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I was trying to think about why in instances like $\cos(0)$ and $\sin(90)$ that the triangle formed in the unit circle becomes degenerate-that is, it collapses into a line.

I am aware that this sounds like a previous question I asked: How would you describe a triangle with a sin 90 based on the definition of sine? However, while that question was asking how to describe a triangle with $\sin(90)$, this question is about explaining why it is that way. Specifically, I thought of a reason to explain this easily using prior knowledge and previous answers to similar questions.

It was helpful to me to visualize coordinates in the unit circle as just being representations of the slopes of lines. The $y$-coordinate would be the rise and the $x$-coordinate would be the run of a given line in the circle. Therefore, as the slope of the line becomes steeper, the rise increases and the run decreases, and likewise, the opposite happens when the line becomes less steep.

enter image description here

At sin(90), the line becomes vertical. You cannot have a triangle with two right angles (this just ends up forming a line), as brought up in Why is the cosine of a right angle, 90 degrees, equal to zero?.

Knowing that at $\sin(90)$, the line formed is a vertical line because it is perpendicular to the $x$-axis, I also started thinking about properties of vertical lines and remembered that vertical lines have an undefined slope-that is, rise but not run.

Combining the two definitions, I concluded that the reason for vertical lines having an undefined slope is because trying to construct a right triangle to find the slope results in a degenerate triangle (that is, there cannot be both rise and run).

As this triangle cannot be constructed from vertical lines, vertical lines evidently have a rise but not a run, so by Pythagorean Theorem $0^2+y^2=c^2$, where $y$=opposite (rise) and $c$=hypotenuse, and $y=c$. Therefore, $\sin(90)=1$, and this sort of thing can also be applied to other angles like $0$ and $180$. I just used $90$ degrees for this example because I feel that it is easier to visualize.

What I was wondering is, am I correct to assume that, for the purpose of explaining this to myself, that there is a correlation between the slope of vertical (or horizontal) lines and the formation of degenerate triangles? In other words, does the fact that triangles cannot have angle measures of $90$-$90$-$0$ mean that vertical lines have an undefined slope (lack of change in the $x$-coordinate or "adjacent" side) and that horizontal lines have a $0$ slope (lack of change in the $y$-coordinate or "opposite" side)? Or am I way off? I was trying to find a different way to explain the concept of $\sin(90)$ and $\cos(0)$ and other angles to myself, especially as someone who enjoys algebra and graphs and can visualize things this way.

Thanks and sorry if this question is vague. Please let me know if I can improve it.

Note: When I say a triangle I am referring to one formed by the rise and run of a line, with the line as the hypotenuse.

enter image description here

As this sort of triangle cannot be formed for vertical or horizontal lines (like those forming $\sin(90)$ or $\cos(0)$), these lines do not have both rise and run, meaning that either the adjacent or opposite side equals the hypotenuse.

Bumblebee
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user386598
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    The use of right triangles to define sine and cosine seems to be a bit of a pedagogical fluke. The original trig table by Ptolemy didn't refer to right triangles and definitions of sine and cosine in higher math don't use right triangles. But you're correct that vertical lines have undefined slope and horizontal lines have zero slope, all due to either the run or rise being zero. – David K Jun 11 '25 at 03:36
  • The "SOH-CAH-TOA" rules of the trigonometric functions only apply for angles greater than 0 degrees and less than 90 degrees. Which is why these functions are not defined by "$\frac {\text{opposite}}{\text{adjacent}}$" rules. It appears that you are familiar with the unit circle and these definitions, so I would suggest it is time to let go of the thing that worked in a special case an emprace the more advanced definition. – user317176 Jun 11 '25 at 03:45
  • Looking at some of your other posts, I see that you are trying to explain the unit circle as a consequence of "SOH-CAH-TOA." Your reasoning is backwards. The unit circle is the definition, and "SOH-CAH-TOA" is the consequence. This isn't your fault; it was probably taught to you in this way. – user317176 Jun 11 '25 at 03:54
  • I was sort of looking at the definition of sine and cosine based off of SOH CAH TOA, where sin(90) is a limit of sorts because you can see the run decreasing and the rise increasing until you can no longer make a right triangle. Is it incorrect to use opposite/hypotenuse=1 for sin(90) because opposite=hypotenuse in this scenario? If so, is my reasoning about degenerate triangles and slope still somewhat correct? – user386598 Jun 11 '25 at 04:10
  • Also, the way I was taught is that the side length ratios in 45-45-90 and 30-60-90 triangles can be used to find the x and y coordinates (cosine and sine, respectively) in the unit circle. I do not know any other ways to derive these coordinates so I always assumed you start with triangles and then extend the definition. – user386598 Jun 11 '25 at 04:34
  • I was taught the unit circle. Then, special triangles could be inscribed inside the unit circle. Some identities fall out of the unit-circle definition, e.g. $\sin (-x) = -\sin x, \sin x = \cos (\frac {pi}{2} - x), etc.$ Eventually, we learn angle addition identities, and we can find the points every 15 degrees. By the way, the 36-72-72 isosceles triangle is special, too, though not usually taught in introductory classes. – user317176 Jun 11 '25 at 04:50
  • Without learning of special right triangles first, how were you taught what the coordinates of the unit circle were (e.g. the sine of 30 is 1/2 because this is the opposite side or y-coordinate in a 30-60-90 triangle with hypotenuse of 1)? – user386598 Jun 11 '25 at 14:31

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