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In trigonometry, how can there be angle thetas larger than 90 degrees, if sine and cosine only work for right triangles? I realize what we do is that we take the sine or cosine of the angle complementary to it. But then, how does that give the ratio between the opposite/adjacent and hypotenuse of the angle larger than 90 degrees? It doesn't! It only gives the ratio between the opposite/adjacent and hypotenuse of the angle complementary to it! Can someone please explain?

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marked as duplicate by N. F. Taussig, Community Jun 26 '18 at 3:42

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    $\begingroup$ It's an extension of the sine and cosine functions using the properties of the unit circle. Yes, it's true that right triangles' angles don't go past 90, but that still doesn't mean that the functions for those angles can work past 90. $\endgroup$ – Christopher Marley Jun 25 '18 at 5:01
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    $\begingroup$ I think sine and cosine is defined particularly on the ratio of the lengths of right triangles and not on non-right triangles. Regarding taking the sine and cosine of obtuse angles, we are not really taking the sine and cosine of the obtuse angle but it is defined to be that of the complement of the obtuse angle. $\endgroup$ – TheLast Cipher Jun 25 '18 at 5:19
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    $\begingroup$ The right-triangle-based definition of the trigonometric functions is defined only for angles $\theta$ with radian measure $0 < \theta < {\large{\frac{\pi}{2}}}$. The more general definition, which works for all values of $\theta$, is based on the coordinates of the point $(x,y)$ where the terminal ray of the angle $\theta$ meets the standard unit circle. For angles $\theta$ such that $0 < \theta < {\large{\frac{\pi}{2}}}$, the general definition yields the same results as the right-triangle-based definition, so for those angles, either definition is applicable. $\endgroup$ – quasi Jun 25 '18 at 5:23
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    $\begingroup$ I don't think it's "going past definitions". What I meant was, it is defined that taking the sine or cosine of any angle past 90 degrees meant you should take the sine or cosine of its complement. It is by definition. The reason textbooks asks you to consider the complement is due to it being defined that way, no more no less. With regards to the motivation, on why it's defined that way, I am not sure why. $\endgroup$ – TheLast Cipher Jun 25 '18 at 5:48
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    $\begingroup$ The trigonometric functions were originally defined on the circle. The right triangle definitions were introduced later. They are a special case of the general definition. $\endgroup$ – N. F. Taussig Jun 25 '18 at 8:40