1
$\begingroup$

How does the unit circle work for trigonometric ratios of obtuse angles? I know that the x coordinate is $cos (\theta )$ and the y coordinate is $sin (\theta )$. But I understand these in context of only acute angles? I don't understand why the unit circle definition works for other than acute angles? Somebody please provide me some good intuition.

$\endgroup$
  • 1
    $\begingroup$ Your question seems a little odd -- part of the reason for the unit circle definition is that it yields a definition in an obvious way for all angles. I suspect you mean to ask something different: can you try to state your question more precisely? e.g. maybe what you mean by "work" needs clarification. $\endgroup$ – Hurkyl Jan 24 '14 at 16:45
  • $\begingroup$ Note that the coordinates of the circle is given by $(\cos\theta,\sin\theta)$ for acute angles. Try understanding now as, that we extend the acute angle's definition to all the angles by the fact that it gives the coordinates. $\endgroup$ – Sawarnik Apr 16 '14 at 20:03
  • $\begingroup$ I had a similar difficulty, while exploring trig functions. But, when I deeply thought about it, I realised these definitions are indeed correct. Yes, they are DEFINITIONS, not proofs of some kind. This convention is flexible with all the triangles, obtuse, acute, any triangle. And the question is , why did we go beyond $\dfrac{\pi}{2}$, and the answer is just because we can. One more important thing to note is the though the name trigonometry, trig functions are not only about triangles, it has a deeper meaning. $\endgroup$ – codetalker Jun 29 '16 at 15:17
  • $\begingroup$ math.stackexchange.com/questions/136543/… $\endgroup$ – David K yesterday
1
$\begingroup$

The "magnitude" of $\cos \theta, \sin \theta$ of an obtuse angles can be measure in terms of the acute angle its terminal arm forms with the $x$-axis, positive or negative. So an angle of, say, $\theta = \frac 34 \pi$ has its terminal arm pointing in the direction of Quadrant II, and the absolute values of $\cos\theta$, $\;\sin \theta$ are the same as the values of $\theta' = \pi/4$, the angle formed by the terminal arm of $\theta$ with respect to the negative $x$-axis: $\theta' = \pi - \frac{3\pi}{4}$.

The sign of that magnitude can be determined by the quadrant in which the angle's terminating ray is directed. (Recall that $\cos \theta$ corresponds with the $x$ coordinate, $\sin \theta$ with $y$):

Quadrant I: $\cos \theta, \sin \theta \gt 0$

Quadrant II: $\cos \theta \lt 0$, $\;\sin\theta > 0$

Quadrant III: $\cos \theta \lt 0,\;\;\sin\theta \lt 0$.

Quadrant IV: $\cos \theta \gt 0,\;\;\sin\theta \lt 0$.

$\endgroup$
  • $\begingroup$ Your answer does tell me how to find the cosine and sine of all sorts of angles, but I don't understand "why" this "how" works. $\endgroup$ – Shaurya Gupta Jan 24 '14 at 17:02
1
$\begingroup$

This intuition can be found in the history of trig. Looking into my history of math notes (a course mandatory for my degree) I learned that once upon a time there was a man Hipparchus. He is regarded as a ofundaing father of what's now known as trig. Apparently there was a desire at that time to have tables that would convert any point on a unit circle expressed in Polar form, into corresponding Cartesian form. Hipparchus made those tables and assigned "functions" to the idea how every angle (measured from the pos. x-axis) of a point P on the circle would "correspond" to (x,y) coordinates. That's how the modern cosine and sine could be defined. Right triangle geometry is then derived from here. In this realm, the functions cosine and sine are born from the desire to have a "translation" from Polar to Cartesian form. In this light, all quandrants of the unit circle are covered and thus the sign of cosine and sine are understood accordingly.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.