Chapter 8.3, Problem 56AYU

To determine

To find: The prove of the given identity.

Explanation of Solution

Given information:

The given identity is,

  $\cos \frac{C}{2}=\sqrt{\frac{s\left(s-c\right)}{ab}}$

Here, $s=\frac{1}{2}\left(a+b+c\right)$ .

Concept used:

Law of Cosines:

If ABC is a triangle with sides $a$ , $b$ , and $c$ , then

  $\begin{array}{l}\cos A=\frac{b^2+c^2-a^2}{2bc}\\ \cos B=\frac{a^2+c^2-b^2}{2ac}\\ \cos C=\frac{a^2+b^2-c^2}{2ab}\end{array}$

Write the expression for the half-angle formula as,

  $\cos \frac{\theta }{2}=\sqrt{\frac{1+\cos \theta }{2}}$

Write the expression for the half-angle formula as,

  $\cos \frac{C}{2}=\sqrt{\frac{1+\cos C}{2}}$

Apply Cosine rule formula as,

  $\begin{array}{c}\cos \frac{C}{2}=\sqrt{\frac{1+\left(\frac{a^2+b^2-c^2}{2ab}\right)}{2}}\\ =\sqrt{\frac{2ab+a^2+b^2-c^2}{4ab}}\\ =\sqrt{\frac{{\left(a+b\right)}^2-c^2}{4ab}}\\ =\sqrt{\frac{{\left(2s-c\right)}^2-c^2}{4ab}}\end{array}$

Simplify further,

  $\begin{array}{c}\cos \frac{C}{2}=\sqrt{\frac{\left(4s^2-4sc+c^2\right)-c^2}{4ab}}\\ =\sqrt{\frac{s\left(s-c\right)}{ab}}\end{array}$

Hence, the given identity is proved.