Convert to Rectangular Form...6

[nycmathguy]

Question 12

r = 4 sin 2θ

r = 4 (2 sin θ cos θ)

r = 8 sin θ cos θ

r • r = 8 r sin θ cos θ

r^2 = 8y cos θ

x^2 + y^2 = 8y cos θ

Stuck here. How do I remove cos θ?

 

[MathLover1]

Apply the formula
sin(2α)=2sin(α)cos(α) with α=θ: r=8sin(θ)cos(θ).

From x=rcos(θ) and y=rsin(θ), we have that cos(θ)= x/r, sin(θ)= y/r, tan(θ)= y/x, and cot(θ)= x/y.

the input now takes the form
r=8xy/r^2
r^3=8xy
r^3-8xy=0

In rectangular coordinates, r= x^2+y^2

(x^2+y^2)^3-8xy=0

 

[nycmathguy]

Apply the formula
sin(2α)=2sin(α)cos(α) with α=θ: r=8sin(θ)cos(θ).

From x=rcos(θ) and y=rsin(θ), we have that cos(θ)= x/r, sin(θ)= y/r, tan(θ)= y/x, and cot(θ)= x/y.

the input now takes the form
r=8xy/r^2
r^3=8xy
r^3-8xy=0

In rectangular coordinates, r= x^2+y^2

(x^2+y^2)^3-8xy=0

I thought you said r = sqrt{x^2 + y^2}.
In this reply, you said:

"In rectangular coordinates, r= x^2+y^2."

So, which is it?

 

[MathLover1]

r = sqrt(x^2 + y^2)

here r= x^2+y^2 made typo, should be , r^2= x^2+y^2 and when you solve it for r you get r = sqrt(x^2 + y^2)

 

[nycmathguy]

r = sqrt(x^2 + y^2)

here r= x^2+y^2 made typo, should be , r^2= x^2+y^2 and when you solve it for r you get r = sqrt(x^2 + y^2)

Understood. Thank you.