# Logic of Step to Convert Rectangular Equation of Circle to Parametric

I recently learned in math class how to convert the rectangular equation of a circle into a parametric equation:

$$x^2+y^2=1$$

$$\cos^2(t)+\sin^2(t)=1$$

$$x^2+y^2=\cos^2(t)+\sin^2(t)$$

$$x^2=\cos^2(t)$$

$$y^2=\sin^2(t)$$

$$x=\cos(t)$$

$$y=\sin(t)$$

I don't understand the logic behind going from $$x^2+y^2=\cos^2(t)+\sin^2(t)$$ to $$x^2=\cos^2(t)$$ and $$y^2=\sin^2(t)$$. My teacher explained through an analogy:

$$4+3=7$$

$$(1+3)+(1+2)=7$$

Therefore, $$1+3=4$$ and $$1+2=3$$.

My teacher also said that I would know to pair $$x^2$$ with $$\cos^2(t)$$ because cosine is related to x in the unit circle and that $$y^2$$ pairs with $$\sin^2(t)$$ for the same reason.

My problem with this explanation is that I can also write these equations:

$$4+3=7$$

$$(1+1)+(2+3)=7$$

In this case, $$4\neq1+1$$ and $$3\neq2+3$$, so the analogy doesn't seem to apply to all cases.

Additionally, the graph of the parametric function is the same (except for direction) when $$x$$ is paired with $$\sin$$ or $$\cos$$, so I don't understand why that matters. The reasoning also doesn't make sense to me because I don't see how the unit circle definition of $$\sin$$ and $$cos$$ can decide how they pair with two variables.

Is the logic behind this step just knowing that $$x^2=\cos^2(t)$$ and $$y^2=\sin^2(t)$$ can be simplified to $$x^2+y^2=\cos^2(t)+\sin^2(t)$$ from the steps to convert a parametric circle equation to rectangular? Or is there a way to do this step without having to already know that it's allowed?

• You are right, the teacher is wrong for presenting that bogus transformation, Also in error is taking the square roots of both sides with disregard to whether the root is positive or negative. – William Elliot Apr 9 at 2:04
• An equally valid parameterization of the circle is x = sin t, y = cos t. Can you transfer to a class with a competent teacher? – William Elliot Apr 9 at 2:11

## 1 Answer

Let
$$x = sin(t + a),$$
$$y = cos(t + a).$$
Clearly $$x^2 + y^2 = 1$$.