Question

Prove the formula:
${r_1} = 4R\sin \left( {\dfrac{A}{2}} \right)\cos \left( {\dfrac{B}{2}} \right)\cos \left( {\dfrac{C}{2}} \right)$ and similar expressions for ${r_2}$ and ${r_3}$
${r_2} = 4R\cos \left( {\dfrac{A}{2}} \right)sin\left( {\dfrac{B}{2}} \right)\cos \left( {\dfrac{C}{2}} \right) \\\ {r_3} = 4R\cos \left( {\dfrac{A}{2}} \right)\cos \left( {\dfrac{B}{2}} \right)\sin \left( {\dfrac{C}{2}} \right) \\\$
Where ${r_1},{r_2},{r_3}$ have their usual meanings.

Answer 1

Hint: Here, we can see there are three formulas which we have to prove. So, we will solve them by one by using the appropriate trigonometric identities.

Complete step-by-step answer:
We have,
(i) ${r_1} = 4R\sin \left( {\dfrac{A}{2}} \right)\cos \left( {\dfrac{B}{2}} \right)\cos \left( {\dfrac{C}{2}} \right)$
Now, we will use the trigonometry half-angle triangle formula to prove the equation i.e. $\sin \left( {\dfrac{A}{2}} \right) = \sqrt {\dfrac{{\left( {s - b} \right)\left( {s - c} \right)}}{{bc}}}$, $\cos \left( {\dfrac{B}{2}} \right) = \sqrt {\dfrac{{s\left( {s - b} \right)}}{{ac}}}$ and $\cos \left( {\dfrac{C}{2}} \right) = \sqrt {\dfrac{{s\left( {s - c} \right)}}{{ab}}}$
So, we will get
${r_1} = 4R\sqrt {\dfrac{{\left( {s - b} \right)\left( {s - c} \right)}}{{bc}}} \sqrt {\dfrac{{s\left( {s - b} \right)}}{{ac}}} \sqrt {\dfrac{{s\left( {s - c} \right)}}{{ab}}}$
By multiplying all the terms

Simplify the above equation
$= \dfrac{{4R}}{{abc}}s\left( {s - b} \right)\left( {s - c} \right)$
Now, we will put the value $R$ in above equation i.e. $R = \dfrac{{abc}}{{4\Delta }}$ ,where $\Delta = \sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)}$
$= \dfrac{4}{{abc}}\left( {\dfrac{{abc}}{{4\Delta }}} \right)s\left( {s - b} \right)\left( {s - c} \right)$
After cancelling out the terms, we will get
$= \dfrac{1}{\Delta }s\left( {s - b} \right)\left( {s - c} \right)$
Multiply the denominator and numerator with $\left( {s - a} \right)$
$= \dfrac{1}{\Delta }s\left( {s - b} \right)\left( {s - c} \right) \times \dfrac{{\left( {s - a} \right)}}{{\left( {s - a} \right)}}$
As we know the value $\Delta$ we can say that,
$= \dfrac{{{\Delta ^2}}}{{\Delta \left( {s - a} \right)}} = \dfrac{\Delta }{{s - a}} = {r_1}$
Hence, it is proved.
(ii) ${r_2} = 4R\cos \left( {\dfrac{A}{2}} \right)sin\left( {\dfrac{B}{2}} \right)\cos \left( {\dfrac{C}{2}} \right)$
Similarly by using the trigonometry half-angle triangle formulae and concepts we will prove the above formula i.e. $\sin \left( {\dfrac{B}{2}} \right) = \sqrt {\dfrac{{\left( {s - a} \right)\left( {s - c} \right)}}{{ac}}}$, $\cos \left( {\dfrac{A}{2}} \right) = \sqrt {\dfrac{{s\left( {s - a} \right)}}{{bc}}}$ and $\cos \left( {\dfrac{C}{2}} \right) = \sqrt {\dfrac{{s\left( {s - c} \right)}}{{ab}}}$
$= 4R\sqrt {\dfrac{{s\left( {s - a} \right)}}{{bc}}} \sqrt {\dfrac{{\left( {s - a} \right)\left( {s - c} \right)}}{{ac}}} \sqrt {\dfrac{{s\left( {s - c} \right)}}{{ab}}} \\\ = \dfrac{{4R}}{{abc}}\sqrt {{s^2}{{\left( {s - a} \right)}^2}{{\left( {s - c} \right)}^2}} \\\$
Again by putting the value of $R$ and Multiply the denominator and numerator with $\left( {s - b} \right)$
$= \dfrac{4}{{abc}}\left( {\dfrac{{abc}}{{4\Delta }}} \right)s\left( {s - a} \right)\left( {s - c} \right) \times \dfrac{{s - b}}{{s - b}}$
After cancelling out the terms and substitute the value of $\Delta$
$= \dfrac{{{\Delta ^2}}}{{\Delta \left( {s - b} \right)}} = \dfrac{\Delta }{{s - b}} = {r_2}$
Hence, it is also proved.
(iii) ${r_3} = 4R\cos \left( {\dfrac{A}{2}} \right)\cos \left( {\dfrac{B}{2}} \right)\sin \left( {\dfrac{C}{2}} \right)$
Similarly, we will use the formula in order to solve this are:
$\sin \left( {\dfrac{C}{2}} \right) = \sqrt {\dfrac{{\left( {s - a} \right)\left( {s - b} \right)}}{{ab}}}$, $\cos \left( {\dfrac{A}{2}} \right) = \sqrt {\dfrac{{s\left( {s - a} \right)}}{{bc}}}$ and $\cos \left( {\dfrac{B}{2}} \right) = \sqrt {\dfrac{{s\left( {s - b} \right)}}{{ac}}}$
$= 4R\sqrt {\dfrac{{s\left( {s - a} \right)}}{{bc}}} \sqrt {\dfrac{{s\left( {s - b} \right)}}{{ac}}} \sqrt {\dfrac{{\left( {s - a} \right)\left( {s - b} \right)}}{{ab}}} \\\ = \dfrac{{4R}}{{abc}}\sqrt {{s^2}{{\left( {s - a} \right)}^2}{{\left( {s - b} \right)}^2}} \\\$
Now, put the value of $R$ and Multiply the denominator and numerator with $\left( {s - c} \right)$
$= \dfrac{4}{{abc}}\left( {\dfrac{{abc}}{{4\Delta }}} \right)s\left( {s - a} \right)\left( {s - b} \right) \times \dfrac{{s - c}}{{s - c}}$
Cancel out the terms in numerator and denominator, also substitute the value of $\Delta$
$= \dfrac{{{\Delta ^2}}}{{\Delta \left( {s - c} \right)}} = \dfrac{\Delta }{{s - c}} = {r_3}$
Hence, this formula is also proved.
Since, we proved all the given formulas, we can conclude the result and tells the value of ${r_1},{r_2},{r_3}$.

Note: It is to be noted that for trigonometry half-angle triangle formulae for sines and cosines, there exists a symmetry which means there are similar expressions involving all the angles A, B and C. The other way to solve this is one should put the value of R and formula at once and try to solve it. Try to make the equations easy as this is a confusing question with all the parts almost similar to each other.