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$ \frac{a\sin A+b\sin B+c\sin C}{a\cos A+b\cos B+c\cos C}=R\left(\frac{a^2+b^2+c^2}{abc}\right) $

This is what I have so far:

I know that $A + B + C = 180^\circ$, so $C = 180^\circ - (A+B)$. Plugging this in, I get that $\sin C = \sin(A+B)$ and $\cos C = -\cos(A+B)$. When I plug this back into the equation, I get:

$\frac{a\sin A + b\sin B + c \sin(A+B)}{a \cos A + b \cos B - c \cos (A+B)}$.

Expanding out $c \sin(A+B)$ and $c \cos(A+B)$ gives me:

$\frac{a\sin A + b\sin B + c \cos A \cos B - c \sin A \sin B}{a \cos A + b \cos B - c \cos A \cos B + c \sin A \sin B}$.

If I use the Extended Law of Sines, then I get: $ a = 2R * \sin A$, $b = 2R * \sin B$, and $c = 2R * \sin C$, and plugging that in gives me:

$\frac{2R*\sin^{2} A + 2R * \sin^{2} B + 2R \sin C \cos A \cos B - 2R \sin C \sin A \sin B}{2R \sin A \cos A + 2R \sin B \cos B - 2R \sin C \cos A \cos B + 2R \sin C \sin A \sin B}$.

I can factor out all of the $2R's$ to get:

$\frac{\sin^{2}A + \sin^{2} B + \sin C \cos A \cos B - \sin C \sin A \sin B)}{\sin A \cos A + \sin B \cos B - \sin C \cos A \cos B + \sin C \sin A \sin B}$.

Now I'm stuck. What do I do next, to get the end result of $R (\frac{a^2+b^2+c^2}{abc})$?

jimjim
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Mathy Person
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1 Answers1

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Your use of Extended Law of Sines is correct. Get rid of all the trigo ratios in the numerator.

For the denominator, try to prove that $$a\cos A+b\cos B+c\cos C=R(\sin(2A)+\sin(2B)+\sin(2C))=4\sin A \sin B \sin C$$

Then, use the Extended Law of Sines again! :D

meta_warrior
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  • Would I plug that into the last long fraction that I have in my original post? – Mathy Person Jan 13 '15 at 01:43
  • No start again. For the numerator, you should have $\frac{a^2+b^2+c^2}{2R}$ – meta_warrior Jan 13 '15 at 01:44
  • Would I simplify the numerator by using this: $a=2R∗\sin A$, $b=2R∗\sin B, and $c=2R∗\sin C$? How did you get $\frac{a^2+b^2+c^2}{2R}$? – Mathy Person Jan 13 '15 at 02:01
  • @MathyPerson $\sin A = \frac{a}{2R}, \sin B = \frac{b}{2R}, \sin C = \frac{c}{2R}$, sub in... – meta_warrior Jan 13 '15 at 02:24
  • Oh, I see now. I was thinking only of $a, b, c$ instead of $\sin A, \sin B, \sin C$. – Mathy Person Jan 13 '15 at 02:37
  • $R(\sin(2A)+\sin(2B)+\sin(2C)) = R(2\sin A \cos A + 2 \sin B \cos B + 2 \sin C \cos C)$. I don't see how it can be related to $a \cos A + b \cos B + c \cos C$ or $4 \sin A \sin B \sin C$. Can you provide an additional hint? – Mathy Person Jan 13 '15 at 04:12
  • Use $$\begin{align} \sin(2A) + \sin(2B) & = 2 \sin(A+B) \cos(A-B)\ & = 2 \sin(\pi-(A+B)) \cos(A-B)\ & = 2 \sin(C) \cos(A-B) \end{align}$$ – meta_warrior Jan 13 '15 at 09:52