Chapter 8.3, Problem 56AYU

To determine

show that $\cos \frac{C}{2}=\sqrt{\frac{s\left(s-c\right)}{ab}}$ Where $s=\frac{1}{2}\left(a+b+c\right)$ .

Answer to Problem 56AYU

  $\boxed{\cos \frac{C}{2}=\sqrt{\frac{s\left(s-c\right)}{ab}}}$ , $\boxed{s=\frac{a+b+c}{2}}$

Explanation of Solution

Given information:

For any triangle, show that

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

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

[hint: Use a Half-angle Formula and the Law of Cosines.]

Calculation:

Let us consider the triangle having sides a,b,c and corresponding opposite angle A, B, C. By the Law of Cosines, we get

  $\cos C=\frac{a^2+b^2+c^2}{2ab}\mathrm{.....}(1)$

As we know that

  $\cos 2\theta =2{\cos }^2\theta -1$ for each value of angle $\theta$

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

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

From (2), we get

  $\cos C=\sqrt{\frac{1+\cos C}{2}}$

Using (1), we get

  $\cos C=\sqrt{1+\frac{a^2+b^2{c}^2}{\frac{2ab}{2}}}$

  $=\sqrt{\frac{2ab+a^2+b^2-c^2}{4ab}}$

  $=\sqrt{\frac{\left(a^2+b^{2}2ab\right)-c^2}{4ab}}$

  $=\sqrt{\frac{{\left(a+b\right)}^2-c^2}{4ab}}\mathrm{......}\left(3\right)$

Since,

  $\frac{a+b+c}{2}=s$

  $a+b=2s-c$

By putting it in equation (3), we get

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

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

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

Hence, $\cos \frac{C}{2}=\sqrt{\frac{s\left(s-c\right)}{ab}}$ , where $s=\frac{a+b+c}{2}$