How do you convert $r = 2 sin 3 (θ)$ $r=2sin3(theta)$ to rectangular form?

Question

3769 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-04-29T14:46:41

$x^{2} + y^{2} = 4 {sin}^{2} ({tan}^{- 1} (\frac{y}{x}))$ $x^2+y^2=4 sin^2(tan^(-1)(y/x))$

Here, $| r | \leq 2, r = \pm \sqrt{x^{2} + y^{2}} and θ = {tan}^{- 1} (\frac{y}{x})$ $|r|<=2, r=+-sqrt(x^2+y^2) and theta=tan^(-1)(y/x)$

So, the rectangular form is $\pm \sqrt{x^{2} + y^{2}} = 2 sin (3 {tan}^{- 1} (\frac{y}{x}))$ $+-sqrt(x^2+y^2)=2 sin (3 tan^(-1)(y/x))$

Remove the ambiguity in sign for r, by squaring both sides.

It is also possible to have a form, sans trigonometric functions, by

expanding $sin 3 θ$ $sin 3 theta$ in powers of $sin θ and cos θ$ $sin theta and cos theta$ and

substituting $sin θ = \frac{y}{r} and cos θ = \frac{x}{r}$ $sin theta =y/r and cos theta=x/r$ However, the form

given as answer appears to be elegant... .

How do you convert r=2sin3(theta)r=2sin3(θ)r=2sin3(theta) to rectangular form?