Question

Prove that for a triangle $ABC$ with sides $a,b,c$ and circumradius and inradius respectively $R$ and $r$, area of the triangle $S = 4Rr\cos \dfrac{A}{2}\cos \dfrac{B}{2}\cos \dfrac{C}{2}$.

Answer 1

Hint: Area of a triangle is expressed using a different formula. Here we use equations using semi-perimeter, circumradius and inradius. Substituting these appropriately and making necessary simplifications we can prove the expression.

Formula used
For a triangle ABC with sides $a,b,c$
Area of the triangle, $S = \sqrt {s(s - a)(s - b)(s - c)} $
where $s = \dfrac{{a + b + c}}{2}$, the semi-perimeter of the triangle.
Also, $\cos \dfrac{A}{2} = \sqrt {\dfrac{{s(s - a)}}{{bc}}} ,\cos \dfrac{B}{2} = \sqrt {\dfrac{{s(s - b)}}{{ac}}} ,\cos \dfrac{C}{2} = \sqrt {\dfrac{{s(s - c)}}{{ab}}} $
If the circumradius and inradius are represented by $R,r$, then
Area of the triangle, $S = \dfrac{{abc}}{{4R}} = rs$, where $s$ is the semi-perimeter.

Complete step-by-step answer:

Given that the triangle has sides $a,b,c$ and circumradius and inradius, $R,r$ respectively.
If $s$ is the semi-perimeter of a triangle, then $s = \dfrac{{a + b + c}}{2}$.
Also we have $\cos \dfrac{A}{2} = \sqrt {\dfrac{{s(s - a)}}{{bc}}} ,\cos \dfrac{B}{2} = \sqrt {\dfrac{{s(s - b)}}{{ac}}} ,\cos \dfrac{C}{2} = \sqrt {\dfrac{{s(s - c)}}{{ab}}} $
To prove the area of the triangle is the expression given, we can start with the RHS.
Substituting for $\cos $ terms in RHS we have,
$ \Rightarrow 4Rr\cos \dfrac{A}{2}\cos \dfrac{B}{2}\cos \dfrac{C}{2} = 4Rr\sqrt {\dfrac{{s(s - a)}}{{bc}}} \sqrt {\dfrac{{s(s - b)}}{{ac}}} \sqrt {\dfrac{{s(s - c)}}{{ab}}} $
Simplifying we get,
$ \Rightarrow 4Rr\cos \dfrac{A}{2}\cos \dfrac{B}{2}\cos \dfrac{C}{2} = 4Rr\sqrt {\dfrac{{{s^3}(s - a)(s - b)(s - c)}}{{{a^2}{b^2}{c^2}}}} $
$ \Rightarrow 4Rr\cos \dfrac{A}{2}\cos \dfrac{B}{2}\cos \dfrac{C}{2} = \dfrac{{4Rrs}}{{abc}}\sqrt {s(s - a)(s - b)(s - c)} $
Now since the expression inside the radical/root symbol is the area of a triangle, we can substitute it with $S$.
$ \Rightarrow 4Rr\cos \dfrac{A}{2}\cos \dfrac{B}{2}\cos \dfrac{C}{2} = \dfrac{{4Rrs}}{{abc}}S$
Also, if the circumradius and inradius are represented by $R,r$, then
Area of the triangle, $S = \dfrac{{abc}}{{4R}} = rs$, where $s$ is the semi-perimeter.
So we get,
$ \Rightarrow 4Rr\cos \dfrac{A}{2}\cos \dfrac{B}{2}\cos \dfrac{C}{2} = \dfrac{{4R}}{{abc}} \times rs \times S$
$ \Rightarrow 4Rr\cos \dfrac{A}{2}\cos \dfrac{B}{2}\cos \dfrac{C}{2} = \dfrac{1}{S} \times S \times S$
Cancelling $S$ from numerator and denominator we have,
$ \Rightarrow 4Rr\cos \dfrac{A}{2}\cos \dfrac{B}{2}\cos \dfrac{C}{2} = S$
Hence we had proved the given expression.

Note: Here to substitute the terms easily we start from the RHS. Starting with LHS and getting RHS is somewhat impossible or will be a difficult task. So in these types of questions we have to observe and decide from where we have to start. After making necessary steps we can get into the solution.