Showing $\cos A\cos B\cos C=\frac{s^2-(2R+r)^2}{4R^2}$ and $\cos A+\cos B+\cos C=1+\frac rR$ in $\triangle ABC$

Question

In a triangle with vertices $A$, $B$, $C$, semiperimeter $s$, inradius $r$ and circumradius $R$, prove that $$\cos A\cos B\cos C=\frac{s^2-(2R+r)^2}{4R^2}$$ and $$\cos A+\cos B+\cos C=1+\frac rR$$

(note: we can also discover the value of $\cos A\cos B+\cos B\cos C+\cos C\cos A$ using the identity $\cos^2A+\cos^2B+\cos^2C+2\cos A\cos B\cos C=1$)

Since the last time I've posted this question (the original thread is now deleted), I've reflected a bit on the suggestions of several users. First, I included relevant informations and defintion and second I did try to use the cosine law, but It did not give me help.

I was referred by a friend to the identities $$\begin{align} a+b+c &= 2s \tag{1} \\[4pt] ab+ac+bc &= s^2+r^2+4rR \tag{2} \\[4pt] abc &= 4Rrs \tag{3} \end{align}$$ The first and third facts are obvious, while the second I do not know for sure to be true (although it probably is) and appears to model the numerator of the first identity in $\cos A\cos B\cos C$.

Any other idea?

Answer 1

The area of the triangle can be expressed as $ \frac12r(a+b+c)= \frac{abc}{4R}$. Then

\begin{align} \frac rR =\frac12 \frac{\frac{abc}{R^3}}{\frac{a+b+c}R} =\frac{2\sin A \sin B \sin C}{\sin A +\sin B +\sin C}\tag1 \end{align} Note $$\sin A +\sin B +\sin C = 2\cos\frac A2\sin\frac A2 + 2\sin\frac {B+C}2\cos\frac{B-C}2\\ = 2 \cos\frac A2 (\cos\frac {B+C}2+\cos\frac{B-C}2) = 4 \cos\frac A2 \cos\frac B2 \cos\frac C2 $$

Substitute into (1) $$\frac rR = 4 \sin\frac A2 \sin\frac B2 \sin\frac C2\tag2 $$ Similarly $$\cos A +\cos B +\cos C = 1-2\sin^2\frac A2 +2\cos\frac {B+C}2\cos\frac{B-C}2\\ =1- 2 \sin\frac A2 (\cos\frac {B+C}2-\cos\frac{B-C}2)=1+ 4 \sin\frac A2 \sin\frac B2 \sin\frac C2 $$ Substitute into (2)

$$\cos A+\cos B+\cos C=1+\frac rR$$

Answer 2

Use the result $1+\frac rR=\cos A+\cos B+\cos C$ in the derivation below

\begin{align} & \frac{s^2-(2R+r)^2}{4R^2}\\ = &\frac14\left(\frac{a+b+c}{2R}\right)^2-\frac14\left(2+\frac rR\right)^2\\ =& \frac14\left[ (\sin A +\sin B +\sin C)^2 -(1+ \cos A +\cos B +\cos C)^2 \right]\\ =& \frac14\left[(\sin^2A-\cos^2 A)+(\sin^2 B-\cos^2 B) + (\sin^2 C-\cos^2 C)\right.\\ & \left.-1 -2( \cos A +\cos B +\cos C)-2(\cos (A+B)+\cos (B+C) +\cos (C+A)) \right]\\ = &\frac14( -\cos2A - \cos2B -\cos2C-1)\\ = &\frac14( -2\cos(A+B)\cos(A-B) -2\cos^2C +1 -1)\\ = &\frac12\cos C(\cos(A-B)+\cos(A+B))\\ = &\>\cos A \cos B\cos C\\ \end{align}