how do I convert $r=6+\sin(4t)$ to rectangular form

Question

I am trying to convert $r=6+\sin(4t)$ to rectangular form. would you please show me how?

Answer 1

First, I'm assuming that $t$ means $\theta$ in $r = 6 + \sin(4t)$.

If you get it into a form where $\theta$ is only present as "$\sin \theta$" or "$\cos \theta$", you can then replace $\sin \theta$ sith $\frac{y}{r}$, $\cos \theta$ with $\frac{x}{r}$, and finally get rid of all the $r$s by replacing them with $\sqrt{x^2 + y^2}$.

So, the goal should be to get $\sin 4 \theta$ written in terms of $\sin \theta$ and $\cos \theta$. Here is a hint: apply the following formulas as many times as you need: \begin{align*} \sin 2 \alpha &= 2 \sin \alpha \cos \alpha \\ \cos 2 \alpha &= \cos^2 \alpha - \sin^2 \alpha. \end{align*}

Answer 2

Hint: Express $\sin 4t$ in terms of powers of $\sin$ and $\cos$ through the use of double angle identities.

Answer 3

Recall from trigonometry that $$ \sin(4t) = 2\sin(2t)\cos(2t) = 2\Big(2\sin t\cos t\Big)\Big(\cos^2 t - \sin^2 t\Big) = 4\sin t\cos^3 t - 4\sin^3 t \cos t. $$

You have $r = 6\sin(4t).$ Multiplying both sides of that by $r^4$ gives you \begin{align} r^5 & = 4(r\sin t)(r\cos t)^3 - 4(r\sin t)^3(r\cos t) \\[10pt] \sqrt{x^2+y^2}^5 & = 4yx^3 - ry^3 x. \end{align}