Question

In $△ABC$, $sin\left(\frac{A+B}{2}\right)$ is equal to

• A. $Sin\left(\frac{C}{2}\right)$
• B. $Cos\left(\frac{C}{2}\right)$
• C. $Tan\left(\frac{C}{2}\right)$
• D. $Cot\left(\frac{C}{2}\right)$

Answer 1

The correct option is B $Cos\left(\frac{C}{2}\right)$

In a triangle, sum of all the internal angles is 180${}^{\circ }$.
$A+B+C={180}^{\circ }$
$⟹\frac{A+B+C}{2}={90}^{\circ }$
$⟹\frac{A+B}{2}={90}^{\circ }-\frac{C}{2}$
$⟹sin\left(\frac{A+B}{2}\right)=sin({90}^{\circ }-\frac{C}{2})$
Since $sin\theta =cos({90}^{\circ }-\theta )$
$⟹sin\left(\frac{A+B}{2}\right)=cos\left(\frac{C}{2}\right)$