Question

If $A,B,C$ are the interior angles of a $△ABC$, then $\cos \left(\frac{A+B}{2}\right)$ equals to

• A. $\cos \frac{C}{2}$
• B. $\sec \frac{C}{2}$
• C. $\text{cosec}\frac{C}{2}$
• D. $\sin \frac{c}{2}$

Answer 1

Answer: (D)

$\sin \frac{c}{2}$

Given, A, B, C are the interior angles of a triangle ABC
$\therefore \angle A+\angle B+\angle C=180$
$\therefore \frac{\angle A+\angle B+\angle C}{2}=\frac{180}{2}$
$\therefore \frac{\angle A+\angle B+\angle C}{2}=90$
$\therefore \frac{\angle A+\angle B}{2}=90-\frac{\angle C}{2}$
Now, taking $\cos$ on both sides. we get,
$\cos \left(\frac{\angle A+\angle B}{2}\right)=\cos (90-\frac{\angle C}{2})$
$\Rightarrow \cos \left(\frac{\angle A+\angle B}{2}\right)=\sin \left(\frac{\angle C}{2}\right)$
Option $D$ is correct.