# How do you convert r = 2 sin theta into cartesian form?

Jul 14, 2016

Make use of a few formulas and do some simplification. See below.

#### Explanation:

When dealing with transformations between polar and Cartesian coordinates, always remember these formulas:

• $x = r \cos \theta$
• $y = r \sin \theta$
• ${r}^{2} = {x}^{2} + {y}^{2}$

From $y = r \sin \theta$, we can see that dividing both sides by $r$ gives us $\frac{y}{r} = \sin \theta$. We can therefore replace $\sin \theta$ in $r = 2 \sin \theta$ with $\frac{y}{r}$:
$r = 2 \sin \theta$
$\to r = 2 \left(\frac{y}{r}\right)$
$\to {r}^{2} = 2 y$

We can also replace ${r}^{2}$ with ${x}^{2} + {y}^{2}$, because ${r}^{2} = {x}^{2} + {y}^{2}$:
${r}^{2} = 2 y$
$\to {x}^{2} + {y}^{2} = 2 y$

We could leave it at that, but if you're interested...

Further Simplification
If we subtract $2 y$ from both sides we end up with this:
${x}^{2} + {y}^{2} - 2 y = 0$

Note that we can complete the square on ${y}^{2} - 2 y$:
${x}^{2} + \left({y}^{2} - 2 y\right) = 0$
$\to {x}^{2} + \left({y}^{2} - 2 y + 1\right) = 0 + 1$
$\to {x}^{2} + {\left(y - 1\right)}^{2} = 1$

And how about that! We end up with the equation of a circle with center $\left(h , k\right) \to \left(0 , 1\right)$ and radius $1$. We know that polar equations of the form $y = a \sin \theta$ form circles, and we just confirmed it using Cartesian coordinates.