Question

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

• A. $cos\left(\frac{C}{2}\right)$
• B. $-cos\left(\frac{C}{2}\right)$
• C. $cos\left(\frac{C}{4}\right)$
• D. $-cos\left(\frac{C}{4}\right)$

Answer 1

The correct option is A $cos\left(\frac{C}{2}\right)$

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
$sin({90}^{\circ }-\theta )=cos\theta$

$⟹sin\left(\frac{A+B}{2}\right)=sin({90}^{\circ }-\frac{C}{2})$

$⟹sin({90}^{\circ }-\frac{C}{2})=cos\frac{C}{2}$