When working in Quadrant I, things make sense to me. Angles are bound between 0 and 90 degrees and we can talk about sin, cos, and tan in terms of ratios of sides of triangles.

But when we increase the angle and start moving into other quadrants I no longer understand how you're supposed to interpret things. I'll see someone draw a triangle in Quadrant III like normal but then somehow it's still reconciling with this huge angle that's sprawling all the way over from 0 on the unit circle.

How am I supposed to interpret these trigonometric functions in other quadrants?

  • $\begingroup$ Said angle forms a right triangle with the x axis still - in quadrant 2, draw a straight line down to the x axis from the terminal side, in quadrant 3 and quadrant 4, a straight line up. $\endgroup$ – TreFox Feb 13 '18 at 17:02
  • $\begingroup$ @TreFox But then wouldn't the "real" angle be whatever's inside the triangle rather than the long angle that's stretching from 0? $\endgroup$ – user525966 Feb 13 '18 at 17:07
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    $\begingroup$ In the second quadrant, the triangle can be drawn with its hypotenuse as its upper right, or as its lower left. Now think alternate interior angles $\endgroup$ – imranfat Feb 13 '18 at 17:11
  • $\begingroup$ @user525966 That's right. Sine (or any trig function) of 140 degrees is going to be the same as sine of 40, since 140 degrees forms what we call a 40 degree "reference angle" with the x axis. $\endgroup$ – TreFox Feb 13 '18 at 17:15

Trigonometric functions have clear definition and geometrical meaning which work good for angles in all quadrants beyond the definition in terms of ratios of sides of triangles which can be viewed as an application.


  • $\cos x$ and $\sin x$ are the coordinates of the point M on the trigonometric circle
  • $\tan x$ is the y coordinate of the intersection between the vertical line from (1,0) and the line OM
  • ...

and so on.

enter image description here

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    $\begingroup$ I'm considering teaching the functions as the coordinate on the unit circle $(\cos \theta, \sin \theta) $ rather than the usual right triangle approach. $\endgroup$ – Karl Feb 13 '18 at 17:22
  • $\begingroup$ @Karl Peronally I prefer this approach for a general introduction in term of functions $\endgroup$ – user Feb 13 '18 at 17:25
  • $\begingroup$ Me too actually. I've always been worried that the students then can't then solve the usual right angle problems and all I've done is solve one problem to create another. $\endgroup$ – Karl Feb 13 '18 at 17:28
  • $\begingroup$ While I'm sure this is fine it is hard to understand when everything in one diagram, if this were a teaching aid I'd recommend splitting it up into multiple diagrams to show each one separately, because all at once is overwhelming for a newbie $\endgroup$ – user525966 Feb 13 '18 at 17:29
  • $\begingroup$ Also I am sure these definitions change depending on quadrant, for example the tangent line would not be from (1, 0) outside of quadrants I and IV I assume, but rather (-1, 0)? $\endgroup$ – user525966 Feb 13 '18 at 17:36

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