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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.

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  • $\begingroup$ Actually, the sine and cosine can also be in other triangles, not just right triangles. $\endgroup$ – Math Lover Feb 5 at 21:54
  • $\begingroup$ When you think of the sine of an angle $\theta$ as the height of a point with radius $1$ and angle $\theta$ it all makes sense. $\endgroup$ – Garmekain Feb 5 at 21:54
  • $\begingroup$ Do you want the sine rule to hold in obtuse triangles as well as in acute triangles? $\endgroup$ – Lord Shark the Unknown Feb 5 at 21:56
  • $\begingroup$ Have a look at some of the related questions in the handy list at right, particularly this one and this one $\endgroup$ – amd Feb 5 at 22:08
  • $\begingroup$ On the figure, $x$ is negative, and so is the cosine. $\endgroup$ – Yves Daoust Feb 5 at 22:18
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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.

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  • $\begingroup$ 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? $\endgroup$ – Kman3 Feb 5 at 22:51
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    $\begingroup$ 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. $\endgroup$ – Michael Rybkin Feb 5 at 22:53
  • $\begingroup$ 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. $\endgroup$ – Kman3 Feb 5 at 22:55
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    $\begingroup$ 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. $\endgroup$ – Michael Rybkin Feb 5 at 22:59
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    $\begingroup$ 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. $\endgroup$ – Michael Rybkin Feb 5 at 23:10

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