5

There is the question (example) of how the power series definitions of sine and cosine relate to their unit-circle definitions. In many answers (example), the first step is usually something like this:

Show that for all $\theta \in \mathbb R$, we have $\sin^2\theta+\cos^2\theta=1$.

Hence conclude that the set $S=\{(\cos\theta,\sin\theta)\in\mathbb{R}^2|\theta\in\mathbb{R}\}$ describes the unit circle.

I have a doubt about the above argument, which seems to say the following:

Let $f:\mathbb R \rightarrow \mathbb R$ and $g:\mathbb R \rightarrow \mathbb R$ be functions.

Show that for all $\theta \in \mathbb R$, we have $[f(\theta)]^2+[g(\theta)]^2=1$.

Hence conclude that $S=\{(f(\theta),g(\theta))\in\mathbb{R}^2|\theta\in\mathbb{R}\}$ describes the unit circle.

But this conclusion does not follow, because it could for example be that $f$ and $g$ are defined by $f(\theta)=1$ and $g(\theta)=0$ for all $\theta \in \mathbb R$. In which case the set $S$ would not describe the unit circle, but would instead just be a single point: $S=\{(1,0)\}$.

Am I perhaps missing or misunderstanding something about the argument here?

Community
  • 1
  • "Hence conclude ..." typically means that there's more work to do. – Blue Aug 27 '19 at 07:32
  • 1
    Showing $\sin^2(x)+\cos^2(x)=1$ only proves that the points $(\sin(x), \cos(x))$ lie on the unit circle. To show that all points on the unit circle are described like that requires a little more work. – quarague Aug 27 '19 at 07:43
  • @Blue is right. Perhaps they could have made it clearer by writing "Hence show". – J.G. Aug 27 '19 at 08:57

3 Answers3

2

I don't think the author is trying to imply that every two functions $f,g$ which satisfy $|f|^2+|g|^2=1$ automatically also describe the unit circle.

However, $|f|^2+|g|^2=1$ is a necessary condition for $f,g$ to describe the unit circle. Besides this, some other conditions must also be met (which your constant functions do not meet), and showing those other conditions is the second part of the task.

Awkwardly worded, I would say, but I see their point.

5xum
  • 123,496
  • 6
  • 128
  • 204
0

No your conclusion is a much stronger claim and actually false.

Take any point $(x,y)$ on the unit circle and construct a right triangle, with base given by $x$, height given by $y$, hypotenuse =$1$ and the angle=$\theta$. Then $x=\cos(\theta)$ and $y = \sin(\theta)$ by trigonometry, now by the identity $\cos(\theta)^2+\sin(\theta)^2 =1$ the parameterisation $(\cos(\theta),\sin(\theta))$ holds for all values of $\theta$

Michael
  • 11
0

Take any point $(x,y)$ on the circle, then the system of equations

$$\begin{cases}x=\cos t,\\y=\sin t\end{cases}$$

has the solution

$$t=k\pi+\begin{cases}x=0\to\dfrac\pi2,\\x\ne0\to\arctan\left(\dfrac yx\right)\end{cases}$$ where $k=0$ or $1$.

As this solution always exists, the whole circle is covered.