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I recently learned about the trigonometric functions of angles greater than $90$ degrees, and I'm having a hard time understanding the concept.

We worked with a diagram like this:

enter image description here

My teacher proceeded to tell the class that the sine of angle $\theta$ was equal to the ratio between the opposite side ($y$) and the hypotenuse ($r$).

I don't quite understand this logic. I'm familiar with trigonometric functions, but as I remember them, they applied to angles inside a right-triangle, not an angle outside of it. This seems counter-intuitive to me.

Shouldn't the sine ratio (or cosine, for that matter) represent the angle $180 - \theta$ located inside the right triangle, rather than outside of it? An explanation would be greatly appreciated, as this is a key part of our current unit.

idriskameni
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Kman3
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When trigonometric functions like sine and cosine are applied to situations where we're dealing with angles that are greater than or equal to 90 degrees, the logic based purely on the right triangle definition of trigonometric functions as we know it breaks because in elementary trigonometry the sum of the angles in a right triangle (or any other triangle, for that matter) can't be greater than $180^\circ$. That's obvious.

Let's say you have a 135-degree angle ($90^\circ+45^\circ=135^\circ$), as in your picture. Basically, you still, kind of, use the right triangle idea, but you now must take into account the fact that the legs of the triangle can have negative values because the whole idea that the angle can be greater than or equal to 90 degrees comes from the unit circle which we work with in the Cartesian coordinate system. Therefore, when $x$ is negative, the length of the corresponding leg has a negative value. When $y$ is negative, the length of the corresponding leg is going to be negative as well. Does this make sense to you?

If you find my explanation confusing, here's a more-to-the-point version: forget about the right triangle definition for angles greater than or equal to 90 degrees. It doesn't really work. Now you work in the Cartesian coordinate system where the hypotenuse equals $1$, $\cos\theta$ is the x-coordinate and $\sin\theta$ is the y-coordinate.

Michael Rybkin
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  • Thanks for your answer; I'm still a little confused. It's only because we're using a right triangle to find the sine of an angle that's outside the triangle. I would understand if $\sin$ $(180-\theta)$ was equal to $\frac yr$, but not $\theta$. It's not even in the triangle! Am I too focused on this particular way of finding sine? – Kman3 Feb 05 '19 at 22:51
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    As I said, the right triangle definition for angles grater than or equal to 90 degrees is no longer valid. It does not work. Period. For angles greater than or equal to 90 degrees, we use the unit circle concept. You attempt to understand whether the angle is inside or outside the triangle is a wild goose chase. – Michael Rybkin Feb 05 '19 at 22:53
  • So is there a better explanation as to why $\sin \theta$ = $\sin (180-\theta)$? My teacher proved it using the method I was telling you about. – Kman3 Feb 05 '19 at 22:55
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    That statement is true because $\sin\theta$ has the same y-coordinate as $\sin(180-\theta)$. I advise you to begin studying the unit circle concept. – Michael Rybkin Feb 05 '19 at 22:59
  • Oh! I see. Since the y coordinates and the hypotenuses are the same, the sine value must be the same! So how come we draw a right triangle underneath the line in the first place? It makes me feel like I have to use the right-angle triangle definition. – Kman3 Feb 05 '19 at 23:02
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    You didn't have to draw that triangle. That triangle is imaginary. Do you see the y written next to its vertical leg? That leg can be shifted to the right to make it align with the y-axis. That $y$ just shows you how far away you are in the vertical direction from the origin. As I said, in order to start understanding all this, you need to come to grips with the unit circle idea. – Michael Rybkin Feb 05 '19 at 23:10
  • Thanks for your help. I'll suggest it to my teacher. – Kman3 Feb 05 '19 at 23:13