# How do you find an equivalent equation in rectangular coordinates r = -3 + 5 cos x ?

Jan 27, 2016

Use ${r}^{2} = {x}^{2} + {y}^{2}$

#### Explanation:

There are a couple of basic relations here that are helpful. The Pythagorean theorem and the unit circle tell us that the square of hypotenuse of a right triangle is equal to the sum of the squares of the of other two sides, or:

${a}^{2} + {b}^{2} = {c}^{2}$

Now throw that right triangle inside a circle (circumscribe it), such that one corner is on the origin, and the other is on the circumference of the circle. This will invariably leave the right angle sitting on the x or y axis.

This makes the hypotenuse the radius of the circle, and the other two sides the x and y axis. So our relation becomes:

${x}^{2} + {y}^{2} = {r}^{2}$

From the unit circle, we can find that the length of the leg on the x axis is determined by the cosine of the angle between the x axis and the hypotenuse of our circumscribed triangle. SOHCAHTOA my friends, SOHCAHTOA. Using this we find that:

$x = r \cos \left(t\right)$
$y = r \sin \left(t\right)$

Where $r$ is the radius, and $\theta$ is the angle.

And since $\tan \left(\theta\right) = \sin \frac{\theta}{\cos} \left(\theta\right)$ we get a bonus (corollary) relation that:

$\tan \left(\theta\right) = \frac{x}{y}$

So putting it all together, you just write $r = \sqrt{{x}^{2} + {y}^{2}}$, substitute this into the given equation, and rearrange the terms.

All this leads me to believe that the x in the above equation should have been a $\theta$, because otherwise this equation is halfway to rectangular already.

If that's the case, you need to solve $x = r \cos \left(t\right)$ for $\cos \left(t\right)$, substitute the given relation for $r$, and then throw it all back in the original equation. Most people like to solve for $y$ after this to put the equation in general form, but with all those square roots kicking around it probably won't be possible.