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We know that $\sin(120^\circ)=\frac{\sqrt{3}}{2}$. But, Why is this true? Let $ABC$ be a triangle, and say $\angle{ACB}=120^\circ=\theta$

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In a right triangle, we can't a construct an angle which is greater than $90^\circ$. Then, how does $\sin(120^\circ)$ or $\sin(150^\circ)$ exist? Also, I am a middle school student.

CiaPan
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Math12
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  • I have a 2nd question too. – Math12 Feb 06 '25 at 13:39
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    The definition by "unit circle" avoids this kind of confusion, refer to for example to https://math.stackexchange.com/q/2736446 and https://math.stackexchange.com/q/2661474 – user Feb 06 '25 at 13:41
  • How do you define sine? – zkutch Feb 06 '25 at 13:47
  • Opposite/hypotenuse – Math12 Feb 06 '25 at 13:48
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    But of course that definition only works for angles between $0$ and $90$. As you know, it is useful to have the trig functions defined more generally than that, for which we typically use coordinates on the unit circle. That coincides with your definition for angles in the confined range, but it makes sense for more general angles as well. – lulu Feb 06 '25 at 13:50
  • If you limit yourself to considering only right triangles, then of course the sine will be defined only on the corresponding interval. But if you want to define the sine on the whole line, what do you do? – zkutch Feb 06 '25 at 13:52
  • I have no idea how you can relate sin to unit circle. – Math12 Feb 06 '25 at 13:54
  • Do you realize that you cannot ask several questions in a single post? – Another User Feb 06 '25 at 13:55
  • Unit circle definition from https://en.wikipedia.org/wiki/Sine_and_cosine – zkutch Feb 06 '25 at 13:55
  • @Another User I have edited the post and it includes only 1 question. – Math12 Feb 06 '25 at 13:57
  • $\sin(120°)=\cos(30°)$, do you accept/understand $\cos(30°)$ being $\frac{\sqrt{3}}{2}$? – Dominique Feb 06 '25 at 14:27
  • Strictly speaking, you mean a sine of $120$ degrees, not just $120$ (which would mean $120$ radians). – CiaPan Feb 06 '25 at 14:28

2 Answers2

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This can be understood by drawing a unit circle

$ x^2 + y^2 = 1 $

In this unit circle every point corresponds to a line drawn from origin making an angle with the x axisrepresentation of a point in unit circle

This can also be proved by power series for sine which is $ \sin x = x - \frac{x^3} {3!} + \frac{x^5} {5!} $ .... By putting $x = 120^\circ$ or $\frac {2π} {3}$ radians

Edit : my answer does not explain why $\sin 120^\circ$ = $ \frac {√3} {2} $ but rather explains that how trigonometric functions for obtuse angle angles exist by representing them in a unit circle.

Ak3.14
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Ved Patel
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    They stated in their question that they're in middle school. I think it's safe to assume that they haven't learned about power series yet. – teadawg1337 Feb 06 '25 at 14:03
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    The unit circle solution should still be good and has tge added benefit of showing that the sines of supplementary angles are equal. – Oscar Lanzi Feb 06 '25 at 14:10
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As noticed we may refer to the more general definition by unit circle in order to avoid this kind of confusion.

As an alternative we can use that

$$\sin (2\theta)=2\cos \theta \sin \theta \implies \sin(120^\circ)=\sin(2\cdot 60^\circ)=2\cos (60^\circ) \sin (60^\circ) =2\cdot\frac{1}{2}\cdot\frac{\sqrt{3}}{2}= \frac{\sqrt{3}}{2}$$

user
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