a question on : why sin(x) = sin(180-x)

I was trying to follow a proof from another question in this forum: Why is $\sin(x) = \sin(180^{\circ}-x)$

I was following the geometric proof given by forum poster : egreg.

I could follow his proof for the cosine. But I think the proof for the sine contains an error.

Egreg stated that from the triangle in the circle you get the following: sin(alpha)/a = 2R

BUT when I analyzed the small triangle with angle alpha in it, I get: sin(alpha)/a = 1/2R

It really doesn't affect the proof much at all. But I am wondering if there was an error, or there is something wrong with my own trig analysis.

Hope that the forum guy Egreg can reply since he is the originator of this proof.

Regards,

P

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Thanks for noting; I fixed the error. –  egreg Nov 24 '13 at 19:11
Hi Egreg, thanks for your response. BUT can you make a picture of the point A' (A prime diametrically opposite A), because I am having difficulty understanding how to construct this new triangle to show that in this new triangle we have the angle "180-alpha". Really appreciate it if you can do that. Sincerely P. –  Palu Nov 24 '13 at 20:31
In a quadrilateral inscribed in a circle, the opposite angles always sum to 180°. –  egreg Nov 24 '13 at 22:48
Hi Egreg, Thanks so much! Yes, that is the geometric fact that I was missing: In quadrilateral inscribed in a circle the opposite angles sum to 180 degrees! After I created the quadrilateral with A' diametrically opposite and the angle at A' being alpha prime, the proof became crystal clear!!! Thanks so much once more! Sincerely Palue –  Palu Nov 25 '13 at 3:08

You are correct. Recall the Law of Sines, which states $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2R$. This means $\dfrac{\sin A}{a} = \dfrac{1}{2R}$. This is probably a typo on his part; it doesn't affect the proof at all. The section of the proof you are referring to is just simply a clever proof of the Law of Sines.