If $A+B+C=180$, then $\frac{\sin2A+\sin2B+\sin2C}{\cos A+\cos B+\cos C-1}=8\cos\frac A2 \cos\frac B2 \cos\frac C2$

Question

Then $2A+2B+2C =360$

So $$\sin 2C=-\sin(2A+2B)$$ Putting that in the equation $$\frac{2\sin(A+B)\sin(A-B)-2\sin(A+B)\cos(A+B)}{\cos A+\cos B-\cos(A+B)+1}$$ $$\frac{2\sin(A+B)[\sin(A-B)-\cos(A+B)]}{\cos A+\cos B-\cos(A+B)+1}$$ I don’t know how to proceed. Please help me continue

Answer 1

We will proceed by simplifying the numerator and denominator separately and repeatedly use some well-known formulae.

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For simplifying the numerator: \begin{align} \sin 2A + \sin 2B + \sin 2C & = 2\sin(A+B)\cos(A-B) + 2\sin C \cos C \\ & = 2\sin(\pi - C) \cos(A - B) + 2\sin C \cos C \\ & = 2\sin C [\cos(A - B) + \cos(\pi - (A + B))] \\ & = 2\sin C[\cos(A - B) - \cos(A + B)] \\ & = 4\sin A \sin B \sin C\end{align}

For simplifying the denominator: \begin{align} \cos A +\cos B + \cos C - 1 & = 2\cos(\frac{A +B}{2}) \cos(\frac{A-B}{2}) - 2 \sin^2 \frac{C}{2} \\ & = 2\cos(\frac{\pi}{2} - \frac{C}{2})\cos(\frac{A-B}{2}) - 2\sin^2\frac{C}{2} \\ & = 2\sin \frac{C}{2}[\cos(\frac{A-B}{2}) - \sin \frac{C}{2}] \\ & =2\sin \frac{C}{2}[\cos(\frac{A-B}{2}) - \cos(\frac{A+B}{2})] \\ & =4\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}\end{align}

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Can you take it from here?

Answer 2

You will need some formulas $$\sin(2\alpha)+\sin(2\beta)+\sin(2\gamma)=4\sin(\alpha)\sin(\beta)\sin(\gamma)$$ $$\sin(\alpha)\sin(\beta)\sin(\gamma)=\frac{abc}{8R^3}$$ $$A=\frac{abc}{4R}$$ $$\cos(\alpha)+\cos(\beta)+\cos(\gamma)-1=4\sin(\alpha/2)\sin(\beta/2)\sin(\gamma/2)$$ $$\sin(\alpha/2)\sin(\beta/2)\sin(\gamma/2)=\frac{(s-a)(s-b)(s-c)}{abc}$$ $$s=\frac{a+b+c}{2}$$ $$A=\sqrt{s(s-a)(s-b)(s-c)}$$

Answer 3

$\begin{align} \sin(2a) + \sin(2b) + \sin(2c) &= 2 \sin(a+b) \cos(a-b) - \sin(2a+2b) \cr &= 2\sin(a+b)\;\{ \cos(a-b) - \cos(a+b) \} \cr &= 2\sin(c)\;\{2 sin(a) sin(b)\} \cr &= 4\sin(a)\sin(b)\sin(c) \cr {\sin(2a) + \sin(2b) + \sin(2c) \over 8 \cos{a\over2}\cos{b\over2}\cos{c\over2}} &= 4 \sin{a\over2} \sin{b\over2} \sin{c\over2} \cr &= 4 \sin{a\over2} \sin{b\over2} \cos{a+b\over2} \cr &= 4 \sin{a\over2} \sin{b\over2} \left(\cos{a\over2}\cos{b\over2} - \sin{a\over2}\sin{b\over2} \right)\cr &= \sin(a)\sin(b) - 4(\sin^2{a\over2})(\sin^2{b\over2}) \cr &= \sin(a)\sin(b) - (1 - \cos(a))(1 - \cos(b)) \cr &= \sin(a)\sin(b) - 1 + \cos(a) + \cos(b) - \cos(a)\cos(b) \cr &= \cos(a) + \cos(b) - \cos(a+b) - 1 \cr &= \cos(a) + \cos(b) + \cos(c) - 1 \cr\cr {\sin(2a) + \sin(2b) + \sin(2c) \over \cos(a) + \cos(b) + \cos(c) - 1 } &= 8 \cos{a\over2}\cos{b\over2}\cos{c\over2} \end{align}$

Answer 4

In a triangle, $$R^2(\sin(2A)+\sin(2B)+\sin(2C))=2\Delta$$ is a consequence of $[ABC]=[OAB]+[OBC]+[OCA]$ with $O$ being the circumcenter.
Carnot's theorem gives $R(\cos A+\cos B+\cos C-1) = r$, hence by $abc=4R\Delta$ the claim is equivalent to

$$ (a+b+c)\Delta = 2abc\cos\tfrac{A}{2}\cos\tfrac{B}{2}\cos\tfrac{C}{2}$$ which is simple to prove by squaring both sides, then exploiting Heron's formula and $$ \cos^2\tfrac{A}{2}=\tfrac{1}{2}\left(1+\cos A\right)=\frac{1}{2}\left(\frac{b^2+c^2-a^2}{2bc}+1\right)=\frac{(a+b+c)(-a+b+c)}{4bc}. $$