers every point in the plane.

Note: In this problem our aim is to convert the given polar coordinate equation which is in the form (r,θ)\left( {r,\theta } \right) into rectangular coordinates, for this we followed some important steps. In that first one we need to write x=rcos⁡θx = r\cos \theta and y=rsin⁡θy = r\sin \theta . Next we need to evaluate cos⁡θ\cos \theta and sin⁡θ\sin \theta . Then we have to multiply cos⁡θ\cos \theta by r to find the x-coordinate of the rectangular form and multiply sin⁡θ\sin \theta by r to find the y-coordinate of the rectangular form.