Show that in any triangle, we have $\frac{a\sin A+b\sin B+c\sin C}{a\cos A+b\cos B+c\cos C}=R\left(\frac{a^2+b^2+c^2}{abc}\right),$

Question

Show that in any triangle, we have $$\frac{a\sin A+b\sin B+c\sin C}{a\cos A+b\cos B+c\cos C}=R\left(\frac{a^2+b^2+c^2}{abc}\right),$$ where $R$ is the circumradius of the triangle.

Here is my work:

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

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

If we expand out $c\sin (A+B)$ and $c\cos (A+B)$, we get

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

Using the Extended Law of Sines, we can use $\sin A=\frac{a}{2R}$, $\sin B=\frac{b}{2R}$, and $\sin C=\frac{c}{2R}$.

How can I continue on?

Answer 1

Since the sine theorem implies: $$\sum_\text{cyc}a\sin A = \frac{1}{2R}\sum_\text{cyc}a^2 \tag{1}$$ we just need to prove: $$ \sum_\text{cyc} a \cos A = \frac{abc}{2R^2}=\frac{2\Delta}{R}\tag 2$$ that is trivial since twice the (signed) area of the triangle made by $B,C$ and the circumcenter $O$ is exactly $aR\cos A$:

Answer 2

Use Law of Sines for the numerator, $\displaystyle\sum_\text{cyc}a\sin A=\dfrac{\sum_\text{cyc} a^2}{2R}$

For the denominator, $\displaystyle\sum_\text{cyc}a\cos A=\sum_\text{cyc}(2R\sin A\cos A)=R\sum_\text{cyc}\sin2A$

Using Prove that $\sin(2A)+\sin(2B)+\sin(2C)=4\sin(A)\sin(B)\sin(C)$ when $A,B,C$ are angles of a triangle ,

$\displaystyle\implies\sum_\text{cyc}a\cos A=R(4\sin A\sin B\sin C)=4R\prod_\text{cyc}\left(\dfrac a{2R}\right)=\cdots$

Can you take it home from here?