In trigonometry, how can we find the sine/cosine of an angle larger than 90 degrees, if sine and cosine only work for right triangles? [duplicate]

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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?

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

• 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. – quasi Jun 25 '18 at 5:23