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So i have $A + B + C = \pi$

$$\frac{A}{2} + \frac {B}{2} + \frac{C}{2} = \frac{\pi}{2}$$

$$4\cos\left(\frac{-B-C + \pi}{2}\right)\cos\left(\frac{-A -C + \pi}{2}\right)\cdots$$ And I doubt this leads to anywhere.

So then I tried, $\sin\left(\frac{-B-C + \pi}{2}\right)\cdots$ and this didn't go anywhere either. I don't know what to try, and I've seen other people's solutions and they do something like: $\sin(C) = \sin(A + B)$, $\cos(C/2) = \sin(\frac{A + B}{2})$ but i don't see where they got this part from. Other people use Euler's formula or whatever but I haven't learned that yet so I can't use it.

Michael Hardy
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Kat
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  • +1. But on the other hand, I know I've answered this one here before. There's an argument from old-fashioned trigonometric identities, but I posted an argument showing that either the right side or the left can be viewed as the area of a certain triangle; hence they must be equal. – Michael Hardy Dec 15 '13 at 21:14
  • @MichaelHardy: Oh, sorry, I did google the problem but I did not find any similar problems on math.stackexchange. – Kat Dec 15 '13 at 21:18
  • Maybe I haven't answered EXACTLY this one..... – Michael Hardy Dec 15 '13 at 21:20
  • ....and maybe this version isn't quite right..... – Michael Hardy Dec 15 '13 at 21:20
  • Duplicate: Proving $\sin A + \sin B + \sin C = 4 \cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}$. – Blue Dec 15 '13 at 21:20
  • In the second post, second-third line of @Blue's link, I don't get how he combined the two? – Kat Dec 15 '13 at 21:25
  • If you need clarification of an answer on a different page, then you should ask for it in a comment below that answer. – Blue Dec 15 '13 at 21:27
  • I've deleted my answer to this version. I had some bad typos, e.g. $\sin A$ and $\sin\dfrac A2$ were confused with each other. See the linked earlier question. – Michael Hardy Dec 15 '13 at 21:30
  • @Kat : You might post separately your other question in a comment under my now-deleted answer. If $A$ is one angle of a triangle inscribed in a circle of unit DIAMETER, then when is $\sin A$ the length of the side opposite the angle $A$? Essentially this is a version of the law of sines. Sometimes that is stated by saying $\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}$, where (capital) $A,B,C$ are the angles and (lower-case) $a,b,c$ are the sides. That version stops short of explicitly saying that that number is the diameter of the circumscribed circle. – Michael Hardy Dec 15 '13 at 21:34
  • I've up-voted the question and also voted to close it as a duplicated. – Michael Hardy Dec 15 '13 at 21:41

1 Answers1

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$$\sin(A) + \sin(B) + \sin(C)= 2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A - B}{2}\right) + 2\sin\left(\frac{C}{2}\right)\cos\left(\frac{C}{2}\right) $$

$$\sin(A) + \sin(B) + \sin(C)= 2\cos\left(\frac{C}{2}\right)\cos\left(\frac{A - B}{2}\right) + 2\sin\left(\frac{C}{2}\right)\cos\left(\frac{C}{2}\right)$$

$$\sin(A) + \sin(B) + \sin(C)= 2\cos\left(\frac{C}{2}\right)\cos\left(\frac{A - B}{2}\right) + \cos\left(\frac{A + B}{2}\right)$$

$$\sin(A) + \sin(B) + \sin(C)= 4\cos\left(\frac{A}{2}\right)\cos\left(\frac{B}{2}\right)\cos\left(\frac{C}{2}\right)$$

DeepSea
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