Question

For any triangle show that $\sin \frac{C}{2}=\sqrt{\frac{(s-a)(s-b)}{ab}}$where $s=\frac{1}{2}(a+b+c)$

Short answer

Hence the result $\sin \frac{C}{2}=\sqrt{\frac{(s-a)(s-b)}{ab}}$has been proved.

Step-by-step solution

Step 1. Given Information

$s=\frac{1}{2}(a+b+c)$

Step 2. Use half-angle formula

By using the half-angle formula,

$\cos C=1-2{\sin }^2\frac{C}{2}$

From the equation,

$$\begin{gathered} {\sin }^2\frac{C}{2}=\frac{1-\cos C}{2} \\ \sin \frac{C}{2}=\sqrt{\frac{1-\cos C}{2}} \end{gathered}$$

Step 3. The law of cosines.

Let us consider a triangle having sides $a,b,c$and the corresponding opposite angles $A,B,C$respectively.

By the law of cosines,

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

Step 4. Solve the equation

Substitute the value of $\cos C$in the first equation and solve,

$$\begin{gathered} \sqrt{\frac{1-\cos C}{2}}=\sqrt{\frac{1-\sqrt{{\displaystyle \frac{a^2+b^2-c^2}{2ab}}}}{2}} \\ =\sqrt{\frac{-(a^2+b^2-2ab)+c^2}{4ab}} \\ =\sqrt{\frac{c^2-{(a-b)}^2}{4ab}} \end{gathered}$$

Step 5. Solve the numerator.

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

Since

$$\begin{gathered} s=\frac{1}{2}(a+b+c) \\ 2s=a+b+c \\ c=2s-a-b \end{gathered}$$

Step 5. Substitute for c

$$\begin{gathered} \sqrt{\frac{(c+a-b)(c-a-b)}{4ab}}=\sqrt{\frac{(2s-a-b+a-b)(2s-a-b-a-b)}{4ab}} \\ =\sqrt{\frac{(2s-2b)(2s-2a)}{4ab}} \\ =\sqrt{\frac{(s-b)(s-a)}{ab}} \end{gathered}$$