We are studying polar and rectangular equations in math. The last question on the homework was:
Write $r = \cos\theta + \sin\theta$ in rectangular form.
I have tried a few ways with no luck. I would love it if someone could tell me if this is possible, and explain the steps to convert it from polar to rectangular form. Thank you!
One example of what I have tried:
- $r = \cos\theta + \sin\theta$
- $r\cdot r = r\cos\theta + r\sin\theta$
Since $r^2 = x^2 + y^2$, $r\cos\theta = x$, and $r\sin\theta = y$,
- $x^2 + y^2 = x + y$
Step 3 is not equivalent to step 1 however, according to desmos here when zoomed in (a lot).
With further insight from the comments section, I have learned Desmos can become inaccurate when zoomed in very (very) closely, which is what caused me to believe my answer was incorrect.
Best Answer
What you tried is perfect, the two equations are almost equivalent, see e.g. the answer I gave here. In your case $$r^2=r\cos\theta+r\sin\theta\Longleftrightarrow\cases{r=\cos\theta+\sin\theta\\\text{or}\\r=0}$$
So the only caveat is what happens when $r=0$: the value of $\theta$ is undefined at the origin, so it does not seem clear whether or not the equation $r=\cos\theta+\sin\theta$ is satisfied.
However, you can look at it this way: the polar curve of equation $r={\color{grey}{(\cos\theta+\sin\theta)}}$ is the set of all points $({\color{grey}{(\cos\theta+\sin\theta)}}\cos\theta,{\color{grey}{(\cos\theta+\sin\theta)}}\sin\theta)$ when $\theta$ runs over $[0,2\pi]$, and this includes the point $\theta=\frac{3\pi}{4}$ where... $r=0$.
From this point of view, one can argue that the equation $r=\cos\theta+\sin\theta$ is satisfied when $r=0$ as well.