0

The Trigonometric Unit Circle is a pretty common thing taught in High-School, yet it doesn't cover some questions I had.

There have been questions in the past about why trigonometric functions can be negative, but none detailing exactly why the centre of the circle has to be on the Origin.

Surely, I can re-draw the same circle alone in the 1st quadrant of a cartesian plane, without overstepping the boundaries. Everything would then simply be positive - yet that would surely break down calculations that yield negative values?

It is very confusing to me why the circle is on the origin, presenting needless problems regarding signs. But Supposing if we do shift the circle, then the values would be different and incorrect.

Any ideas?

Blue
  • 75,673
neel g
  • 113
  • sine and cosine are defined this way. If you move the circle, you'd get something like $\sin(\theta) + 1$ instead. That would have non-negative values, but it's a different function. – A. Thomas Yerger Jan 13 '22 at 19:11
  • 1
    This old answer of mine might provide some insights. – Blue Jan 13 '22 at 19:11
  • 2
    We have $\sin^2 \theta +\cos^2 \theta =1$, hence the parallel with $x^2+y^2=1$. the unit circle centered at the origin. – lulu Jan 13 '22 at 19:12
  • I recommend watching Herb Gross's: Calculus Revisited series, particularly to address your questions here Unit III: Lec 1 - The Circular Functions. I find his explanations excellent, particularly about these and for hyperbolic sine and cosine etc... It is worth pointing out and reminding of the other ways to develop and define the circle functions. $\sin$ and $\cos$ are solutions to the differential equation $f = -f''$. It is convenient to have the geometric/trigonometric definitions and the calculus definitions to refer to the same things here – JMoravitz Jan 13 '22 at 19:27
  • 2
    @lulu we have that formula precisely because the circle is the unit circle. – A. Thomas Yerger Jan 13 '22 at 19:33
  • 1
    I think there is a hint in your question that you find negative values of $\sin$ and $\cos$ problematic because of "needless problems regarding signs". I would like to know more - what sort of problems do you normally encounter that the negative sign makes more complicated? Presumably a lot of problems become less complicated when the circle is centered at the origin, but I would be interested - what sort of problems gets more complicated? –  Jan 13 '22 at 19:33
  • 1
    What’s wrong with negative numbers? – littleO Jan 13 '22 at 19:45
  • Well if the center is off the origin you'd have a lot more problems dealing with the offset. To begin with $\cos^2\theta + \sin^2 \theta \ne 1$ instead if $(a,b)$ is our center of the circle we'd have $(\cos \theta -b)^2 + (\sin\theta -a)^2 = 1$ and $\cos \theta = a+K_c$ and $\sin \theta = b+ K_s$ where $K_c^2+K_s^2 = 1$ corespond to the angle $\theta$. That is in no way easier. But more importantly it begs any question of basic properties with trig is invented to express. – fleablood Jan 13 '22 at 20:15
  • "Surely, I can re-draw the same circle alone in the 1st quadrant of a cartesian plane" If the origin of the circle is $(a,b)$ then your new definition of sin/cos will be $\cos_{new}\theta = a+K_c$ and $\sin_{new}=b + K_s$ for some values of $K_c$ and $K_s$. All work we do will turn out that $a,b$ are utterly irrelevant and $K_c$ and $K_s$ will be all that is important. Want to take a guess as to the values of $K_s$ and $K_c$ are in terms of $\theta$? Well turns out $K_c = \cos_{old} \theta$ and $K_s=\sin_{old}\theta$. This only begs to question for no other purpose than avoiding neg numbers. – fleablood Jan 13 '22 at 20:20
  • 1
    If not the origin, than what principled point would you choose as the center? – David G. Stork Jan 13 '22 at 20:40
  • Any point honestly - my question is simple why the origin is the main choice. Why not something in the 1st qaudrant alone? why is it on centre in the first place? – neel g Jan 14 '22 at 22:49
  • "Any point, honestly" The point is that no other point would be as principled... – ryang Jul 07 '22 at 12:43

1 Answers1

1

Notice that the meaning of negative values in the coordinate system essentially is 'left to' a point (on the x-axis) and 'down to' a point (on the y-axis). So for example $\cos\left(\frac{3\pi}{4}\right)=-\frac{\sqrt{2}}{2}$ means that on a unit circle, regardless where that centre is, the x-coordinate of the point of the triangle on the circle is $\frac{\sqrt{2}}{2}$ units left to the centre of the circle. The only reason why we use the centre as the origin of a unit circle because it allows for easier reading (i.e. the x-coordinate essentially dictates how far away in the horizontal direction it is from the origin, same argument applies for y-coordinate). It does not depend on the location of the unit circle.

Vis37
  • 71