Question

Show that $a{{\cos }^{2}}\dfrac{A}{2}+b{{\cos }^{2}}\dfrac{B}{2}+c{{\cos }^{2}}\dfrac{C}{2}=s+\dfrac{\Delta }{R}$

Answer 1

Hint: In this problem we need to show that $a{{\cos }^{2}}\dfrac{A}{2}+b{{\cos }^{2}}\dfrac{B}{2}+c{{\cos }^{2}}\dfrac{C}{2}=s+\dfrac{\Delta }{R}$. First, we will consider the LHS part of the given equation. In LHS we will consider the term ${{\cos }^{2}}\dfrac{A}{2}$. From the trigonometric formula $\cos 2A=2{{\cos }^{2}}A-1$, we will write ${{\cos }^{2}}\dfrac{A}{2}=\dfrac{1+\cos A}{2}$. Similarly, we can write this for ${{\cos }^{2}}\dfrac{B}{2}$, ${{\cos }^{2}}\dfrac{C}{2}$ and substitute those values in the given equation. Here we will apply the formula for the properties of the triangle which is $s=\dfrac{a+b+c}{2}$ and from sine rule $\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=2R$ we will use $a=2R\sin A$, $b=2R\sin B$, $c=2R\sin C$ in the obtained equation. Now we will use $\sin 2A=2\sin A\cos A$ in the equation. Now we will get $\sin 2A+\sin 2B+\sin 2C$ in the equation and we will use the value of $\sin 2A+\sin 2B+\sin 2C$ as $4\sin A\sin B\sin C$ and then we will use the formula $\Delta =2{{R}^{2}}\sin A\sin B\sin C$ and simplify the equation to get the required result.

Complete step by step solution:
Given equation, $a{{\cos }^{2}}\dfrac{A}{2}+b{{\cos }^{2}}\dfrac{B}{2}+c{{\cos }^{2}}\dfrac{C}{2}=s+\dfrac{\Delta }{R}$.
Considering the LHS part of the above equation, then we will get
$LHS=a{{\cos }^{2}}\dfrac{A}{2}+b{{\cos }^{2}}\dfrac{B}{2}+c{{\cos }^{2}}\dfrac{C}{2}$
From the trigonometric formula $\cos 2A=2{{\cos }^{2}}A-1$, we can write ${{\cos }^{2}}\dfrac{A}{2}=\dfrac{1+\cos A}{2}$. Using this formula in the above equation, then we will get
$LHS=a\left[ \dfrac{1+\cos A}{2} \right]+b\left[ \dfrac{1+\cos B}{2} \right]+c\left[ \dfrac{1+\cos C}{2} \right]$
Simplifying the above equation by using mathematical operations, then we will have
$\begin{align}
  & LHS=\dfrac{a}{2}+\dfrac{a\cos A}{2}+\dfrac{b}{2}+\dfrac{b\cos B}{2}+\dfrac{c}{2}+\dfrac{c\cos C}{2} \\
 & \Rightarrow LHS=\dfrac{a+b+c}{2}+\dfrac{1}{2}\left[ a\cos A+b\cos B+c\cos C \right] \\
\end{align}$
Using the formula $s=\dfrac{a+b+c}{2}$ and from sine rule $\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=2R$ using the values $a=2R\sin A$, $b=2R\sin B$, $c=2R\sin C$ in the above equation, then we will get
$LHS=s+\dfrac{1}{2}\left[ 2R\sin A\cos A+2R\sin B\cos B+2R\sin C\cos C \right]$
Taking $R$ as common from the above equation and using the trigonometric formula $\sin 2A=2\sin A\cos A$ in the above equation, then we will have
$LHS=s+\dfrac{R}{2}\left[ \sin 2A+\sin 2B+\sin 2C \right]$
We have the trigonometric formula $\sin 2A+\sin 2B+\sin 2C=4\sin A\sin B\sin C$ in the above equation, then we will get
$LHS=s+\dfrac{R}{2}\left( 4\sin A\sin B\sin C \right)$
Divide and multiply the term $4\sin A\sin B\sin C$ with $R$, then we will get
$LHS=s+\dfrac{2{{R}^{2}}\sin A\sin B\sin C}{R}$
We have the trigonometric formula $\Delta =2{{R}^{2}}\sin A\sin B\sin C$. Substituting this formula in the above equation, then we will get
$\begin{align}
  & LHS=s+\dfrac{\Delta }{R} \\
 & \therefore LHS=RHS \\
\end{align}$

Note: In trigonometry this kind of problem very often. So, we should be clear about all the formulas and properties of the triangle to use them at a suitable step. Some of the properties of triangle which are used in this kind of problems is given by
$\sin \dfrac{A}{2}=\sqrt{\dfrac{\left( s-b \right)\left( s-c \right)}{bc}}$, $\cos \dfrac{A}{2}=\sqrt{\dfrac{s\left( s-a \right)}{bc}}$, $\tan \dfrac{A}{2}=\sqrt{\dfrac{\left( s-b \right)\left( s-c \right)}{s\left( s-a \right)}}$. We can use these formulas for remaining angles with a minor change.